Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 9 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Math Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.   [CBSE 2014]

Mean of given observations =

Hence, the value of x is 7.

#### Question 2:

If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?

​Mean of given observations =

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500

Then, new mean = $\frac{500}{25}=20$

Thus, the new mean will be 20.

#### Question 3:

Compute the mean of following data:           [CBSE 2013]

 Class 1−3 3−5 5−7 7−9 Frequency 12 22 27 19

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fixi 1−3 12 2 24 3−5 22 4 88 5−7 27 6 162 7−9 19 8 152 Total ∑ fi = 80 ∑ fixi = 426

The mean of given data is given by

Thus, the mean of the following data is 5.325.

#### Question 4:

Find the mean, using direct method:

 Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 7 5 6 12 8 2

 Class Frequency $\left({f}_{i}\right)$ Mid Values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-10 7 5 35 10-20 5 15 75 20-30 6 25 150 30-40 12 35 420 40-50 8 45 360 50-60 2 55 110 ${\sum }_{{f}_{i}}=40$ $\sum \left({f}_{i}×{x}_{i}\right)=1150$

#### Question 5:

Find the mean, using direct method:

 Class 25-35 35-45 45-55 55-65 65-75 Frequency 6 10 8 12 4

 Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 25-35 6 30 180 35-45 10 40 400 45-55 8 50 400 55-65 12 60 720 65-75 4 70 280 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{x}_{i}\right)=1980$

#### Question 6:

Find the mean, using direct method:

 Class 0-100 100-200 200-300 300-400 400-500 Frequency 6 9 15 12 8

 Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-100 6 50 300 100-200 9 150 1350 200-300 15 250 3750 300-400 12 350 4200 400-500 8 450 3600 $\sum {f}_{i}=50$ $\sum \left({f}_{i}×{x}_{i}\right)=13200$

#### Question 7:

Using an appropriate method, find the mean of following frequency distribution:

 Class interval 84−90 90−96 96−102 102−108 108−114 114−120 Frequency 8 10 16 23 12 11

Which method did you use, and why?

The given data is shown as follows:

 Class interval Frequency (fi) Class mark (xi) fixi 84−90 8 87 696 90−96 10 93 930 96−102 16 99 1584 102−108 23 105 2415 108−114 12 111 1332 114−120 11 117 1287 Total ∑ fi = 80 ∑ fixi = 8244

The mean of given data is given by

​Thus, the mean of the following data is 103.05 .

#### Question 8:

If the mean of the following frequency distribution is 24, find the value of p.  [CBSE 2013]

 Class 0−10 10−20 20−30 30−40 40−50 Frequency 3 4 p 3 2

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fixi 0−10 3 5 15 10−20 4 15 60 20−30 p 25 25p 30−40 3 35 105 40−50 2 45 90 Total ∑ fi = 12 + p ∑ fixi = 270 + 25p

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒24=\frac{270+25p}{12+p}\phantom{\rule{0ex}{0ex}}⇒24\left(12+p\right)=270+25p\phantom{\rule{0ex}{0ex}}⇒288+24p=270+25p\phantom{\rule{0ex}{0ex}}⇒25p-24p=288-270\phantom{\rule{0ex}{0ex}}⇒p=18$

Hence, the value of p is 18.

#### Question 9:

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹18, find the missing frequency f.

 Daily pocket allowance (in ₹) 11−13 13−15 15−17 17−19 19−21 21−23 23−25 Number of children 7 6 9 13 f 5 4

The given data is shown as follows:

 Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fixi 11−13 7 12 84 13−15 6 14 84 15−17 9 16 144 17−19 13 18 234 19−21 f 20 20f 21−23 5 22 110 23−25 4 24 96 Total ∑ fi = 44 + f ∑ fixi = 752 + 20f

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒18=\frac{752+20f}{44+f}\phantom{\rule{0ex}{0ex}}⇒18\left(44+f\right)=752+20f\phantom{\rule{0ex}{0ex}}⇒792+18f=752+20f\phantom{\rule{0ex}{0ex}}⇒20f-18f=792-752\phantom{\rule{0ex}{0ex}}⇒2f=40\phantom{\rule{0ex}{0ex}}⇒f=20$

​Hence, the value of f is 20.

#### Question 10:

If the mean of the following frequency distribution is 54, find the value of p.   [CBSE 2006C]

 Class 0−20 20−40 40−60 60−80 80−100 Frequency 7 p 10 9 13

The given data is shown as follows:

 Class Frequency (fi) Class mark (xi) fixi 0−20 7 10 70 20−40 p 30 30p 40−60 10 50 500 60−80 9 70 630 80−100 13 90 1170 Total ∑ fi = 39 + p ∑ fixi = 2370 + 30p

The mean of given data is given by

$\overline{x}=\frac{\sum _{i}{f}_{i}{x}_{i}}{\sum _{i}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒54=\frac{2370+30p}{39+p}\phantom{\rule{0ex}{0ex}}⇒54\left(39+p\right)=2370+30p\phantom{\rule{0ex}{0ex}}⇒2106+54p=2370+30p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106\phantom{\rule{0ex}{0ex}}⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=11$

​Thus, the value of is 11.

#### Question 11:

The mean of the following data is 42, find the missing frequencies and y if the sum of the frequencies is 100.

 Class interval 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 7 10 x 13 y 10 14 9

The given data is shown as follows:

 Class interval Frequency (fi) Class mark (xi) fixi 0−10 7 5 35 10−20 10 15 150 20−30 x 25 25x 30−40 13 35 455 40−50 y 45 45y 50−60 10 55 550 60−70 14 65 910 70−80 9 75 675 Total ∑ fi = 63 + x + y ∑ fixi = 2775 + 25x + 45y

Sum of the frequencies = 100

Now, The mean of given data is given by

If x = 12, then y = 37 − 12 = 25

​Thus, the value of is 12 and y is 25.

#### Question 12:

The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹188.

 Expenditure (in ₹) 140−160 160−180 180−200 200−220 220−240 Number of families 5 25 f1 f2 5

The given data is shown as follows:

 Expenditure (in ₹) Number of families (fi) Class mark (xi) fixi 140−160 5 150 750 160−180 25 170 4250 180−200 f1 190 190f1 200−220 f2 210 210f2 220−240 5 230 1150 Total ∑ fi = 35 + f1 + f2 ∑ fixi = 6150 + 190f1 + 210f2

Sum of the frequencies = 100

Now, The mean of given data is given by

If f1 = 50, then f2 = 65 − 50 = 15

​Thus, the value of f1 is 50 and f2 is 15.

#### Question 13:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

 Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 f1 12 f2 8 5

 Class Frequency $\left({f}_{i}\right)$ Mid values $\left({x}_{i}\right)$ $\left({f}_{i}×{x}_{i}\right)$ 0-20 7 10 70 20-40 f1 30 30 f1 40-60 12 50 600 60-80 18- f1 70 1260-70 f1 80-100 8 90 720 100-120 5 110 550 $\sum {f}_{i}=50$ $\sum \left({f}_{i}×{x}_{i}\right)=3200-40{f}_{1}$

#### Question 14:

During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarised as follows:

 Number of heartbeats per minute 65−68 68−71 71−74 74−77 77−80 80−83 83−86 Number of patients 2 4 3 8 7 4 2

Find the mean heartbeats per minute for these patients, choosing a suitable method.

Using Direct method, the given data is shown as follows:

 Number of heartbeats per minute Number of patients (fi) Class mark (xi) fixi 65−68 2 66.5 133 68−71 4 69.5 278 71−74 3 72.5 217.5 74−77 8 75.5 604 77−80 7 78.5 549.5 80−83 4 81.5 326 83−86 2 84.5 169 Total ∑ fi = 30 ∑ fixi = 2277

The mean of given data is given by

​Thus, the mean heartbeats per minute for these patients is 75.9.

#### Question 15:

Find the mean, using assumed-mean method:

 Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of students 12 18 27 20 17 6

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-25\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-10 12 5 -20 -240 10-20 18 15 -10 -180 20-30 27 25=A 0 0 30-40 20 35 10 200 40-50 17 45 20 340 50-60 6 55 30 180 $\sum {f}_{i}=100$ $\sum \left({f}_{i}×{d}_{i}\right)=300$

#### Question 16:

Find the mean, using assumed-mean method:

 Class 100-120 120-140 140-160 160-180 180-200 Frequency 10 20 30 15 5

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-150\right)$ $\left({f}_{i}×{d}_{i}\right)$ 100-120 10 110 -40 -400 120-140 20 130 -20 -400 140-160 30 150=A 0 0 160-180 15 170 20 300 180-200 5 190 40 200 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{d}_{i}\right)=-\text{3}00$

#### Question 17:

Find the mean, using assumed-mean method:

 Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 20 35 52 44 38 31

 Class Frequency $\left({f}_{i}\right)$ Mid Values$\left({x}_{i}\right)$ Deviation $\left({d}_{i}\right)$ ${d}_{i}=\left({x}_{i}-50\right)$ $\left({f}_{i}×{d}_{i}\right)$ 0-20 20 10 -40 -800 20-40 35 30 -20 -700 40-60 52 50=A 0 0 60-80 44 70 20 880 80-100 38 90 40 1520 100-120 31 110 60 1860 $\sum {f}_{i}=220$ $\sum \left({f}_{i}×{d}_{i}\right)=2760$

#### Question 18:

The following table gives the literacy rate (in percentage) in 40 cities. Find the mean literacy rate, choosing a suitable method.

 Literacy rate (%) 45−55 55−65 65−75 75−85 85−95 Number of cities 4 11 12 9 4

Using Direct method, the given data is shown as follows:

 Literacy rate (%) Number of cities (fi) Class mark (xi) fixi 45−55 4 50 200 55−65 11 60 660 65−75 12 70 840 75−85 9 80 720 85−95 4 90 360 Total ∑ fi = 40 ∑ fixi = 2780

The mean of given data is given by

​Thus, the mean literacy rate is 69.5%.

#### Question 19:

Find the mean of the following frequency distribution using step-deviation method.

 Class 0−10 10−20 20−30 30−40 40−50 Frequency 7 10 15 8 10

Let us choose a = 25, h = 10, then di = xi − 25 and ui$\frac{{x}_{i}-25}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 25 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{25}}{\mathbf{10}}$ fiui 0−10 7 5 −20 −2 −14 10−20 10 15 −10 −1 −10 20−30 15 25 0 0 0 30−40 8 35 10 1 8 40−50 10 45 20 2 20 Total ∑ fi = 50 ∑ fiui = 4

The mean of given data is given by

​Thus, the mean is 25.8.

#### Question 20:

Find the mean of the following data, using step-deviation method.

 Class 5−15 15−25 25−35 35−45 45−55 55−65 65−75 Frequency 6 10 16 15 24 8 7

Let us choose a = 40, h = 10, then di = xi − 40 and ui = $\frac{{x}_{i}-40}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 40 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{40}}{\mathbf{10}}$ fiui 5−15 6 10 −30 −3 −18 15−25 10 20 −20 −2 −20 25−35 16 30 −10 −1 −16 35−45 15 40 0 0 0 45−55 24 50 10 1 24 55−65 8 60 20 2 16 65−75 7 70 30 3 21 Total ∑ fi = 86 ∑ fiui = 7

The mean of given data is given by

​Thus, the mean is 40.81.

#### Question 21:

​The weights of tea in 70 packets are shown in the following table:

 Weight (in grams) 200−201 201−202 202−203 203−204 204−205 205−206 Number of packets 13 27 18 10 1 1

Find the mean weight of packets using step-deviation method.                     [CBSE 2013]

​Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = $\frac{{x}_{i}-202.5}{1}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of packets (fi) Class mark (xi) di = xi − 202.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{202}\mathbf{.}\mathbf{5}}{\mathbf{1}}$ fiui 200−201 13 200.5 −2 −2 −26 201−202 27 201.5 −1 −1 −27 202−203 18 202.5 0 0 0 203−204 10 203.5 1 1 10 204−205 1 204.5 2 2 2 205−206 1 205.5 3 3 3 Total ∑ fi = 70 ∑ fiui = −38

The mean of given data is given by

​Hence, the mean is 201.96 g.

#### Question 22:

Find the mean of the following frequency distribution using a suitable method.

 Class 20−30 30−40 40−50 50−60 60−70 Frequency 25 40 42 33 10

Let us choose a = 45, h = 10, then di = xi − 45 and ui = $\frac{{x}_{i}-45}{10}$.

Using Step-deviation method, the given data is shown as follows:

 Class Frequency (fi) Class mark (xi) di = xi − 45 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ fiui 20−30 25 25 −20 −2 −50 30−40 40 35 −10 −1 −40 40−50 42 45 0 0 0 50−60 33 55 10 1 33 60−70 10 65 20 2 20 Total ∑ fi = 150 ∑ fiui = −37

The mean of given data is given by

​Thus, the mean is 42.534.

#### Question 23:

In an annual examination, marks (out of 90) obtained by students of class X in mathematics are given below:

 Marks obtained 0−15 15−30 30−45 45−60 60−75 75−90 Number of students 2 4 5 20 9 10

Find the mean marks.

​Let us choose a = 52.5, h = 15, then di = xi − 52.5 and ui = $\frac{{x}_{i}-52.5}{15}$.

Using Step-deviation method, the given data is shown as follows:

 Marks obtained Number of students (fi) Class mark (xi) di = xi − 37.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{52}\mathbf{.}\mathbf{5}}{\mathbf{15}}$ fiui 0−15 2 7.5 −45 −3 −6 15−30 4 22.5 −30 −2 −8 30−45 5 37.5 −15 −1 −5 45−60 20 52.5 0 0 0 60−75 9 67.5 15 1 9 75−90 10 82.5 30 2 20 Total ∑ fi = 50 ∑ fiui = 10

The mean of given data is given by

​Thus, the mean is 55.5.

#### Question 24:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

 Age (in years) 18-24 24-30 30-36 36-42 42-48 48-54 Number of workers 6 8 12 8 4 2

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-33\right)}{6}$ $\left({f}_{i}×{u}_{i}\right)$ 18-24 6 21 −2 −12 24-30 8 27 −1 −8 30-36 12 33 = A 0 0 36-42 8 39 1 8 42-48 4 45 2 8 48-54 2 51 3 6 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=2$

#### Question 25:

Find the arithmetic mean of each of the following frequency distributions using step-deviation method:

 Class 500-520 520-540 540-560 560-580 580-600 600-620 Frequency 14 9 5 4 3 5

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-550\right)}{20}$ $\left({f}_{i}×{u}_{i}\right)$ 500-520 14 510 −2 −28 520-540 9 530 −1 −9 540-560 5 550 = A 0 0 560-580 4 570 1 4 580-600 3 590 2 6 600-620 5 610 3 15 $\sum {f}_{i}=40$ $\sum \left({f}_{i}×{u}_{i}\right)=-12$

#### Question 26:

Find the mean age from the following frequency distribution:

 Age (in years) 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of person 4 14 22 16 6 5 3

Converting the series into exclusive form, we get:

 Class Frequency$\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-42\right)}{5}$ $\left({f}_{i}×{u}_{i}\right)$ 24.5-29.5 4 27 −3 −12 29.5-34.5 14 32 −2 −28 34.5-39.5 22 37 −1 −22 39.5-44.5 16 42 = A 0 0 44.5-49.5 6 47 1 6 49.5-54.5 5 52 2 10 54.5-59.5 3 57 3 9 $\sum {f}_{i}=70$ $\sum \left({f}_{i}×{u}_{i}\right)=-37$

#### Question 27:

The following table shows the age distribution of patients of malaria in a village during a particular month.

 Age (in years) 5-14 15-24 25-34 35-44 45-54 55-64 No. of cases 6 11 21 23 14 5

Find the average age of the patients.

Converting the series into exclusive form, we get:

 Class Frequency $\left({f}_{i}\right)$ Mid values$\left({x}_{i}\right)$ ${u}_{i}=\frac{\left({x}_{i}-A\right)}{h}\phantom{\rule{0ex}{0ex}}\text{=}\frac{\left({x}_{i}-29.5\right)}{10}$ $\left({f}_{i}×{u}_{i}\right)$ 4.5-14.5 6 9.5 −2 −12 14.5-24.5 11 19.5 −1 −11 24.5-34.5 21 29.5 = A 0 0 34.5-44.5 23 39.5 1 23 44.5-54.5 14 49.5 2 28 54.5-64.5 5 59.5 3 15 $\sum {f}_{i}=80$ $\sum \left({f}_{i}×{u}_{i}\right)=43$

#### Question 28:

Weight of 60 eggs were recorded as given below:

 Weight (in grams) 75−79 80−84 85−89 90−94 95−99 100−104 105−109 Number of eggs 4 9 13 17 12 3 2

Calculate their mean weight to the nearest gram.

Let us choose a = 92, h = 5, then di = xi − 92 and ui = $\frac{{x}_{i}-92}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Weight (in grams) Number of eggs (fi) Class mark (xi) di = xi − 92 ui = $\frac{{\mathbf{x}}_{\mathbf{i}}\mathbf{-}\mathbf{92}}{\mathbf{5}}$ fiui 74.5−79.5 4 77 −15 −3 −12 79.5−84.5 9 82 −10 −2 −18 84.5−89.5 13 87 −5 −1 −13 89.5−94.5 17 92 0 0 0 94.5−99.5 12 97 5 1 12 99.5−104.5 3 102 10 2 6 104.5−109.5 2 107 15 3 6 Total ∑ fi = 60 ∑ fiui = −19

The mean of given data is given by

​Thus, the mean weight to the nearest gram is 90 g.

#### Question 29:

​The following table shows the marks scored by 80 students in an examination:

 Marks Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Number of students 3 10 25 49 65 73 78 80

Calculate the mean marks correct to 2 decimal places.

Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = $\frac{{x}_{i}-17.5}{5}$.

Using Step-deviation method, the given data is shown as follows:

 Marks Number of students (cf) Frequency (fi) Class mark (xi) di = xi − 17.5 ui = $\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{5}}{\mathbf{5}}$ fiui 0−5 3 3 2.5 −15 −3 −9 5−10 10 7 7.5 −10 −2 −14 10−15 25 15 12.5 −5 −1 −15 15−20 49 24 17.5 0 0 0 20−25 65 16 22.5 5 1 16 25−30 73 8 27.5 10 2 16 30−35 78 5 32.5 15 3 15 35−40 80 2 37.5 20 4 8 Total ∑ fi = 80 ∑ fiui = 17

The mean of given data is given by

​Thus, the mean marks correct to 2 decimal places is 18.56.

#### Question 1:

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.    [CBSE 2014]

 Age (in years) 0−15 15−30 30−45 45−60 60−75 Number of patients 5 20 40 50 25

We prepare the cumulative frequency table, as shown below:

 Age (in years) Number of patients (fi) Cumulative Frequency (cf) 0−15 5 5 15−30 20 25 30−45 40 65 45−60 50 115 60−75 25 140 Total N = ∑ fi = 140

Now, N = 140 $⇒\frac{N}{2}=70$.

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.

Thus, the median class is 45−60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Hence, the median age is 46.5 years.

#### Question 2:

Compute the median from the following data:

 Marks 0-7 7-14 14-21 21-28 28-35 35-42 42-49 Number of students 3 4 7 11 0 16 9

 Class Frequency (f) Cumulative frequency 0-7 3 3 7-14 4 7 14-21 7 14 21-28 11 25 28-35 0 25 35-42 16 41 42-49 9 50 N=∑f=50

#### Question 3:

The following table shows the daily wages of workers in a factory:

 Daily wages (in Rs) 0-100 100-200 200-300 300-400 400-500 Number of workers 40 32 48 22 8

Find the median daily wage income of the workers.

 Class Frequency(f) Cumulative frequency 0-100 40 40 100-200 32 72 200-300 48 120 300-400 22 142 400-500 8 150 N=∑f=150

#### Question 4:

Calculate the median from the following frequency distribution:

 Class 5-10 10-15 15-20 20-25 20-30 30-35 35-40 40-45 Frequency 5 6 15 10 5 4 2 2

 Class Frequency(f) Cumulative frequency 5-10 5 5 10-15 6 11 15-20 15 26 20-25 10 36 25-30 5 41 30-35 4 45 35-40 2 47 40-45 2 49 N=∑f=49

#### Question 5:

Given below is the number of units of electricity consumed in a week in a certain locality:

 Consumption (in units) 65-85 85-105 105-125 125-145 145-165 165-185 195-205 Number of consumers 4 5 13 20 14 7 4

Calculate the median

 Class Frequency(f) 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 N=∑f=67

#### Question 6:

Calculate the median from the following data:

 Height (in cm) 135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 No. of boys 6 10 18 22 20 15 6 3

 Class Frequency(f) Cumulative frequency 135-140 6 6 140=145 10 16 145-150 18 34 150-155 22 56 155-160 20 76 160-165 15 91 165-170 6 97 170-175 3 100 N=∑f=100

#### Question 7:

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 25 ? 18 7

 Class Frequency (fi) c.f 0-10 5 5 10-20 25 30 20-30 x x+30 30-40 18 x+48 40-50 7 x+55

#### Question 8:

The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.

 Class 0−5 5−10 10−15 15−20 20−25 25−30 30−35 35−40 Frequency 12 a 12 15 b 6 6 4

We prepare the cumulative frequency table, as shown below:

 Class Frequency (fi) Cumulative frequency (cf) 0−5 12 12 5−10 a 12 + a 10−15 12 24 + a 15−20 15 39 + a 20−25 b 39 + a + b 25−30 6 45 + a + b 30−35 6 51 + a + b 35−40 4 55 + a + b Total N = ∑fi = 70

Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,

55 a + = 70 ⇒ a + = 15     ...(1)

Median is 16, which lies in 15−20. So, the median class is 15−20.

l = 15, h = 5, N = 70, f = 15 and cf = 24 + a

Now,

$\mathrm{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)×h\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{\frac{70}{2}-\left(24+a\right)}{15}\right)×5\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{35-24-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16=15+\left(\frac{11-a}{3}\right)\phantom{\rule{0ex}{0ex}}⇒16-15=\frac{11-a}{3}\phantom{\rule{0ex}{0ex}}⇒1×3=11-a\phantom{\rule{0ex}{0ex}}⇒a=11-3\phantom{\rule{0ex}{0ex}}⇒a=8$

b = 15 a    [From (1)]
b = 15 − 8
b = 7

Hence, a = 8 and b = 7.

#### Question 9:

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.

 Runs scored 2500−3500 3500−4500 4500−5500 5500−6500 6500−7500 7500−8500 Number of batsmen 5 x y 12 6 2

We prepare the cumulative frequency table, as shown below:

 Runs scored Number of batsmen (fi) Cumulative frequency (cf) 2500−3500 5 5 3500−4500 x 5 + x 4500−5500 y 5 + x + y 5500−6500 12 17 + x + y 6500−7500 6 23 + x + y 7500−8500 2 25 + x + y Total N = ∑fi = 60

Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,

25 x + = 60 ⇒ x + = 35     ...(1)

Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x

Now,

∴ y = 35 − x    [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.

#### Question 10:

If the median of the following frequency distribution is 32.5, find the values of f1 and f2.

 Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total Frequency f1 5 9 12 f2 3 2 40

 Class Frequency(f) Cumulative frequency 0-10 ${f}_{1}$ ${f}_{1}$ 10-20 5 ${f}_{1}$+5 20-30 9 ${f}_{1}$+14 30-40 12 ${f}_{1}$+26 40-50 ${f}_{2}$ ${f}_{1}+{f}_{2}$+26 50-60 3 ${f}_{1}+{f}_{2}$+29 60-70 2 ${f}_{1}+{f}_{2}$+31 N=∑f=40

#### Question 11:

Calculate the median for the following data:

 Age (in years) 19-25 26-32 33-39 40-46 47-53 54-60 Frequency 35 96 68 102 35 4

First, we will convert the data into exclusive form.

 Class Frequency(f) Cumulative frequency 18.5-25.5 35 35 25.5-32.5 96 131 32.5-39.5 68 199 39.5-46.5 102 301 46.5-53.5 35 336 53.5-60.5 4 340 N=∑f=340

#### Question 12:

Find the median wages for the following frequencies distribution:

 Wages per day (in Rs) 61-70 71-80 81-90 91-100 101-110 111-120 No. of women workers 5 15 20 30 20 8

Converting the given data into exclusive form, we get:

 Class Frequency(f) Cumulative frequency 60.5-70.5 5 5 70.5-80.5 15 20 80.5-90.5 20 40 90.5-100.5 30 70 100.5-110.5 20 90 110.5-120.5 8 98 N=∑f=98

#### Question 13:

Find the median from the following data:

 Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35 35-40 41-45 Frequency 7 10 16 32 24 16 11 5 2