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Page No 15.13:
Question 1:
The number of telephone calls received at an exchange per interval for 250 successive oneminute intervals are given in the following frequency table:
No. of calls (x):  0  1  2  3  4  5  6 
No. of intervals (f):  15  24  29  49  54  43  39 
Compute the mean number of calls per intervals.
Answer:
Let the assume mean be.
We know that mean,
Here, we have .
Putting the values in the formula, we get
Hence, the mean number of calls per interval is 3.54.
Page No 15.13:
Question 2:
Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below: Find the mean number of heads per toss
No. of heads per toss (x):  0  1  2  3  4  5 
No. of tosses (f):  38  144  342  287  164  25 
Answer:
Let the assume mean be.
We know that mean
Now, we have
Putting the values above in formula, we have
Hence, the mean number of heads per toss is 2.47.
Page No 15.13:
Question 3:
The following table given the number of branches and number of plants in the garden of a school.
No. of branches (x):  2  3  4  5  6 
No. of plants (f):  49  43  57  38  13 
Answer:
Let the assume mean be.
We know that mean,
Now, we have
Putting the values in the above formula, we get
Hence, the mean number of branches per plant is approximately 3.62.
Page No 15.13:
Question 4:
The following table gives the number of children of 150 families in a village
No. of children (x):  0  1  2  3  4  5 
No. of families (f):  10  21  55  42  15  7 
Find the average number of children per family
Answer:
Let the assume mean be.
We know that mean,
Now, we have
Putting the values above in formula, we get
Hence, the average number of children per family is 2.35.
Page No 15.13:
Question 5:
The marks obtained out of 50, by 102 students in a Physics test are given in the frequency table below:
Marks (x):  15  20  22  24  25  30  33  38  45 
Frequency (f):  5  8  11  20  23  18  13  3  1 
Find the average number of marks.
Answer:
Let the assume mean be.
We know that mean,
Now, we have .
Putting the values in the above formula, we get
Hence, the average number of marks is 26.08.
Page No 15.13:
Question 6:
The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:
No. of students absent (x):  0  1  2  3  4  5  6  7 
No. of days (f):  1  4  10  50  34  15  4  2 
Find the mean number of students absent per day.
Answer:
Let the assume mean be.
We know that mean,
Now, we have .
Putting the values in the above formula,
Hence, the mean number of students absent per day is approximately 3.53.
Page No 15.13:
Question 7:
In the first proof reading of a book containing 300 pages the following distribution of misprints was obtained:
No. of misprints per page (x):  0  1  2  3  4  5 
No. of pages (f):  154  95  36  9  5  1 
Find the average number of misprints per page.
Answer:
Let the assume mean be.
We know that mean,
Now, we have .
Putting the values in above formula, we have
Hence, the mean number of students absent per day is 0.73.
Page No 15.13:
Question 8:
The following distribution gives the number of accidents met by 160 workers in a factory during a month.
No. of accidents (x):  0  1  2  3  4 
No. of workers (f):  70  52  34  3  1 
Find the average number of accidents per worker.
Answer:
Let the assume mean be.
We know that mean,
Now, we have .
Putting the values in the above formula, we get
Hence, the average number of accidents per worker is 0.83.
Page No 15.14:
Question 9:
Find the mean from the following frequency distribution of marks at a test in statistics:
Marks (x):  5  10  15  20  25  30  35  40  45  50 
No. of students (f):  15  50  80  76  72  45  39  9  8  6 
Answer:
Let the assumed mean beand .
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the mean marks is 22.075.
Page No 15.22:
Question 1:
The following tables gives the distribution of total household expenditure (in rupees) of manual workers in a city.
Expenditure (in rupees) (x_{i}) 
Frequency (f_{i}) 
Expenditure (in rupees) (x_{i}) 
Frequency (f_{i}) 
100150 150200 200250 250300 
24 40 33 28 
300350 350400 400450 450500 
30 22 16 7 
Find the average expenditure (in rupees) per household
Answer:
Given:
Let the assumed mean be A = 275 and h = 50.
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the average expenditureper household is 266.25.
Page No 15.22:
Question 2:
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of a plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants:  02  24  46  68  810  1012  1214 
Number of houses:  1  2  1  5  6  2  3 
Which method did you use for finding the mean, and why?
Answer:
We may prepare the table as shown:
We know that mean,
Hence, mean = 8.1
Direct method is easier than other methods. Therefore, we used direct method.
Page No 15.22:
Question 3:
Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)  100120  120140  140160  160180  180200 
Number of workers  12  14  8  6  10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Let the assumed mean be A = 120 and h = 20.
We know that mean,
Now, we have .
Putting the values in the above formula, we have
Hence, the mean daily wage of the workers is Rs 145.20.
Page No 15.22:
Question 4:
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarised as follows. Find the mean heat beats per minute for these women, choosing a suitable method.
Number of heart beats per minute:  6558  6871  7174  7477  7780  8083  8386 
Number of women:  2  4  3  8  7  4  2 
Answer:
Let the assumed mean be A = 75.5 and h = 3.
No. of Heart Beats per Min:  Mid Value $\left({x}_{i}\right)$  No. of Women $\left({f}_{i}\right)$:  ${d}_{i}={x}_{i}A$  ${u}_{i}=\frac{1}{h}\times {d}_{i}$  ${f}_{i}{u}_{i}$ 
6568  66.5  2  $$9  $$3  $$6 
6871  69.5  4  $$6  $$2  $$8 
7174  72.5  3  $$3  $$1  $$3 
7477  75.5 = A  8  0  0  0 
7780  78.5  7  3  1  7 
8083  81.5  4  6  2  8 
8386  84.5  2  9  3  6 
$\sum {f}_{i}=30$  $\sum {f}_{i}{u}_{i}=4$ 
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean heart beats per minute for women is 75.9.
Page No 15.22:
Question 5:
Find the mean of each of the following frequency distributions :
Class interval:  06  612  1218  1824  2430 
Frequency:  6  8  10  9  7 
Answer:
Let the assumed mean be A = 15 and h = 6.
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the mean is 15.45.
Page No 15.22:
Question 6:
Find the mean of each of the following frequency distributions :
Class interval:  5070  7090  90110  110130  130150  150170 
Frequency:  18  12  13  27  8  22 
Answer:
Let the assumed mean be A = 100 and h = 20.
We know that mean
Now we have
Putting the values in the above formula, we get
Hence, the mean is 112.20.
Page No 15.23:
Question 7:
Find the mean of each of the following frequency distributions :
Class interval:  0−8  8−16  16−24  24−32  32−40 
Frequency:  6  7  10  8  9 
Answer:
Let the assumed mean be A = 20 and h = 8.
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the mean is 21.4.
Page No 15.23:
Question 8:
Find the mean of each of the following frequency distributions :
Class interval:  0−6  6−12  12−18  18−24  24−30 
Frequency:  7  5  10  12  6 
Answer:
Let the assumed mean be A = 15 and h = 6.
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the mean is 15.75.
Page No 15.23:
Question 9:
Find the mean of each of the following frequency distributions :
Class interval:  0−10  10−20  20−30  30−40  40−50 
Frequency:  9  12  15  10  14 
Answer:
Let the assumed mean be A = 25 and h = 10.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 26.333.
Page No 15.23:
Question 10:
Find the mean of each of the following frequency distributions :
Class interval:  0−8  8−16  16−24  24−32  32−40 
Frequency:  5  9  10  8  8 
Answer:
Let the assumed mean be A = 20 and h = 8.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 21.
Page No 15.23:
Question 11:
Find the mean of each of the following frequency distributions :
Class interval:  0−8  8−16  16−24  24−32  32−40 
Frequency:  5  6  4  3  2 
Answer:
Let the assumed mean be A = 20 and h = 8.
We know that mean,
Now, we have.
Putting the values in the above formula, we get
Hence, the mean is 16.4.
Page No 15.23:
Question 12:
Find the mean of each of the following frequency distributions :
Class interval:  10−30  30−50  50−70  70−90  90−110  110−130 
Frequency:  5  8  12  20  3  2 
Answer:
Let the assumed mean be A = 60 and h = 20.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 65.6.
Page No 15.23:
Question 13:
Find the mean of each of the following frequency distributions :
Class interval:  25−35  35−45  45−55  55−65  65−75 
Frequency:  6  10  8  12  4 
Answer:
Let the assumed mean be A = 50 and h = 10.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 49.5.
Page No 15.23:
Question 14:
Find the mean of each of the following frequency distributions :
Classes:  25−29  30−34  35−39  40−44  45−49  50−54  55−59 
Frequency:  14  22  16  6  5  3  4 
Answer:
The given series is an inclusive series. Firstly, make it an exclusive series.
Class Interval  MidValue(x_{i})  Frequency(f_{i})  ${d}_{\mathit{i}}={x}_{i}A\phantom{\rule{0ex}{0ex}}={x}_{i}42$  ${u}_{i}=\frac{{d}_{i}}{h}=\frac{{d}_{i}}{5}$  ${f}_{i}{u}_{i}$ 
24.5$$29.5  27  14  −15  −3  −42 
29.5$$34.5  32  22  −10  −2  −44 
34.5$$39.5  37  16  −5  −1  −16 
39.5$$44.5  A = 42  6  0  0  0 
44.5$$49.5  47  5  5  1  5 
49.5$$54.5  52  3  10  2  6 
54.5$$59.5  57  4  15  3  12 
$\sum _{}^{}{f}_{i}=70$  $\sum _{}^{}{f}_{i}{u}_{i}=79$ 
Let the assumed mean be A = 42 and h = 5.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 36.357.
Page No 15.23:
Question 15:
For the following distribution, calculate mean using all suitable methods:
Size of item:  1−4  4−9  9−16  16−27 
Frequency:  6  12  26  20 
Answer:
Direct Method:
We may prepare the table as shown:
We know that mean,
Hence, the mean is 13.25.
ShortCut Method:
We may prepare the table as shown:
Size of Item  Mid value(x_{i})  ${d}_{i}={x}_{i}A\phantom{\rule{0ex}{0ex}}={x}_{i}12.5$  Frequency(f_{i})  ${f}_{i}{d}_{i}$ 
1–4  2.5  −10  6  −60 
4–9  6.5  −6  12  −72 
9–16  12.5 = A  0  26  0 
16–27  21.5  9  20  180 
N = $\sum _{}{f}_{i}=64$  $\sum _{}{f}_{i}{d}_{i}=48$ 
Let the assumed mean be A = 12.5.
We know that mean, $\overline{)X}=A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}$
$=12.5+\frac{48}{64}\phantom{\rule{0ex}{0ex}}=12.5+0.75\phantom{\rule{0ex}{0ex}}=13.25$
Hence, the mean is 13.25.
Stepdeviation method cannot be used to evaluate the mean of the distribution as the width of the class intervals are not equal. Here, h is not fixed.
Page No 15.23:
Question 16:
The weekly observations on cost of living index in a certain city for the year 2004 − 2005 are given below. Compute the weekly cost of living index.
Cost of living Index  Number of Students  Cost of living Index  Number of Students 
1400−1500 1500−1600 1600−1700 
5 10 20 
1700−1800 1800−1900 1900−2000 
9 6 2 
Answer:
Let the assumed mean be A = 1650 and h = 100.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 1663.46.
Page No 15.23:
Question 17:
The following table shows the marks scored by 140 students in an examination of a certain paper.
Marks:  0−10  10−20  20−30  30−40  40−50 
Number of students:  20  24  40  36  20 
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Answer:
We may prepare the table as shown:
(i) Direct method
We know that mean,
$=\frac{3620}{140}\phantom{\rule{0ex}{0ex}}=25.857$
Hence, the mean is 25.857.
(ii) Shortcut method
Let the assumed mean A = 25.
We know that mean,
Hence, the mean is 25.857.
(iii) Step deviation method
Let the assumed mean A = 25 and h = 10.
We know that mean,
Hence, the mean is 25.857.
Page No 15.23:
Question 18:
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f_{1} and f_{2}.
Class:  0−20  20−40  40−60  60−80  80−100  100−120 
Frequency:  5  f_{1}  10  f_{2}  7  9 
Answer:
It is given that mean = 62.8 and .
Let the assumed mean A = 50 and h = 20.
.....(1)
We know that mean,
Now, we have , , .
Putting the values in the above formula, we have
.....(2)
Putting the value of in (2), we get
Putting the value of in (1), we get
Hence, the missing frequency f_{1 }= 8 and f_{2 }= 12.
Page No 15.24:
Question 19:
The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find the missing frequency.
Class:  11−13  13−15  15−17  17−19  19−21  21−23  23−25 
Frequency:  7  6  9  13  −  5  4 
Answer:
Given: Mean = 18
Suppose the missing frequency is x.
Let the assumed mean A = 18 and h = 2.
We know that mean,
Now, we have,,.
Putting the values in the above formula, we have
$18=18+2\left(\frac{x20}{x+44}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(\frac{x20}{x+44}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x20=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=20$
Page No 15.24:
Question 20:
If the mean of the following distribution is 27, find the value of p.
Class:  0−10  10−20  20−30  30−40  40−50 
Frequency:  8  p  12  13  10 
Answer:
Given: Mean = 27
Let the assumed mean A = 25 and h = 10.We know that mean,
Now, we have,,
Putting the values in the above formula, we have
Thus, the value of p is 7.
Page No 15.24:
Question 21:
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes:  50−52  53−55  56−58  59−61  62−64 
Number of boxes:  15  110  135  115  25 
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer:
The given series is an inclusive series. Firstly, make it an exclusive series.
Class Interval  MidValue(x_{i})  Frequency(f_{i})  ${d}_{i}={x}_{i}A\phantom{\rule{0ex}{0ex}}={x}_{i}57$  ${u}_{i}=\frac{{d}_{i}}{h}=\frac{{d}_{i}}{3}$  ${f}_{i}{u}_{i}$ 
49.5$$52.5  51  15  −6  −2  −30 
52.5$$55.5  54  110  −3  −1  −110 
55.5$$58.5  57  135  0  0  0 
58.5$$61.5  60  115  3  1  115 
61.5$$64.5  63  25  6  2  50 
$\sum _{}^{}{f}_{i}=400$  $\sum _{}^{}{f}_{i}{u}_{i}=25$ 
Let the assumed mean be A = 57 and h = 3.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is approximately 57.19.
Page No 15.24:
Question 22:
The table below shows the daily expenditure on food of 25 household in a locality
Daily expenditure (in Rs):  100−150  150−200  200−250  250−300  300−350 
Number of households:  4  5  12  2  2 
Find the mean daily expenditure on food by a suitable method.
Answer:
Let the assumed mean a = 225 and h = 50.
Daily expenditure (in Rs)  f_{i}  x_{i}  d_{i} = x_{i }− 225  f_{i}u_{i}  
100−150  4  125  − 100  − 2  − 8 
150−200  5  175  − 50  − 1  − 5 
200−250  12  225  0  0  0 
250−300  2  275  50  1  2 
300−350  2  325  100  2  4 
Total  $\sum _{}{f}_{i}=$25  $\sum _{}{f}_{i}{u}_{i}=$− 7 
Now,
Therefore, mean daily expenditure on food is Rs 211.
Page No 15.24:
Question 23:
To find out the concentration of SO_{2} in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Concentration of SO_{2} (in ppm)  Frequency 
0.00−0.04 0.04−0.08 0.08−0.12 0.12−0.16 0.16−0.20 0.20−0.24 
4 9 9 2 4 2 
Find the mean of concentration of SO_{2} in the air.
Answer:
Let the assumed mean A = 0.1 and h = 0.04.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
$=0.10+0.04\left[\frac{1}{30}\times \left(1\right)\right]\phantom{\rule{0ex}{0ex}}=0.10\frac{0.04}{30}\phantom{\rule{0ex}{0ex}}=0.100.001\phantom{\rule{0ex}{0ex}}=0.099$
Hence, the mean concentration of SO_{2}_{ }in the air is 0.099 ppm.
Page No 15.24:
Question 24:
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days:  0−6  6−10  10−14  14−20  20−28  28−38  38−40 
Number of students:  11  10  7  4  4  3  1 
Answer:
Let the assume mean A = 17.
We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean number of days a student was absent is 12.475.
Page No 15.24:
Question 25:
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %):  45−55  55−65  65−75  75−85  85−95 
Number of cities:  3  10  11  8  3 
Answer:
Let the assumed mean A = 70 and h = 10.
We know that mean,
Now, we have
Putting the values in the above formula, we have
Hence, the mean literacy rate is approximately 69.43%.
Page No 15.24:
Question 26:
The following is the cumulative frequency distribution ( of less than type ) of 1000 persons each of age 20 years and above . Determine the mean age .
Age below (in years) : 30 40 50 60 70 80
Number of persons: 100 220 350 750 950 1000
Answer:
Let the assumed mean A = 55 and h = 10.
Marks  MidValue  Frequency  ${\mathit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathit{x}}_{\mathit{i}}\mathbf{}\mathbf{55}}{\mathbf{10}}$  ${\mathit{f}}_{\mathit{i}}{\mathit{u}}_{\mathit{i}}$ 
20–30  25  100  −3  −300 
30–40  35  120  −2  −240 
40–50  45  130  −1  −130 
50–60  55  400  0  0 
60–70  65  200  1  200 
70–80  75  50  2  100 
N = 1000  $\sum _{}{f}_{i}{u}_{i}=370$ 
We know that mean,
Now, we have $N=\sum _{}{f}_{i}=1000,h=10,A=55,\sum _{}{f}_{i}{u}_{i}=370$
$\overline{X}=55+10\left[\frac{1}{1000}\times \left(370\right)\right]\phantom{\rule{0ex}{0ex}}=553.7\phantom{\rule{0ex}{0ex}}=51.3\mathrm{years}$
Page No 15.25:
Question 27:
If the mean of the following frequency distribution is 18 , find the missing frequency .
Class interval : 1113 1315 1517 1719 1921 2123 2325
Frequency: 3 6 9 13 f 5 4
Answer:
It is given that mean of the data is 18.
Let the assumed mean A = 18 and h = 2.
Marks  MidValue(x_{i})  Frequency  ${\mathit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathit{x}}_{\mathit{i}}\mathbf{}\mathbf{18}}{\mathbf{2}}$  ${\mathit{f}}_{\mathit{i}}{\mathit{u}}_{\mathit{i}}$ 
11–13  12  3  −3  −9 
13–15  14  6  −2  −12 
15–17  16  9  −1  −9 
17–19  18  13  0  0 
19–21  20  f  1  f 
21–23  22  5  2  10 
23–25  24  4  3  12 
N = 40 + f  $\sum _{}{f}_{i}{u}_{i}=8+f$ 
We know that mean,
Now, we have $N=\sum _{}{f}_{i}=40+f,h=2,A=18,\sum _{}{f}_{i}{u}_{i}=8+f$
$18=18+2\left[\frac{1}{\left(40+f\right)}\times \left(8+f\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 0=\frac{16+2f}{40+f}\phantom{\rule{0ex}{0ex}}\Rightarrow 16+2f=0\phantom{\rule{0ex}{0ex}}\Rightarrow f=8$
Page No 15.25:
Question 28:
Find the missing frequencies in the following distribution , if the sum of the frequencies is 120 and the mean is 50 .
Class: 020 2040 4060 6080 80100
Frequency: 17 f_{1 } 32 f_{2} 19
Answer:
It is given that mean of the data is 50.
Let the assumed mean A = 50 and h = 10.
Marks  MidValue(x_{i})  Frequency  ${\mathit{u}}_{\mathit{i}}\mathbf{=}\frac{{\mathit{x}}_{\mathit{i}}\mathbf{}\mathbf{50}}{\mathbf{20}}$  ${\mathit{f}}_{\mathit{i}}{\mathit{u}}_{\mathit{i}}$ 
0–20  10  17  −2  −34 
20–40  30  f_{1}  −1  −f_{1} 
40–60  50  32  0  0 
60–80  70  f_{2}  1  f_{2} 
80–100  90  19  2  38 
N = 68 + f_{1 }+ f_{2}  $\sum _{}{f}_{i}{u}_{i}=4{f}_{1}+{f}_{2}$ 
We know that mean,
Now, we have N = 68 + f_{1 }+ f_{2}, $h=20,A=50,$ $\sum _{}{f}_{i}{u}_{i}=4{f}_{1}+{f}_{2}$.
$50=50+20\left(\frac{4{f}_{1}+{f}_{2}}{68+{f}_{1}+{f}_{2}}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{1}{f}_{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{1}=4+{f}_{2}.....\left(1\right)$
Now, N = 68 + f_{1 }+ f_{2} = 120
${f}_{1}+{f}_{2}=12068=52\phantom{\rule{0ex}{0ex}}\Rightarrow 4+{f}_{2}+{f}_{2}=52\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {f}_{2}=24\phantom{\rule{0ex}{0ex}}\mathrm{So},\phantom{\rule{0ex}{0ex}}{f}_{1}=4+24=28$
Page No 15.25:
Question 29:
The daily income of a sample of 50 employees are tabulated as follows:
Income (in â‚ą): 1200 201400 401600 601800
No.of employees : 14 15 14 7
Find the mean daily income of employees.
Answer:
Let the assumed mean A = 500.5 and h = 200.
Marks  MidValue(x_{i})  Frequency  ${\mathit{u}}_{\mathbf{i}}\mathbf{=}\frac{{\mathit{x}}_{\mathit{i}}\mathbf{}\mathbf{500}\mathbf{.}\mathbf{5}}{\mathbf{200}}$  ${\mathit{f}}_{\mathit{i}}{\mathit{u}}_{\mathit{i}}$ 
1–200  100.5  14  −2  −28 
201–400  300.5  15  −1  −15 
401–600  500.5  14  0  0 
601–800  700.5  7  1  7 
N = 50  $\sum _{}{f}_{i}{u}_{i}=36$ 
We know that mean,
Now, we have $N=\sum _{}{f}_{i}=50,h=200,A=500.5,\sum _{}{f}_{i}{u}_{i}=36$
$\overline{X}=500.5+200\left[\frac{1}{50}\times \left(36\right)\right]\phantom{\rule{0ex}{0ex}}=500.5144\phantom{\rule{0ex}{0ex}}=356.5$
Page No 15.34:
Question 1:
Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719
Answer:
First of all arranging the data in ascending order of magnitude, we have
Here,, which is an odd number
Therefore, median is the value of
Page No 15.34:
Question 2:
The following is the distribution of height of students of a certain class in a certain city:
Height (in cms):  160−162  163−165  166−168  169−171  172−174 
No. of students:  15  118  142  127  18 
Find the median height.
Answer:
First we prepare the following cummulative table to compute the median.
Now,
∴
Thus, the cumulative frequency just greater than 210 is 275 and the corresponding class is .
Therefore, is the median class.
We know that,
Hence, the median height is approximately 167.1 cm.
Page No 15.34:
Question 3:
Following is the distribution of I.Q. of 100 students. Find the median I.Q.
I.Q.:  55−64  65−74  75−84  85−94  95−104  105−114  115−124  125−134  135144 
No. of students:  1  2  9  22  33  22  8  2  1 
Answer:
Here, the frequency table is given in inclusive form. Transforming the given table into exclusive form and prepare the cumulative frequency table.
IQ  Frequency(f_{i})  Cummulative Frequency(c.f.) 
54.5−64.5  1  1 
64.5−74.5  2  3 
74.5−84.5  9  12 
84.5−94.5  22  34 
94.5−104.5  33  67 
104.5−114.5  22  89 
114.5−124.5  8  97 
124.5−134.5  2  99 
134.5−144.5  1  100 
N = 100 
Here,
So,
Thus, the cumulative frequency just greater than 50 is 67 and the corresponding class is 94.5−104.5.
Therefore, 94.5−104.5 is the median class.
Here, l = 94.5, f = 33, F = 34 and h = 9
We know that,
$=94.5+\left(\frac{5034}{33}\right)\times 10\phantom{\rule{0ex}{0ex}}=94.5+\frac{160}{33}\phantom{\rule{0ex}{0ex}}=94.5+4.85\phantom{\rule{0ex}{0ex}}=99.35$
Hence, the median is 99.35.
Page No 15.34:
Question 4:
Calculate the median salary of the following data giving salaries of 280 persons:
Salary (in thousands)  5 – 10  10 – 15  15 – 20  20 – 25  25 – 30  30 – 35  35 – 40  40 – 45  45 – 50 
No. of Persons  49  133  63  15  6  7  4  2  1 
Answer:
Salary (In thousand Rs)  Frequency  CF 
5 – 10  49  49 
10 – 15  133  182 
15 – 20  63  245 
20 – 25  15  260 
25 – 30  6  266 
30 – 35  7  273 
35 – 40  4  277 
40 – 45  2  279 
45 – 50  1  280 
$\frac{N}{2}=\frac{280}{2}=140$
The cumulative frequency which is greater than and nearest to 140 is 182.
Median class = 10 – 15
We also have,
l (lower limit of median class) = 10
h (class size) = 5
n (number of observations) = 280
cf = (cumulative frequency of the class preceding the median class) = 49
f (frequency of median class) = 133
Median for grouped data is given by the formula :
Median = $l+\left(\frac{{\displaystyle \frac{n}{2}}cf}{f}\right)\times h$
where f is frequency of median class and $cf$ is cumulative frequency of previous class.
Median = 10 + $\frac{\left(14049\right)}{133}\times 5$ = Rs 13.42.
Disclaimer: The obtained answer does not match the answer in the book.
Page No 15.34:
Question 5:
Calculate the median from the following data:
Marks below:  10  20  30  40  50  60  70  80 
No. of students:  15  35  60  84  96  127  198  250 
Answer:
First we prepare the following cummulative table to compute the median.
Here,
So,
Thus, the cumulative frequency just greater than 125 is 127 and the corresponding class is .
Therefore, is the median class.
Here,
We know that
Hence, the median is 59.35.
Page No 15.34:
Question 6:
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Age in years:  0−10  10−20  20−30  30−40  40−50 
No. of persons:  5  25  ?  18  7 
Answer:
Let the frequency of the class 20−30 be.It is given that median is 35 which lies in the class 20−30. So 20−30 is the median class.
Now, lower limit of median class
Height of the class
Frequency of median class
Cumulative frequency of preceding median class
Total frequency
Formula to be used to calculate median,
Where,
 Lower limit of median class
 Height of the class
 Frequency of median class
 Cumulative frequency of preceding median class
 Total frequency
Put the values in the above,
Hence, the required frequency is 25.
Page No 15.34:
Question 7:
The following table gives the frequency distribution of married women by age at marriage:
Age (in years) 
Frequency  Age (in years) 
Frequency 
15−19 20−24 25−29 30−34 35−39 
53 140 98 32 12 
40−44 45−49 50−54 55−59 60 and above 
9 5 3 3 2 
Calculate the median and interpret the results.
Answer:
Here, the frequency table is given in inclusive form. So, we first transform it into exclusive form by subtracting and adding h/2 to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and upper limit of the previous class.
We have, N = 357
So, N/2 = 178.5
Thus, the cumulative frequency just greater than 178.5 is 193 and the corresponding class is 19.5−24.5.
Therefore, 19.5−24.5 is the median class.
Here, l = 19.5, f = 140, F = 193 and h = 5
We know that
$=19.5+\left(\frac{178.553}{140}\right)\times 5\phantom{\rule{0ex}{0ex}}=19.5+\frac{125.5}{140}\times 5\phantom{\rule{0ex}{0ex}}=19.5+\frac{125.5}{28}\phantom{\rule{0ex}{0ex}}=\frac{546+125.5}{28}\phantom{\rule{0ex}{0ex}}=23.98$
Hence, the median age 23.98 years.
Thus, nearly half the women were married between the age of 19.5 years and 24.5 years.
Page No 15.34:
Question 8:
The following table gives the distribution of the life time of 400 neon lamps:
Lite time: (in hours) 
Number of lamps 
1500−2000 2000−2500 2500−3000 3000−3500 3500−4000 4000−4500 4500−5000 
14 56 60 86 74 62 48 
Find the median life.
Answer:
We prepare the cumulative frequency table, as given below.
We have,
So,
Now, the cumulative frequency just greater than 200 is 216 and the corresponding class is .
Therefore, is the median class.
Here,
We know that
Hence, the median life of the lamps is approximately 3406.98 hours.
Page No 15.35:
Question 9:
The distribution below gives the weight of 30 students in a class. Find the median weight of students:
Weight (in kg):  40−45  45−50  50−55  55−60  60−65  65−70  70−75 
No. of students:  2  3  8  6  6  3  2 
Answer:
We prepare the cumulative frequency table, as given below.
We have,
So,
Now, the cumulative frequency just greater than 15 is 19 and the corresponding class is .
Therefore, is the median class.
Here,
We know that
Hence, the median weight of students is 56.67 kg.
Page No 15.35:
Question 10:
Find the missing frequencies and the median for the following distribution if the mean is 1.46.
No. of accidents:  0  1  2  3  4  5  Total 
Frequency:  46  ?  ?  25  10  5  200 
Answer:
(1) Let the missing frequencies be x and y.
Given:
.....(1)
We know that mean,
.....(2)
Solving (1) and (2), we get
Therefore,
Hence, the missing frequencies are 38 and 76.
(2) Calculation of median.
Now, we have.
So,
Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.
Hence, the median is 1.
Page No 15.35:
Question 11:
An incomplete distribution is given below :
Variable:  10−20  20−30  30−40  40−50  50−60  60−70  70−80 
Frequency:  12  30  −  65  −  25  18 
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Answer:
(i) Let the missing frequencies be x and y.
First we prepare the following cummulative table.
Here,
So,
It is given that the median is 46.
Therefore, is the median class.
We know that
Also,
Putting the value of x, we get
Hence, the missing frequencies are 34 and 46.
(ii)
We may prepare the table as shown.We know that mean,
Now, we have.
Putting the values in the above formula, we have
Hence, the mean is 45.87.
Page No 15.35:
Question 12:
If the median of the following frequency distribution is 28.5 find the missing frequencies:
Class interval:  0−10  10−20  20−30  30−40  40−50  50−60  Total 
Frequency:  5  f_{1}  20  15  f_{2}  5  60 
Answer:
Given: Median = 28.5
We prepare the cumulative frequency table, as given below.
Now, we have
.....(1)
Also,
Since the median = 28.5 so the median class is 20−30.
Here,
We know that
Putting the value of in (1), we get
Hence, the missing frequencies are 7 and 8.
Page No 15.35:
Question 13:
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observation in the data:
Class interval  Frequency  Class interval  Frequency 
0−100 100−200 200−300 300−400 400−500 
2 5 f_{1} 12 17 
500−600 600−700 700−800 800−900 900−1000 
20 f_{2} 9 7 4 
Answer:
Given: Median = 525
We prepare the cumulative frequency table, as given below.
Now, we have
.....(1)
So,
Since median = 525 so the median class is .
Here,
We know that
Putting the value of in (1), we get
Hence, the missing frequencies are 9 and 15.
Page No 15.35:
Question 14:
If the median of the following data is 32.5, find the missing frequencies.
Class interval:  0−10  10−20  20−30  30−40  40−50  50−60  60−70  Total 
Frequency:  f_{1}  5  9  12  f_{2}  3  2  40 
Answer:
Given: Median = 32.5
We prepare the cumulative frequency table, as given below.
Now, we have
.....(1)
Also,
Since median = 32.5 so the median class is .
Here,
We know that
Putting the value of in (1), we get
Hence, the missing frequencies are 3 and 6.
Page No 15.35:
Question 15:
Compute the median for each of the following data:
(i)
Marks  No. of students 
Less than 10 Less than 30 Less than 50 Less than 70 Less than 90 Less than 110 Less than 130 Less than 150 
0 10 25 43 65 87 96 100 
(ii)
Marks  No. of students 
More than 150 More than 140 More than 130 More than 120 More than 110 More than 100 More than 90 More than 80 
0 12 27 60 105 124 141 150 
Answer:
(i)
We prepare the cumulative frequency table, as given below.
Now, we have
So,
Now, the cumulative frequency just greater than 50 is 65 and the corresponding class is .
Therefore, is the median class.
Here,
We know that
Hence, the median is 76.36.
Note: The first class in the table can be omitted also.
(ii)
We prepare the cumulative frequency table, as given below.
Now, we have
So.
Thus, the cumulative frequency just greater than 75 is 105 and the corresponding class is .
Therefore, is the median class.
(Because class interval given in descending order)
We know that
= 120 − 3.333
= 116.67 (approx)
Hence, the median is 116.67.
Page No 15.36:
Question 16:
A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:
Height in cm  Number of Girls 
Less than 140 Less than 145 Less than 150 Less than 155 Less than 160 Less than 165 
4 11 29 40 46 51 
Find the median height.
Answer:
We prepare the cumulative frequency, table as given below.
Now, we have
So,
Now, the cumulative frequency just greater than 25.5 is 40 and the corresponding class is .
Therefore, is the median class.
We know that
= 150 − 1.59
= 148.41
Hence, the median height is 148.41 cm.
Page No 15.36:
Question 17:
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onward but less than 60 years.
Age in years  Number of policy holders 
Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60 
2 6 24 45 78 89 92 98 100 
Answer:
We prepare the cumulative frequency, table as given below.
Now, we have
So,
Now, the cumulative frequency just greater than 50 is 78 and the corresponding class is .
Therefore, is the median class.
Here,
We know that
Hence, the median age is 35.76 years.
Page No 15.36:
Question 18:
The length of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length (in mm):  118−126  127−135  136−144  145−153  154−162  163−171  172−180 
No. of leaves  3  5  9  12  5  4  2 
Find the mean length of leaf.
Answer:
Calculation for mean.
Length (in mm)  MidValues(x_{i})  Number of Leaves(f_{i}_{ })  f_{i} x_{i} 
117.5–126.5  122  3  366 
126.5–135.5  131  5  655 
135.5–144.5  140  9  1260 
144.5–153.5  149  12  1788 
153.5–162.5  158  5  790 
162.5–171.5  167  4  668 
171.5–180.5  176  2  353 
N = 40  $\sum _{}{f}_{i}{x}_{i}=$5880 
Mean length of the leaf = $\frac{1}{N}\sum _{}^{}{f}_{i}{x}_{i}=\frac{1}{40}\times 5880=147$
Calculation for median.
The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have
Length (in mm)  Number of Leaves(f_{i}_{ })  Cumulative Frequency (c.f.) 
117.5–126.5  3  3 
126.5–135.5  5  8 
135.5–144.5  9  17 
144.5–153.5  12  29 
153.5–162.5  5  34 
162.5–171.5  4  38 
171.5–180.5  2  40 
N = 40 
Now, we have
So,
Now, the cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5–153.5.
Therefore, 144.5–153.5 is the median class.
Here,
We know that
$=144.5+\left(\frac{2017}{12}\right)\times 9\phantom{\rule{0ex}{0ex}}=144.5+\frac{27}{12}\phantom{\rule{0ex}{0ex}}=144.5+2.25\phantom{\rule{0ex}{0ex}}=146.75$
Hence, the median length of leaf is 146.75 mm.
Disclaimer: If the question asks for the mean length of the leaf, then the answer is 147 mm whereas if the question asks fro the median length of the leaf, then the answer is 146.75 mm, which is same as the answer given in the book.
Page No 15.36:
Question 19:
An incomplete distribution is given as follows:
Variable:  0−10  10−20  20−30  30−40  40−50  50 − 60  60 − 70 
Frequency:  10  20  ?  40  ?  25  15 
You are given that the median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies.
Answer:
Let the frequency of the class 20−30 be and that of class 40−50 be. The total frequency is 170. So,
So, .....(1)
It is given that median is 35 which lies in the class 3040. So 3040 is the median class.
Now, lower limit of median class
Height of the class
Frequency of median class
Cumulative frequency of preceding median class
Total frequency
Formula to be used to calculate median,
Where,
 Lower limit of median class
 Height of the class
 Frequency of median class
 Cumulative frequency of preceding median class
 Total frequency
Put the values in the above,
Using equation (1), we have
Therefore,
Page No 15.36:
Question 20:
The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20.
Class interval: 06 612 1218 1824 2430
Frequency: 4 x 5 y 1
Answer:
The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have
Class interval  Frequency(f_{i}_{ })  Cumulative Frequency(c.f.) 
0–6  4  4 
6–12  x  4 + x 
12–18  5  9 + x 
18–24  y  9 + x + y 
24–30  1  10 + x + y 
10 + x + y = 20 
Median = 14.4
It lies in the interval 12–18, so the median class is 12–18.
Now, we have
$l=12,h=6,f=5,F=4+x,N=20$
We know that
$14.4=12+\frac{6\times \left(104x\right)}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 12=366x\phantom{\rule{0ex}{0ex}}\Rightarrow 6x=24\phantom{\rule{0ex}{0ex}}\Rightarrow x=4$
Now,
10 + x + y = 20
$\Rightarrow x+y=10\phantom{\rule{0ex}{0ex}}\Rightarrow y=104=6$
Page No 15.36:
Question 21:
The median of the following data is 50. Find the values of p and q , if the sum of all the frequencies is 90 .
Marks: 2030 3040 4050 5060 6070 7080 8090
Frequency: p 15 25 20 q 8 10
Answer:
The given series is in inclusive form. Converting it to exclusive form and preparing the cumulative frequency table, we have
Class interval  Frequency(f_{i}_{ })  Cumulative Frequency (c.f.) 
20–30  p  p 
30–40  15  p + 15 
40–50  25  p + 40 
50–60  20  p + 60 
60–70  q  p + q + 60 
70–80  8  p + q + 68 
80–90  10  p + q + 78 
78 + p + q = 90 
Median = 50
It lies in the interval 50–60, so the median class is 50–60.
Now, we have
$l=50,h=10,f=20,F=p+40,N=90$
We know that
$50=50+\frac{45\left(p+40\right)}{20}\times 10\phantom{\rule{0ex}{0ex}}\Rightarrow 0=\frac{5p}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow p=5\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}p+q+78=90\phantom{\rule{0ex}{0ex}}\Rightarrow p+q=12\phantom{\rule{0ex}{0ex}}\Rightarrow q=125=7$
Page No 15.45:
Question 1:
Find the mode of the following data:
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
Answer:
(i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
The frequency table for the given data
Value x  3  4  5  6  7  8  9 
Frequency f  4  2  5  2  2  1  2 
We observe that the value 5 has the maximum frequency.
Hence, the mode of data is 5.
(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4
The frequency table for the given data
Value x  3  4  5  6  7  8  9 
Frequency f  5  2  4  2  2  1  2 
We observe that the value 3 has the maximum frequency.
Hence, the mode of data is 3.
(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15
The frequency table for the given data
Value x  8  15  18  19  20  24  25  26 
Frequency f  1  4  1  1  1  2  1  1 
We observe that the value 15 has the maximum frequency.
Hence, the mode of data is 15.
Page No 15.45:
Question 2:
The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:
Shirt Size:  37  38  39  40  41  42  43  44 
Number of persons:  15  25  39  41  36  17  15  12 
Find the modal shirt size worn by the group.
Answer:
Shirt Size  37  38  39  40  41  42  43  44 
Number of persons  15  25  39  41  36  17  15  12 
Here, shirt size 40 has the maximum number of persons.
Hence, the mode shirt size is 40.
Page No 15.45:
Question 3:
Find the mode of the following distribution.
(i)
Classinterval:  0−10  10−20  20−30  30−40  40−50  50−60  60−70  70−80 
Frequency:  5  8  7  12  28  20  10  10 
(ii)
Classinterval:  10−15  15−20  20−25  25−30  30−35  35−40 
Frequency:  30  45  75  35  25  15 
(iii)
Classinterval:  25−30  30−35  35−40  40−45  45−50  50−55 
Frequency:  25  34  50  42  38  14 
Answer:
(i) Here, maximum frequency is 28 so the modal class is 40−50.
Therefore,
(ii) Here, maximum frequency is 75 so the modal class is 20−25.
Therefore,(iii) Here, maximum frequency is 50 so the modal class is 35−40.
Therefore,Page No 15.45:
Question 4:
Compare the modal ages of two groups of students appearing for an entrance test:
Age (in years):  16−18  18−20  20−22  22−24  24−26 
Group A:  50  78  46  28  23 
Group B:  54  89  40  25  17 
Answer:
Age (in years)  Group ‘A’  Group ‘B’ 
16–18  50  54 
18–20  78  89 
20–22  46  40 
22–24  28  25 
24–25  23  17 
For group “A”
The maximum frequency is 78 so the modal class is 18–20.
Therefore,For group “B”
The maximum frequency 89 so modal class 18–20.
Therefore,Thus, the modal age of group A is 18.93 years whereas the modal age of group B is 18.83 years.
Page No 15.46:
Question 5:
The marks in science of 80 students of class X are given below: Find the mode of the marks obtained by the students in science.
Marks:  0−10  10−20  20−30  30−40  40−50  50−60  60−70  70−80  80−90  90−100 
Frequency:  3  5  16  12  13  20  5  4  1  1 
Answer:
Marks  0−10  10−20  20−30  30−40  40−50  50−60  60−70  70−80  80−90  90−100 
Frequency  3  5  16  12  13  20  5  4  1  1 
Here, the maximum frequency is 20 so the modal class is 50−60.
Therefore,
Now,
Thus, the mode of the marks obtained by the students in science is 53.17.
Page No 15.46:
Question 6:
The following is the distribution of height of students of a certain class in a certain city:
Height (in cms):  160−162  163−165  166−168  169−171  172−174 
Number of students:  15  118  142  127  18 
Find the average height of maximum number of students.
Answer:
The given data is an inclusive series. So, firstly convert it into an exclusive series as given below.
Height (in cm)  159.5−162.5  162.5−165.5  165.5−168.5  168.5−171.5  171.5−174.5 
Number of students  15  118  142  127  18 
Here, the maximum frequency is 142 so the modal class is 165.5−168.5.
Therefore,
Now,
= 165.5 + 1.85
= 167.35
Thus, the average height of maximum number of students is 167.35 cm.
Page No 15.46:
Question 7:
The following table show the ages of the patients admitted in a hospital during a year:
Age (in years):  5−15  15−25  25−35  35−45  45−55  55−65 
Number of patients:  6  11  21  23  14  5 
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
Age (in years)  5−15  15−25  25−35  35−45  45−55  55−65 
Number of patients  6  11  21  23  14  5 
Here, the maximum frequency is 23 so the modal class is 35−45.
Therefore,
Thus, the mode of the ages of the patients is 36.8 years.
Calculation for mean.
Age (in years)  MidValues(x)  Number of patients(f)  fx 
5−15  10  6  60 
15−25  20  11  220 
25−35  30  21  630 
35−45  40  23  920 
45−55  50  14  700 
55−65  60  5  300 
Mean = $\frac{\sum _{}fx}{\sum _{}f}=\frac{2830}{80}=35.37$
Thus, the mean age of the patients is 35.37 years.
The mean age of the patients is less than the modal age of the patients.
Page No 15.46:
Question 8:
The following data is gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in hours)  0−20  20−40  40−60  60−80  80−100  100−120 
No. of components  10  35  52  61  38  29 
Determine the modal lifetimes of the components.
Answer:
Lifetimes (in hours)  0−20  20−40  40−60  60−80  80−100  100−120 
No. of components  10  35  52  61  38  29 
Here, the maximum frequency of electrical components is 61 so the modal class is 60−80.
Therefore,
Thus, the modal lifetimes of the components is 65.625 hours.
Page No 15.46:
Question 9:
The following table gives the daily income of 50 workers of a factory :
Daily income (in â‚ą) 100120 120140 140160 160180 180200
Number of workers : 12 14 8 6 10
Find the mean, mode and median of the above data .
Answer:
Let the assumed mean a be 150.
The table for the given data can be drawn as

Class Interval
Number of Workers(f_{i})
Classmark(x_{i})
d_{i} = x_{i} − 150
f_{i}d_{i}
Cumulative Frequency(c.f.)
100−120
12
110
−40
−480
12
120−140
14
130
−20
−280
26
140−160
8
150
0
0
34
160−180
6
170
20
120
40
180−200
10
190
40
400
50
Total
50
−240
Mean is given by .
Thus, the mean of the given data is 145.2.
It can be seen in the data table that the maximum frequency is 14. The class corresponding to this frequency is 120−140.
∴ Modal class = 120 − 140
Lower limit of modal class (l) = 120
Class size (h) = 140 − 120 = 20
Frequency of modal class (f_{1}) = 14
Frequency of class preceding the modal class (f_{1}) = 12
Frequency of class succeeding the modal class (f_{1}) = 8
Mode is given by
Thus, the mode of the given data is 125.
Here, number of observations (n) = 50
This observation lies in class interval 120−140.
Therefore, the median class is 120−140.
Lower limit of median class (l) = 120
Cumulative frequency of class preceding the median class(c.f.) = 12
Frequency of median class(f) = 14
Median of the data is given by
Thus, the median of the given data is 138.57.
Page No 15.46:
Question 10:
The following distribution gives the statewise teacherstudent ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:
Number of students per Teacher  Number of States/U.T  Number of students per Teacher  Number of State/ U.T 
15−20 20−25 25−30 30−35 
3 8 9 10 
35−40 40−45 45−50 50−55 
3 0 0 2 
Answer:
Number of students per teacher 
Number of states/U.T. (f_{i}) 
x_{i} 
f_{ }_{i} x_{i} 
15−20  3  17.5  52.5 
20−25  8  22.5  180.0 
25−30  9  27.5  247.5 
30−35  10  32.5  325.0 
35−40  3  37.5  112.5 
40−45  0  42.5  0 
45−50  0  47.5  0 
50−55  2  52.5  105.0 
Here, the maximum frequency is 10 so the modal class is 30−35.
Therefore,
Thus, the mode of the data is 30.6.
Mean of the data = $\frac{\sum _{}{f}_{i}{x}_{i}}{\sum _{}{f}_{i}}=\frac{1022.5}{35}=29.2$
Thus, the mean of the data is 29.2.
The mode of the number of students per teacher in the states is more than the mean of the number of students per teacher in the states.
Page No 15.46:
Question 11:
Find the mean, median and mode of the following data:
Classes:  0−50  50−100  100−150  150−200  200−250  250−300  300−350 
Frequency:  2  3  5  6  5  3  1 
Answer:
Classes  Frequency (f_{i})  x_{i}  f_{i} x_{i}  C.f. 
0−50  2  25  50  2 
50−100  3  75  225  5 
100−150  5  125  625  10 
150−200  6  175  1050  16 
200−250  5  225  1125  21 
250−300  3  275  825  24 
300−350  1  325  325  25 
$\sum _{}{f}_{i}{x}_{i}=4225$ 
Here, the maximum frequency is 6 so the modal class 150−200.
Therefore,
Thus, the mean of the data is 169.
Thus, the median of the data is 170.83.
= 175
Thus, the mode of the data is 175.
Page No 15.46:
Question 12:
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data:
Number of cars:  0−10  10−20  20−30  30−40  40−50  50−60  60−70  70−80 
Frequency:  7  14  13  12  20  11  15  8 
Answer:
The given data is shown below.
Number of cars  Frequency 
0−10  7 
10−20  14 
20−30  13 
30−40  12 
40−50  20 
50−60  11 
60−70  15 
70−80  8 
Here, the maximum frequency is 20 so the modal class is 40−50.
Therefore,
$\therefore \mathrm{Mode}=l+\frac{f{f}_{1}}{2f{f}_{1}{f}_{2}}\times h\phantom{\rule{0ex}{0ex}}=40+\frac{2012}{2\times 201211}\times 10\phantom{\rule{0ex}{0ex}}=40+\frac{8}{17}\times 10\phantom{\rule{0ex}{0ex}}=40+4.7\phantom{\rule{0ex}{0ex}}=44.7$
Thus, the mode of the data is 44.7.
Page No 15.46:
Question 13:
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption: (in units)  65−85  85−105  105−125  125−145  145−165  165−185  185−205 
No. of consumers:  4  5  13  20  14  8  4 
Answer:
The given data is shown below.
Monthly Consumption (in units)  No. of consumers (f_{ i})  x_{i}  f_{i} x_{i}  C.f. 
65−85  4  75  300  4 
85−105  5  95  475  9 
105−125  13  115  1495  22 
125−145  20  135  2700  42 
145−165  14  155  2170  56 
165−185  8  175  1400  64 
185−205  4  195  780  68 
Here, the maximum frequency is 20 so the modal class is 125−145.
Therefore,
Thus, the mode of the monthly consumption of electricity is 135.76 units.
Mean = = 137.05
Thus, the mean of the monthly consumption of electricity is 137.05 units.
Here,
Total number of consumers, N = 68 (even)
Then, $\frac{N}{2}=34$
∴ Median
= 137
Thus, the median of the monthly consumption of electricity is 137 units.
Page No 15.47:
Question 14:
100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters:  1−4  4−7  7−10  10−13  13−16  16−19 
Number of surnames:  6  30  40  16  4  4 
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, fund the modal size of the surnames.
Answer:
Consider the following table.
Number of letters  No. of surname  x_{i}  f_{i} x_{i}  C.f. 
1−4  6  2.5  15  6 
4−7  30  5.5  165  36 
7−10  40  8.5  340  76 
10−13  16  11.5  184  92 
13−16  4  14.5  58  96 
16−19  4  17.5  70  100 
Here, the maximum frequency is 40 so the modal class is 7−10.
Therefore,
Thus, the modal sizes of the surnames is 7.88.
Thus, the mean number of letters in the surnames is 8.32.
Median
Thus, the median number of letters in the surnames is 8.05.
Page No 15.47:
Question 15:
Find the mean, median and mode of the following data:
Classes:  0−20  20−40  40−60  60−80  80−100  100−120  120−140 
Frequency:  6  8  10  12  6  5  3 
Answer:
Consider the following data.
Class  Frequency (f_{i})  x_{i}  f_{i} x_{i}  C.f. 
0−20  6  10  60  6 
20−40  8  30  240  14 
40−60  10  50  500  24 
60−80  12  70  840  36 
80−100  6  90  540  42 
100−120  5  110  550  47 
120−140  3  130  390  50 
Here, the maximum frequency is 12 so the modal class is 60−80.
Therefore,
Thus, the median of the data is 61.66.
Thus, the mean of the data is 62.4.
Mode
Thus, the mode of the data is 65.
Page No 15.47:
Question 16:
The following data gives the distribution of total monthly household expenditure of 200 families of a villages. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
Expenditure (in Rs.) 
Frequency  Expenditure (in Rs.) 
Frequency 
1000−1500 1500−2000 2000−2500 2500−3000 
24 40 33 28 
3000−3500 3500−4000 4000−4500 4500−5000 
30 22 16 7 
Answer:
Expenditure  Frequency (f_{ i})  x_{i}  f_{ }_{i}x_{i} 
1000−1500  24  1250  30000 
1500−2000  40  1750  70000 
2000−2500  33  2250  74250 
2500−3000  28  2750  77000 
3000−3500  30  3250  97500 
3500−4000  22  3750  82500 
4000−4500  16  4250  68000 
4500−5000  7  4750  33250 
Here, the maximum frequency is 40 so the modal class is 1500−2000.
Therefore,
= 1500 + 347.83
= Rs 1847.83
Thus, the modal monthly expenditure of the families is Rs 1847.83.
Now,
Mean monthly expenditure of the families
Thus, the mean monthly expenditure of the families is Rs 2662.50.
Page No 15.47:
Question 17:
The given distribution shows the number of runs scored by some top batsmen of the world in oneday international cricket matches.
Runs scored  Number of batsman  Runs scored  Number of Batsman 
3000−4000 4000−5000 5000−6000 6000−7000 
4 18 9 7 
7000−8000 8000−9000 9000−10000 10000−11000 
6 3 1 1 
Find the mode of data.
Answer:
The given data is shown below.
Runs scored  Number of batsmen 
3000−4000  4 
4000−5000  18 
5000−6000  9 
6000−7000  7 
7000−8000  6 
8000−9000  3 
9000−10000  1 
10000−11000  1 
Here, the maximum frequency is 18 so the modal class is 40005000.
Therefore,
$\therefore \mathrm{Mode}=l+\frac{f{f}_{1}}{2f{f}_{1}{f}_{2}}\times h\phantom{\rule{0ex}{0ex}}=4000+\frac{184}{2\times 1849}\times 1000\phantom{\rule{0ex}{0ex}}=4000+\frac{14}{23}\times 1000\phantom{\rule{0ex}{0ex}}=4000+608.7\phantom{\rule{0ex}{0ex}}=4608.7$
Thus, the mode of the data (or runs scored) is 4608.7.
Page No 15.47:
Question 18:
The frequency distribution table of agriculture holdings in a village is given below:
Area of land (in hectares): 13 35 57 79 911 1113
Number of families: 20 45 80 55 40 12
Find the modal agricultural holdings of the village .
Answer:
The maximum class frequency is 80. The class corresponding to this frequency is 5–7.
So, the modal class is 5–7.
l (the lower limit of modal class) = 5
f_{1} (frequency of the modal class) = 80
f_{0 }(frequency of the class preceding the modal class) = 45
f_{2 }(frequency of the class succeeding the modal class) = 55
h (class size) = 2
$\mathrm{Mode}=l+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h\phantom{\rule{0ex}{0ex}}=5+\left(\frac{8045}{2\times 804555}\right)\times 2\phantom{\rule{0ex}{0ex}}=5+\frac{35}{60}\times 2\phantom{\rule{0ex}{0ex}}=5+\frac{35}{30}\phantom{\rule{0ex}{0ex}}=6.2$
Hence, the modal agricultural holdings of the village is 6.2 hectares.
Page No 15.48:
Question 19:
The monthly income of 100 families are given as below :
Income in ( in â‚ą) Number of families
05000 8
500010000 26
1000015000 41
1500020000 16
2000025000 3
2500030000 3
3000035000 2
3500040000 1
Calculate the modal income.
Answer:
The maximum class frequency is 41. The class corresponding to this frequency is 10000–15000.
So, the modal class is 10000–15000.
l (the lower limit of modal class) = 10000
f_{1} (frequency of the modal class) = 41
f_{0 }(frequency of the class preceding the modal class) = 26
f_{2 }(frequency of the class succeeding the modal class) = 16
h (class size) = 5000
$\mathrm{Mode}=l+\left(\frac{{f}_{1}{f}_{0}}{2{f}_{1}{f}_{0}{f}_{2}}\right)\times h\phantom{\rule{0ex}{0ex}}=10000+\left(\frac{4126}{2\times 412616}\right)\times 5000\phantom{\rule{0ex}{0ex}}=10000+\frac{15}{40}\times 5000\phantom{\rule{0ex}{0ex}}=10000+1875\phantom{\rule{0ex}{0ex}}=11875$
Thus, the modal income is â‚ą11875.
Page No 15.5:
Question 1:
Calculate the mean for the following distribution :
x:  5  6  7  8  9 
f:  4  8  14  11  3 
Answer:
Given:
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
Hence, mean
Page No 15.5:
Question 2:
Find the mean of the following data:
x:  19  21  23  25  27  29  31 
f:  13  15  16  18  16  15  13 
Answer:
Given:
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
Hence, mean
Page No 15.5:
Question 3:
If the mean of the following data is 20.6. Find the value of p.
x:  10  15  p  25  35 
f:  3  10  25  7  5 
Answer:
Given:
Also, mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method,
Hence, p
Page No 15.5:
Question 4:
If the mean of the following data is 15, find p.
x:  5  10  15  20  25 
f:  6  p  6  10  5 
Answer:
Given:
Also, mean = 15
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method
Hence, p
Page No 15.5:
Question 5:
Find the value of p for the following distribution whose mean is 16.6.
x:  8  12  15  p  20  25  30 
f:  12  16  20  24  16  8  4 
Answer:
Given:
Mean = 16.6
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method
Hence, p
Page No 15.5:
Question 6:
Find the missing value of p for the following distribution whose mean is 12.58
x:  5  8  10  12  p  20  25 
f:  2  5  8  22  7  4  2 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method
Hence, p
Page No 15.5:
Question 7:
Find the missing frequency (p) for the following distribution whose mean is 7.68.
x:  3  5  7  9  11  13 
f:  6  8  15  p  8  4 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method,
$303+9p=314.88+7.68p\phantom{\rule{0ex}{0ex}}9p7.68p=314.88303$
Hence, p
Page No 15.5:
Question 8:
The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students
Age (in years):  15  16  17  18  19  20 
No. of students:  3  8  10  10  5  4 
Answer:
Given:
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
Hence, the mean age of the students
Page No 15.6:
Question 9:
Candidate of four schools appear in a mathematics test. The data were as follows:
Schools  No. of Candidates  Average Score 
I II III IV 
60 48 Not available 40 
75 80 55 50 
If the average score of the candidates of all the four schools is 66, find the number of candidates that appeared from school III.
Answer:
Given:
Mean score of the candidates = 66
Let the number of candidates that appeared from school III be x.
First of all prepare the frequency table in such a way that its first column consists of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
$66=\frac{10340+55x}{260}$
By using cross multiplication method,
Hence, the number of candidates that appeared from school III is 124.
Page No 15.6:
Question 10:
Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.
No. of heads per toss  No. of tosses 
0 1 2 3 4 5 
38 144 342 287 164 25 
Total  1000 
Answer:
Given:
First of all prepare the frequency table in such a way that its first column consist of the numnber of heads per tosses and the second column the corresponding number of tosses .
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denote by and in the third column to obtain.
We know that mean,
$=\frac{2470}{1000}\phantom{\rule{0ex}{0ex}}=2.47$
Hence, the mean number of heads per toss is 2.47.
Page No 15.6:
Question 11:
The arithmetic mean of the following data is 14. Find the value of k.
x_{i}:  5  10  15  20  25 
f_{i}:  7  k  8  4  5 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method,
Hence, k = 6
Page No 15.6:
Question 12:
The arithmetic mean of the following data is 25, find the value of k.
x_{i}:  5  15  25  35  45 
f_{i}:  3  k  3  6  2 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method,
Hence, k = 4
Page No 15.6:
Question 13:
If the mean of the following data is 18.75. Find the value of p.
x_{i}:  10  15  p  25  30 
f_{i}:  5  10  7  8  2 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
By using cross multiplication method,
Hence, p = 20
Page No 15.6:
Question 14:
Find the value of p, if the mean of the following distribution is 20.
x:  15  17  19  20+p  23 
f:  2  3  4  5p  6 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consist of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
We know that mean,
$20=\frac{295+5p\left(20+p\right)}{15+5p}$
By using cross multiplication method,
$5{p}^{2}+100p+295=300+100p\phantom{\rule{0ex}{0ex}}\Rightarrow 5{p}^{2}=300295=5\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}=1\phantom{\rule{0ex}{0ex}}\Rightarrow p=1$
Hence, p
Page No 15.6:
Question 15:
Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 50.
x:  10  30  50  70  90  
f:  17  f_{1}  32  f_{2}  19  Total 120 
Answer:
Given:
Mean
First of all prepare the frequency table in such a way that its first column consists of the values of the variate and the second column the corresponding frequencies.
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing.
Then, sum of all entries in the column second and denoted by and in the third column to obtain.
Now,..... (1)
We know that mean,
By using cross multiplication method,
..... (2)
Putting the value of from equation (1) in (2), we get
Therefore,
Putting the value of in equation (1), we get
Hence,
Page No 15.62:
Question 1:
Draw an ogive by less than method for the following data:
No. of rooms:  1  2  3  4  5  6  7  8  9  10 
No. of houses:  4  9  22  28  24  12  8  6  5  2 
Answer:
Firstly, we prepare the cumulative frequency table by less than method.
No. of rooms  No. of houses (f)  Cumulative frequency 
1  4  4 
2  9  13 
3  22  35 
4  28  63 
5  24  87 
6  12  99 
7  8  107 
8  6  113 
9  5  118 
10  2  120 
Now, plot the less than ogive using the points (1, 4), (2, 13), (3, 35), (4, 63), (5, 87), (6, 99), (7, 107), (8, 113), (9, 118) and (10, 120).
Page No 15.62:
Question 2:
The marks scored by 750 students in an examination are given in the form of a frequency distribution table:
Marks  No. of student  Marks  No. of students 
600−640 640−680 680−720 720−760 
16 45 156 284 
760−800 800−840 840−880 
172 59 18 
Prepare a cumulative frequency table by less than method and draw an ogive.
Answer:
Marks  No. of students  Marks less then  Cumulative frequency  Suitable Points 
600640  16  640  16  (640,16) 
640680  45  680  61  (680,61) 
680720  156  720  217  ( 720, 217) 
720760  284  470  501  (760, 501) 
760800  172  800  673  ( 800, 673) 
800840  59  840  732  ( 840,732) 
840880  18  880  750  ( 880,750) 
Now, we draw the less than ogive with suitable points.
Page No 15.63:
Question 3:
Draw an ogive to represent the following frequency distribution:
Classinterval:  0−4  5−9  10−14  15−19  20−24 
Frequency:  2  6  10  5  3 
Answer:
Firstly, prepare the cumulative frequency table.
Class Interval  No. of students  Less than  Cumulative frequency  Suitable points 
04  2  4  2  (4, 2) 
59  6  9  8  (9, 8) 
1014  10  14  18  (14, 18) 
1519  5  19  23  (19, 23) 
2024  3  24  26  (24, 26) 
Now, plot the less than ogive using the suitable points.
Page No 15.63:
Question 4:
The monthly profits (in Rs.) of 100 shops are distributed as follows:
Profits per shop:  0−50  50−100  100−150  150−200  200−250  250−300 
No. of shops:  12  18  27  20  17  6 
Draw the frequency polygon for it.
Answer:
Firstly, we make a cumulative frequency table.
Profit per shop  No. of shop  More than profit  Cumulative frequency  Suitable points 
050  12  0  100  (0, 100) 
50100  18  50  88  (50, 88) 
100150  27  100  70  (100, 70) 
150200  20  150  43  (150, 43) 
200250  17  200  23  (200, 23) 
250300  6  250  6  (250, 6) 
Now, plot the frequency polygon (or more than ogive) using suitable points.
Page No 15.63:
Question 5:
The following distribution gives the daily income of 50 workers of a factory:
Daily income in (Rs):  100−120  120−140  140−160  160−180  180−200 
Number of workers:  12  14  8  6  10 
Convert the above distribution to a less than type cumulative frequency distribution and draw its ogive.
Answer:
Daily income (in Rs.) 
No. of workers (f)  Daily income (less than)  Cumulative frequency  Suitable points 
100−120  12  120  12  (120, 12) 
120−140  14  140  26  (140, 26) 
140−160  8  160  34  (160, 34) 
160−180  6  180  40  (180, 40) 
180−200  10  200  50  (200, 50) 
Now, plot the less than ogive with suitable points.
Page No 15.63:
Question 6:
The following table gives production yield per hectare of wheat of 100 farms of a village:
Production yield in kg per hectare:  50−55  55−60  60−65  65−70  70−75  75−80  
Number of farms:  2  8  12  24  38  16 
Draw 'less than' ogive and 'more than' ogive.
Answer:
Prepare a table for less than type.
Production yield 
No. of farms 
Production yield (less than) 
Cumulative frequency 
Suitable points 
50−55  2  55  2  (55, 2) 
55−60  8  60  10  (60, 10) 
60−65  12  65  22  (65, 22) 
65−70  24  70  46  (70, 46) 
70−75  38  75  84  (75, 84) 
75−80  16  80  100  (80, 100) 
Now, plot the less than ogive using suitable points.
Again, prepare a table for more than type.
Production yield 
No. of farms 
Production yield (more than) 
Cumulative frequency 
Suitable points 
5055  2  50  100  (50, 100) 
5560  8  55  98  (55, 98) 
6065  12  60  90  (60, 90) 
6570  24  65  78  (65, 78) 
7075  38  70  54  (70, 54) 
7580  16  75  16  (75, 16) 
Now, plot the more than ogive with suitable points.
Page No 15.63:
Question 7:
During the medical check up of 35 students of a class, their weights were recorded as follows:
Weight (in kg)  Number of students 
Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52 
0 3 5 9 14 28 32 35 
Draw a less than type ogive for the given data. Hence, obtain the median weight from the graph and verify the result by using the formula.
Answer:
Prepare the table for cumulative frequency for less than type.
Weight (in kg) 
No. of students 
Weight (less than) 
Cumulative frequency 
Suitable points 
36–38  0  38  0  (38, 0) 
38–40  3  40  3  (40, 3) 
40–42  2  42  5  (42, 5) 
42–44  4  44  9  (44, 9) 
44–46  5  46  14  (46, 14) 
46–48  14  48  28  (48, 28) 
48–50  4  50  32  (50, 32) 
50–52  3  52  35  (52, 35) 
Now, draw the less than ogive using suitable points.
Here, =
From (0, 17.5) draw a line parallel to horizontal axis, which intersects the graph at the point M having xcoordinate as 46.5.
Therefore, the median is 46.5 kg.
Calculation for median using formula.
Now,
So, 46–48 is the median class.
Here,
Hence, both the methods give same result.
Page No 15.63:
Question 8:
The annual rainfall record of a city for 66 days is given in the following table :
Rainfall (in cm ): 010 1020 2030 3040 4050 5060
Number of days : 22 10 8 15 5 6
Calculate the median rainfall using ogives of more than type and less than type.
Answer:
Prepare a table for less than type.
Rainfall (in cm) 
Number of Days 
Rainfall (Less Than) 
Cumulative Frequency 
Suitable Points 
0−10  22  10  22  (10, 22) 
10−20  10  20  32  (20, 32) 
20−30  8  30  40  (30, 40) 
30−40  15  40  55  (40, 55) 
40−50  5  50  60  (50, 60) 
50−60  6  60  66  (60, 66) 
Now, plot the less than ogive using suitable points.
Here, N = 66.
$\therefore \frac{N}{2}=33$
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the yaxis. Let the point be P. From this point draw a line parallel to the xaxis cutting the curve at Q. From this point Q, draw a line parallel to the yaxis and meeting the xaxis at the point R. The xcoordinate of R is 21.25.
Thus, the median rainfall is 21.25 cm.
Let us now prepare a table for more than type.
Rainfall (in cm)  Number of Days 
Rainfall (More Than) 
Cumulative Frequency 
Suitable Points 
0−10  22  0  66  (0, 66) 
10−20  10  10  44  (10, 44) 
20−30  8  20  34  (20, 34) 
30−40  15  30  26  (30, 26) 
40−50  5  40  11  (40, 11) 
50−60  6  50  6  (50, 6) 
Now, plot the more than ogive with suitable points.
Here, N = 66.
$\therefore \frac{N}{2}=33$
In order to find the median rainfall, we first locate the point corresponding to 33rd day on the yaxis. Let the point be P. From this point draw a line parallel to the xaxis cutting the curve at Q. From this point Q, draw a line parallel to the yaxis and meeting the xaxis at the point R. The xcoordinate of R is 21.25.
Thus, the median rainfall is 21.25 cm.
Page No 15.64:
Question 9:
The following table gives the height of trees:
Height  No. of trees 
Less than 7 Less than 14 Less than 21 Less than 28 Less than 35 Less than 42 Less than 49 Less than 56 
26 57 92 134 216 287 341 360 
Draw 'less than' ogive and 'more than' ogive.
Answer:
Consider the following table.
Height (less than)  Heightclass  No. of trees  Cumulative frequency  Suitable points 
7  07  26  26  (7, 26) 
14  714  31  57  (14, 57) 
21  1421  35  92  (21, 92) 
28  2128  42  134  (28, 134) 
35  2835  82  216  (35, 216) 
42  3542  71  287  (42, 287) 
49  4249  54  341  (49, 341) 
56  4956  19  360  (56, 360) 
Now, draw the less than ogive using suitable points.
Now, prepare the cumulative frequency table for more than series.
Height  No. of trees  Height (more than)  Cumulative frequency  Suitable points 
07  26  0  360  (0, 360) 
714  31  7  334  (7, 334) 
1421  35  14  303  (14, 303) 
2128  42  21  268  (21, 268) 
2835  82  28  226  (28, 226) 
3542  71  35  144  (35, 144) 
4249  54  42  73  (42, 73) 
4956  19  49  54  (49, 54) 
Now, draw the more than ogive using suitable points.
Page No 15.64:
Question 10:
The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:
Profit (in lakhs in Rs)  Number of shops (frequency) 
More than or equal to 5 More than or equal to 10 More than or equal to 15 More than or equal to 20 More than or equal to 25 More than or equal to 30 More than or equal to 35 
30 28 16 14 10 7 3 
Draw both ogives for the above data and hence obtain the median.
Answer:
Firstly, we prepare the cumulative frequency table for less than type.
Profit (In lakh in Rs.) 
No. of shops (frequency) 
Profit (less than) 
Cumulative frequency 
Suitable points 
510  2  10  2  (10, 2) 
1015  12  15  14  (15, 14) 
1520  2  20  16  (20, 16) 
2025  4  25  20  (25, 20) 
2530  3  30  23  (30, 23) 
3035  4  35  27  (35, 27) 
3540  3  40  30  (40, 30) 
Again, prepare the cumulative frequency table for more than type.
Profit (In lakh in Rs.) 
No. of shops (frequency) 
Profit (more than) 
Cumulative 
Suitable 
510  2  5  30  (5, 30) 
1015  12  10  28  (10, 28) 
1520  2  15  16  (15, 16) 
2025  4  20  14  (20, 14) 
2530  3  25  10  (25, 10) 
3035  4  30  7  (30, 7) 
3540  3  35  3  (35, 3) 
Now, “more than ogive” and “less than ogive” can be drawn as follows:
The xcoordinate of the point of intersection of the “morethan ogive” and “lessthan ogive” gives the median of the given distribution..
So, the corresponding median is Rs 17.5 lakh.
Page No 15.65:
Question 1:
Define mean.
Answer:
Mean
The mean of a set of observation is equal to sum of observations divided by the number of observations.
(Where = sum of the observation and n = number of observations)
Page No 15.65:
Question 2:
What is the algebraic sum of deviation of a frequency distribution about its mean?
Answer:
The algebraic sum of deviation of a frequency distribution about its mean is zero.
Page No 15.65:
Question 3:
Which measure of central tendency is given by the xcoordinate of the point of intersection of the 'more than' ogive and 'less than' ogive?
Answer:
Median
Page No 15.65:
Question 4:
What is the value of the median of the data using the graph in the following figure of less than ogive and more than ogive?
Answer:
We know that the abscissa of the point of intersection of two ogives gives the median.
From the given figure, it can be seen that both the ogives intersect at the point (4, 15).
∴ Median of the data = 4
Page No 15.65:
Question 5:
Write the empirical relation between mean, mode and median.
Answer:
The empirical relation between mean, median and mode is
Mode = 3 Median − 2 Mean
Page No 15.65:
Question 6:
Which measure of central tendency can be determine graphically?
Answer:
Median can be determined graphically.
Page No 15.65:
Question 7:
Write the modal class for the following frequency distribution:
Classinterval:  10−15  15−20  20−25  25−30  30−35  35−40 
Frequency:  30  35  75  40  30  15 
Answer:
Class Interval  10−15  15−20  20−25  25−30  30−35  35−40 
Frequency  30  35  75  40  30  15 
Here, the maximum frequency is 75 and the corresponding classinterval is 20−25.
Therefore, 20−25 is the modal class.
Page No 15.65:
Question 8:
A student draws a cumulative frequency curve for the marks obtained by 40 students of a class as shown below. Find the median marks obtained by the students of the class.
Answer:
Here, N = 40
So,
Draw a line parallel to xaxis from the point (0, 20), intersecting the graph at point P.
Now, draw PM from P on the xaxis. The xcoordinate of M gives us the median.
∴ Median = 50
Page No 15.65:
Question 9:
Write the median class for the following frequency distribution:
Classinterval:  0−10  10−20  20−30  30−40  40−50  50−60  60−70  70−80 
Frequency:  5  8  7  12  28  20  10  10 
Answer:
We are given the following table.
Class Interval  Frequency  Cumulative Frequency 
0−10  5  5 
10−20  8  13 
20−30  7  20 
30−40  12  32 
40−50  28  60 
50−60  20  80 
60−70  10  90 
70−80  10  100 
N = 100 
Here, N = 100
$\therefore \frac{N}{2}=50$
The cumulative frequency just greater than 50 is 60.
So, the median class is 40−50.
Page No 15.66:
Question 10:
In the graphical representation of a frequency distribution, if the distance between mode and mean is k times the distance between median and mean, then write the value of k.
Answer:
We have,
Mode = 3 Median − 2 Mean
⇒ Mode − Mean = 3 Median − 3 Mean
⇒ Mode − Mean = 3(Median − Mean) .....(1)
Mode − Mean = k (Median − Mean) .....(2)
From (1) and (2), we get
k = 3
Page No 15.66:
Question 11:
Find the class marks of classes 10−25 and 35−55.
Answer:
Class mark of 10−25
Class mark of 35−55
Page No 15.66:
Question 12:
Write the median class of the following distribution:
Classinterval:  0−10  10−20  20−30  30−40  40−50  50−60  60−70 
Frequency:  4  4  8  10  12  8  4 
Answer:
Consider the following distribution table.
Class  Frequency  C.F. 
0−10  4  4 
10−20  4  8 
20−30  8  16 
30−40  10  26 
40−50  12  38 
50−60  8  46 
60−70  4  50 
N = 50 
Here,
The cumulative frequency just greater than 25 is 26.
So, the median class is 30−40.Page No 15.66:
Question 1:
Which of the following is not a measure of central tendency?
(a) Mean
(b) Median
(c) Mode
(d) Standard deviation
Answer:
Standard deviation is not a measure of central tendency.
Hence, correct option is (d).
Page No 15.66:
Question 2:
The algebraic sum of the deviations of a frequency distribution from its mean is
(a) always positive
(b) always negative
(c) 0
(d) a nonzero number
Answer:
The algebraic sum of the deviations of a frequency distribution from its mean is zero.
Hence, the correct option is (c).
Page No 15.66:
Question 3:
The arithmetic mean of 1, 2, 3, ... , n is
(a) $\frac{n+1}{2}$
(b) $\frac{n1}{2}$
(c) $\frac{n}{2}$
(d) $\frac{n}{2}+1$
Answer:
Arithmetic mean of 1, 2, 3, ... , n
$=\frac{1+2+3+...+n}{n}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}\phantom{\rule{0ex}{0ex}}=\frac{n+1}{2}$
Hence, the correct option is (a).
Page No 15.66:
Question 4:
For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean − 2 Median
(b) Mode = 2 Median − 3 Mean
(c) Mode = 3 Median − 2 Mean
(d) Mode = 3 Median + 2 Mean
Answer:
The relation between mean, median and mode is
Mode = 3 Median − 2 Mean
Hence, the correct option is (c).
Page No 15.66:
Question 5:
Which of the following cannot be determined graphically?
(a) Mean
(b) Median
(c) Mode
(d) None of these
Answer:
‘Mean’ cannot be determined by graphically.
Hence, the correct option is (a).
Page No 15.66:
Question 6:
The median of a given frequency distribution is found graphically with the help of
(a) Histogram
(b) Frequency curve
(c) Frequency polygon
(d) Ogive
Answer:
The median of a given frequency distribution is found graphically with the help of ‘Ogive’.
Hence, the correct option is (d).
Page No 15.66:
Question 7:
The mode of a frequency distribution can be determined graphically from
(a) Histogram
(b) Frequency polygon
(c) Ogive
(d) Frequency curve
Answer:
The mode of frequency distribution can be determined graphically from “Histogram”.
Hence, the correct option is (a).
Page No 15.66:
Question 8:
Mode is
(a) least frequency value
(b) middle most value
(c) most frequent value
(d) None of these
Answer:
Mode is “Most frequent value”.
Hence, the correct option is (c).
Page No 15.66:
Question 9:
The mean of n observation is $\overline{)\mathrm{X}}$. If the first item is increased by 1, second by 2 and so on, then the new mean is
(a) $\overline{)\mathrm{X}}+n$
(b) $\overline{)\mathrm{X}}+\frac{n}{2}$
(c) $\overline{)\mathrm{X}}+\frac{n+1}{2}$
(d) None of these
Answer:
Let ${x}_{1},{x}_{2},{x}_{3},...,{x}_{n}$ be the n observations.
Mean$=\overline{)X}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{n}$
$\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)X}$
If the first item is increased by 1, second by 2 and so on.
Then, the new observations are ${x}_{1}+1,{x}_{2}+2,{x}_{3}+3,...,{x}_{n}+n$.
New mean = $\frac{\left({x}_{1}+1\right)+\left({x}_{2}+2\right)+\left({x}_{3}+3\right)+...+\left({x}_{n}+n\right)}{n}$
$=\frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}+\left(1+2+3+...+n\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)X}+{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}\phantom{\rule{0ex}{0ex}}=\overline{)X}+\frac{n+1}{2}$
Hence, the correct answer is (c).
Page No 15.66:
Question 10:
One of the methods of determining mode is
(a) Mode = 2 Median − 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median − 2 Mean
(d) Mode = 3 Median + 2 Mean
Answer:
We have,
Mode = 3 Median − 2 Mean
Hence, the correct option is (c).
Page No 15.67:
Question 11:
If the mean of the following distribution is 2.6, then the value of y is
Variable (x):  1  2  3  4  5 
Frequency  4  5  y  1  2 
(a) 3
(b) 8
(c) 13
(d) 24
Answer:
Consider the following table.
Variable (x)  f  fx 
1  4  4 
2  5  10 
3  y  3y 
4  1  4 
5  2  10 
Now,
y = 8
Hence, the correct option is (b).
Page No 15.67:
Question 12:
The relationship between mean, median and mode for a moderately skewed distribution is
(a) Mode = 2 Median − 3 Mean
(b) Mode = Median − 2 Mean
(c) Mode = 2 Median − Mean
(d) Mode = 3 Median −2 Mean
Answer:
Mode = 3Median − 2Mean
Hence, the correct option is (d).
Page No 15.67:
Question 13:
The mean of a discrete frequency distribution x_{i} / f_{i}, i = 1, 2, ......, n is given by
(a) $\frac{\mathrm{\Sigma}{f}_{i}{x}_{i}}{\mathrm{\Sigma}{f}_{i}}$
(b) $\frac{1}{n}\sum _{i=1}^{n}{f}_{i}{x}_{i}$
(c) $\frac{{\displaystyle \sum _{i=1}^{n}}{f}_{i}{x}_{i}}{{\displaystyle \sum _{i=1}^{n}}{x}_{i}}$
(d) $\frac{{\displaystyle \sum _{i=1}^{n}}{f}_{i}{x}_{i}}{{\displaystyle \sum _{1=1}^{n}}i}$
Answer:
The mean of discrete frequency distribution is
Hence, the correct option is (a).
Page No 15.67:
Question 14:
If the arithmetic mean of x, x + 3, x + 6, x + 9, and x + 12 is 10, the x =
(a) 1
(b) 2
(c) 6
(d) 4
Answer:
The given observations are x, x + 3, x + 6, x + 9, and x + 12.
$\therefore \sum _{}x=5x+30,n=5,\overline{)X}=10$
Now,
Hence, the correct option is (d).
Page No 15.67:
Question 15:
If the median of the data: 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5, then x =
(a) 27
(b) 25
(c) 28
(d) 30
Answer:
The given observations are 24, 25, 26, x + 2, x + 3, 30, 31, 34.
Median = 27.5
Here, n = 8
Hence, the correct option is (b).
Page No 15.67:
Question 16:
If the median of the data: 6, 7, x − 2, x, 17, 20, written in ascending order, is 16. Then x =
(a) 15
(b) 16
(c) 17
(d) 18
Answer:
The given observations arranged in ascending order are
Hence, the correct option is (c).
Page No 15.67:
Question 17:
The median of first 10 prime numbers is
(a) 11
(b) 12
(c) 13
(d) 14
Answer:
First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
n = 10 (even)
Hence, the correct option is (b).
Page No 15.67:
Question 18:
If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
(a) 44
(b) 45
(c) 46
(d) 48
Answer:
Value  34  43  48  60  64  x 
Frequency  1  2  2  1  1  1 
It is given that the mode of the given date is 43. So, it is the value with the maximum frequency.
Now, this is possible only when x = 43. In this case, the frequency of the observation 43 would be 3.
Hence,
x + 3 = 46
Hence, the correct option is (c).
Page No 15.67:
Question 19:
If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
(a) 15
(b) 16
(c) 17
(d) 19
Answer:
Value  14  15  16  17  19  x 
Frequency  1  2  2  2  1  1 
It is given that the mode of the data is 15. So, it is the observation with the maximum frequency.
This is possible only when x = 15. In this case, the frequency of 15 would be 3.
Hence, the correct answer is (a).
Page No 15.67:
Question 20:
The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m − 1 and median q. Then, p + q =
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
Consider the numbers 3, 2, 2, 4, 3, 3, p.
Mean = $\frac{3+2+3+4+3+3+p}{7}$
$\Rightarrow 7\times \left(41\right)=17+p\phantom{\rule{0ex}{0ex}}\Rightarrow 21=17+p\phantom{\rule{0ex}{0ex}}\Rightarrow p=4$
2, 2, 3, 3, 3, 4, 4
∴ q = 3
So,
Hence, the correct option is (d).
Page No 15.67:
Question 21:
If the mean of frequency distribution is 8.1 and Σf_{i}x_{i} = 132 + 5k, Σf_{i} = 20, then k =
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
Given:
Σf_{i}x_{i} = 132 + 5k, Σf_{i} = 20 and mean = 8.1.
Then,
Hence, the correct option is (d).
Page No 15.67:
Question 22:
If the mean of 6, 7, x, 8, y, 14 is 9, then
(a) x + y = 21
(b) x + y = 19
(c) x − y = 19
(d) x − y = 21
Answer:
The given observations are 6, 7, x, 8, y, 14.
Mean = 9 (Given)
$\Rightarrow \frac{6+7+x+8+y+14}{6}=9\phantom{\rule{0ex}{0ex}}\Rightarrow 35+x+y=54\phantom{\rule{0ex}{0ex}}\Rightarrow x+y=5435=19$
Hence, the correct option is (b).
Page No 15.67:
Question 23:
The mean of n observation is $\overline{)x}$. If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is
(a) $\overline{)x}+\left(2n+1\right)$
(b) $\overline{)x}+\frac{n+1}{2}$
(c) $\overline{)x}+\left(n+1\right)$
(d) $\overline{)x}\frac{n+1}{2}$
Answer:
Let ${x}_{1},{x}_{2},{x}_{3},...,{x}_{n}$ be the n observations.
Mean$=\overline{)x}=\frac{{x}_{1}+{x}_{2}+...+{x}_{n}}{n}$
$\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)x}$
If the first item is increased by 1, the second by 2, the third by 3 and so on.
Then, the new observations are ${x}_{1}+1,{x}_{2}+2,{x}_{3}+3,...,{x}_{n}+n$.
New mean = $\frac{\left({x}_{1}+1\right)+\left({x}_{2}+2\right)+\left({x}_{3}+3\right)+...+\left({x}_{n}+n\right)}{n}$
$=\frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}+\left(1+2+3+...+n\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)x}+{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}\phantom{\rule{0ex}{0ex}}=\overline{)x}+\frac{n+1}{2}$
Hence, the correct answer is (c).
Page No 15.67:
Question 24:
If the mean of first n natural numbers is $\frac{5n}{9}$, then n =
(a) 5
(b) 4
(c) 9
(d) 10
Answer:
Given:
Mean of first n natural number =
$\Rightarrow \frac{1+2+3+...+n}{n}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{n+1}{2}=\frac{5n}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow 9n+9=10n\phantom{\rule{0ex}{0ex}}\Rightarrow n=9$
Hence, the correct option is (c).
Page No 15.68:
Question 25:
The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is
(a) 25
(b) 18
(c) 20
(d) 22
Answer:
Given: Mean = 24 and Mode = 12
We know that
Mode = 3Median − 2Mean
⇒ 12 = 3Median − 2 × 24
⇒ 3Median = 12 + 48 = 60
⇒ Median = 20
Hence, the correct option is (c).
Page No 15.68:
Question 26:
The mean of first n odd natural number is
(a) $\frac{n+1}{2}$
(b) $\frac{n}{2}$
(c) n
(d) n^{2}
Answer:
The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).
∴ Mean of first n odd natural numbers
$=\frac{1+3+5+...+\left(2n1\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{n}{2}}\left(1+2n1\right)}{n}\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{2n}{2}\phantom{\rule{0ex}{0ex}}=n$
Hence, the correct answer is (c).
Page No 15.68:
Question 27:
The mean of first n odd natural numbers is $\frac{{n}^{2}}{81}$, then n =
(a) 9
(b) 81
(c) 27
(d) 18
Answer:
The first n odd natural numbers are 1, 3, 5, ... , (2n − 1).
∴ Mean of first n odd natural numbers
$=\frac{1+3+5+...+\left(2n1\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{n}{2}}\left(1+2n1\right)}{n}\left[{S}_{n}=\frac{n}{2}\left(a+l\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{2n}{2}\phantom{\rule{0ex}{0ex}}=n$
Now,
Mean of first n odd natural numbers = $\frac{{n}^{2}}{81}$ (Given)
$\therefore n=\frac{{n}^{2}}{81}\phantom{\rule{0ex}{0ex}}\Rightarrow n=81$
Hence, the correct option is (b).
Page No 15.68:
Question 28:
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36
Answer:
Given: Mode − Median = 24
We know that
Mode = 3Median − 2Mean
Now,
Mode − Median = 2(Median − Mean)
⇒ 24 = 2(Median − Mean)
⇒ Median − Mean = 12
Hence, the correct option is (a).
Page No 15.68:
Question 29:
If the arithmetic mean, 7, 8, x, 11, 14 is x, then x =
(a) 9
(b) 9.5
(c) 10
(d) 10.5
Answer:
The given observations are 7, 8, x, 11, 14.
Mean = x (Given)
Now,
$\mathrm{Mean}=\frac{7+8+x+11+14}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{40+x}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 5x=40+x\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=40\phantom{\rule{0ex}{0ex}}\Rightarrow x=10$
Hence, the correct option is (c).
Page No 15.68:
Question 30:
If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10
Answer:
Given: Mode − Mean = 12
We know that
Mode = 3Median − 2Mean
∴ Mode − Mean = 3(Median − Mean)
⇒ 12 = 3(Median − Mean)
⇒ Median − Mean = 4 .....(1)
Again,
Mode = 3Median − 2Mean
⇒ 2Mode = 6Median − 4Mean
⇒ Mode − Mean + Mode = 6Median − 5Mean
⇒ 12 + (Mode − Median) = 5(Median − Mean)
⇒12 + (Mode − Median) = 20 [Using (1)]
⇒ Mode − Median = 20 − 12 = 8
Hence, the correct option is (b).
Page No 15.68:
Question 31:
If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29
Answer:
Given:
Mean of first n natural numbers = 15
$\Rightarrow \frac{1+2+3+...+n}{n}=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{n\left(n+1\right)}{2}}}{n}=15\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{n+1}{2}=15\phantom{\rule{0ex}{0ex}}\Rightarrow n+1=30\phantom{\rule{0ex}{0ex}}\Rightarrow n=29$
Hence, the correct option is (d).
Page No 15.68:
Question 32:
If the mean of observation ${x}_{1},{x}_{2},....,{x}_{n}\mathrm{is}\overline{)x}$, then the mean of x_{1} + a, x_{2} + a, ....., x_{n} + a is
(a) $a\overline{)x}$
(b) $\overline{)x}a$
(c) $\overline{)x}+a$
(d) $\frac{\overline{)x}}{a}$
Answer:
The mean of ${x}_{1},{x}_{2},...,{x}_{n}\mathrm{is}\overline{)x}$.
$\therefore \frac{{x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}}{n}=\overline{)x}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}=n\overline{)x}$
Mean of x_{1} + a, x_{2} + a, ... , x_{n} + a
$=\frac{\left({x}_{1}+a\right)+\left({x}_{2}+a\right)+\left({x}_{3}+a\right)+...+\left({x}_{n}+a\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{\left({x}_{1}+{x}_{2}+{x}_{3}+...+{x}_{n}\right)+\left(a+a+a+...+a\right)}{n}\phantom{\rule{0ex}{0ex}}=\frac{n\overline{)x}+na}{n}\phantom{\rule{0ex}{0ex}}=\overline{)x}+a$
Hence, the correct option is (c).
Page No 15.68:
Question 33:
Mean of a certain number of observation is $\overline{)x}$. If each observation is divided by m(m ≠ 0) and increased by n, then the mean of new observation is
(a) $\frac{\overline{)x}}{m}+n$
(b) $\frac{\overline{)x}}{n}+m$
(c) $\overline{)x}+\frac{n}{m}$
(d) $\overline{)x}+\frac{m}{n}$
Answer:
Let ${y}_{1},{y}_{2},{y}_{3},...,{y}_{k}$ be k observations.
Mean of the observations = $\overline{)x}$
$\Rightarrow \frac{{y}_{1}+{y}_{2}+{y}_{3}+...+{y}_{k}}{k}=\overline{)x}\phantom{\rule{0ex}{0ex}}\Rightarrow {y}_{1}+{y}_{2}+{y}_{3}+...+{y}_{k}=k\overline{)x}.....\left(1\right)$
If each observation is divided by m and increased by n, then the new observations are
$\frac{{y}_{1}}{m}+n,\frac{{y}_{2}}{m}+n,\frac{{y}_{3}}{m}+n,...,\frac{{y}_{k}}{m}+n$
∴ Mean of new observations
$=\frac{\left({\displaystyle \frac{{y}_{1}}{m}}+n\right)+\left({\displaystyle \frac{{y}_{2}}{m}}+n\right)+...+\left({\displaystyle \frac{{y}_{k}}{m}}+n\right)}{k}\phantom{\rule{0ex}{0ex}}=\frac{\left({\displaystyle \frac{{y}_{1}}{m}+\frac{{y}_{2}}{m}+...+\frac{{y}_{k}}{m}}\right)+\left({\displaystyle n+n+...+n}\right)}{k}\phantom{\rule{0ex}{0ex}}=\frac{{y}_{1}+{y}_{2}+...+{y}_{k}}{mk}+\frac{nk}{k}\phantom{\rule{0ex}{0ex}}=\frac{k\overline{)x}}{mk}+\frac{nk}{k}\phantom{\rule{0ex}{0ex}}=\frac{\overline{)x}}{m}+n$
Hence, the correct option is (a).
Page No 15.68:
Question 34:
If ${u}_{i}=\frac{{x}_{i}25}{10},\mathrm{\Sigma}{f}_{i}{u}_{i}=20,\mathrm{\Sigma}{f}_{i}=100,\mathrm{then}\overline{)x}$ =
(a) 23
(b) 24
(c) 27
(d) 25
Answer:
Given: ${u}_{i}=\frac{{x}_{i}25}{10},\mathrm{\Sigma}{f}_{i}{u}_{i}=20\mathrm{and}\mathrm{\Sigma}{f}_{i}=100$
Now, = $\frac{{x}_{i}25}{10}$
Therefore, h = 10 and A = 25
We know that
Hence, the correct option is (c).
Page No 15.68:
Question 35:
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increased by
(a) 2
(b) 1.5
(c) 1
(d) 0.5
Answer:
The given data set is
If 35 is removed, then the new data set is
30, 34, 36, 37, 38, 39, 40
n = 7 (odd)
Therefore,
Increase in median
Hence, the correct option is (d).
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Question 36:
While computing mean of grouped data , we assume that the frequencies are
(a) evenly distributed over all the classes.
(b) centred at the class marks of the classes .
(c) centred at the upper limit of the classes.
(d) centred at the lower limit of the classes.
Answer:
We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option (b).
Page No 15.68:
Question 37:
In the formula $\overline{\mathrm{X}}=a+h\left(\frac{1}{N}\Sigma {f}_{i}{u}_{i}\right)$ , for finding the mean of grouped frequency distribution ${u}_{i}$ =
(a) $\frac{{x}_{i}+a}{h}$ (b) $h\left({x}_{i}a\right)$ (c) $\frac{{x}_{i}a}{h}$ (d) $\frac{a{x}_{i}}{h}$
Answer:
It is given that .
This is the formula to find mean of any data by step deviation method.
Here,
${u}_{i}=\frac{{x}_{i}A}{h}\phantom{\rule{0ex}{0ex}}\mathrm{Where}{x}_{i}\mathrm{are}\mathrm{the}\mathrm{mid}\mathrm{values},A\mathrm{is}\mathrm{assumed}\mathrm{mean}\mathrm{and}h\mathrm{is}\mathrm{class}\mathrm{size}.$
Hence, the correct answer is option (c).
Page No 15.69:
Question 38:
For the following distribution :
Class: 05 510 1015 1520 2025
Frequency: 10 15 12 20 9
the sum of the lower limits of the median and modal class is
(a) 15 (b) 25 (c) 30 (d) 35
Answer:
Class  Frequency  Cumulative Frequency 
0–5  10  10 
5–10  15  25 
10–15  12  37 
15–20  20  57 
20–25  9  66 
$\therefore \frac{N}{2}=33$, which lies in the interval 10–15.
So, the lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15–20.
Therefore, the lower limit of modal class is 15.
So, the required sum is 10 + 15 = 25.
Hence, the correct answer is option (b).
Page No 15.69:
Question 39:
For the following distribution:
Below: 10 20 30 40 50 60
Number of students: 3 12 27 57 75 80
the modal class is
(a) 1020 (b) 2030 (c) 3040 (d) 5060
Answer:
Below  Class Interval  Cumulative Frequency  Frequency 
10  0–10  3  3 
20  10–20  12  9 
30  20–30  27  15 
40  30–40  57  30 
50  40–50  75  18 
60  50–60  80  5 
Here, N = 80.
$\therefore \frac{N}{2}=40$, which lies in the interval 30–40.
Therefore, the modal class is 30–40.
Hence, the correct answer is option (c).
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Question 40:
Consider the following frequency distributions:
Class: 6585 85105 105125 125145 145165 165185  185205
Frequency: 4 5 13 20 14 7 4
The difference of the upper limit of the median class and the lower limit of the modal class is
(a) 0 (b) 19 (c) 20 (d) 38
Answer:
Class  Frequency  Cumulative frequency 
65–85  4  4 
85–105  5  9 
105–125  13  22 
125–145  20  42 
145–165  14  56 
165–185  7  63 
185–205  4  67 
Here, N = 67.
$\therefore \frac{N}{2}=33.5$, which lies in the interval 125–145.
Therefore, the lower limit of the median class is 125.
The highest frequency is 20, which lies in the interval 125–145.
Therefore, the upper limit of modal class is 145.
So, the required difference is 145 − 125 = 20.
Hence, the correct answer is option (c).
Page No 15.69:
Question 41:
In the formula $\overline{\mathrm{X}}=\mathrm{a}+\frac{\mathrm{\Sigma}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}}{\mathrm{\Sigma}{\mathrm{f}}_{\mathrm{i}}}$, for finding the mean of grouped data ${{d}_{i}}^{{\text{'}}^{s}}$ are derivations from $a$ of
(a) lower limits of classes (b) upper limits ofclasses
(c) midpoints of classes (d) frequency of the class marks
Answer:
The given formula represents the formula to find the mean by assumed mean method.
Here, d_{i }= ${x}_{i}a$ where x_{i }is the ith observation and a is assumed mean.
So, d_{i}'s are the deviation from a of midpoints of the classes.
Hence, the correct answer is option (c).
Page No 15.69:
Question 42:
The abscissa of the point of intersection of less than type and of the more than types cumulative frequency curves of a grouped data gives its
(a) mean (b) median (c) mode (d) all the three above
Answer:
The less than ogive and more than ogive when drawn on the same graph intersect at a point. From this point, if we draw a perpendicular on the xaxis, the point at which it cuts the xaxis gives us the median.
Thus, the abscissa of the point of intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.
Hence, the correct answer is option (b).
Page No 15.69:
Question 43:
Consider the following frequency distribution :
Class: 05 611 1217 1823 2429
Frequency: 13 10 15 8 11
The upper limit of the median class is
(a) 17 (b) 17.5 (c) 18 (d) 18.5
Answer:
The given classes in the table are noncontinuous. So, we first make the classes continuous by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit in each class.
Class  Frequency  Cumulative Frequency 
0.5–5.5  13  13 
5.5–11.5  10  23 
11.5–17.5  15  38 
17.5–23.5  8  46 
23.5–29.5  11  57 
Now, from the table we see that N = 57.
So, $\frac{N}{2}=\frac{57}{2}=28.5$
28.5 lies in the class 11.5–17.5.
The upper limit of the interval 11.5–17.5 is 17.5.
Hence, the correct answer is option (b).
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