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Page No 10.83:
Question 1:
Calculate figures and coefficient of range of the following series, which gives the monthly expenditure (in â¹) of seven students: 22, 35, 32, 45, 42, 48, 39
Answer:
Given:
Highest Value (H) =48
Lowest Value (L) =22
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 48 − 22 = 26
Thus, range is 26 and coefficient of range is 0.37
Page No 10.83:
Question 2:
Find range and coefficient of range from the weekly wage (in â¹) of 10 workers of a factory: 310, 350, 420, 105, 115, 290, 245, 450, 300, 375.
Answer:
Given:
Highest Value (H) = 450
Lowest Value (L) = 105
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 450 − 105 = 345
Thus, range is 345 and coefficient of range is 0.62
Page No 10.83:
Question 3:
Find the range and coefficient of range of the following:
Size of shoes | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
No. of shoes sold | 8 | 12 | 14 | 20 | 15 | 8 | 7 | 10 |
Answer:
Size of Shoes | No. of Shoes Sold |
6 | 8 |
7 | 12 |
8 | 14 |
9 | 20 |
10 | 15 |
11 | 8 |
12 | 7 |
13 | 10 |
Highest value (H) = 13
Lowest value (L) = 6
Range = Highest value − Lowest value
i.e. R = H − L
or, R = 13 − 6=7
Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.
Page No 10.83:
Question 4:
From the following data calculate range and coefficient of range:
Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of Studends | 8 | 12 | 7 | 30 | 10 | 5 | 2 |
Answer:
Marks | No. of Students |
10 | 8 |
20 | 12 |
30 | 7 |
40 | 30 |
50 | 10 |
60 | 5 |
70 | 2 |
Highest value (H) = 70
Lowest value (L) = 10
Range = Highest value − Lowest value
i.e. R = H−L
or, R = 70 − 10= 60 marks
Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.
Page No 10.84:
Question 5:
Find out range and coefficient range.
Marks | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
No. of students | 12 | 18 | 14 | 63 | 19 |
Answer:
Marks | No. of Students |
10 − 20 | 12 |
20 − 30 | 18 |
30 − 40 | 14 |
40 − 50 | 63 |
50 − 60 | 19 |
Hence, the range is 50 marks and coefficient of range is 0.715
Page No 10.84:
Question 6:
Following are the marks obtained by 25 students of class XI in an exam. Find out range and coefficient of range of the marks.
Marks | 5−9 | 10−14 | 15−19 | 20−24 | 25−29 | 30−34 |
No. Students | 4 | 6 | 3 | 2 | 6 | 4 |
Answer:
In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.
Class Interval | Exclusive Class Interval | Frequency |
5 − 9 10 −14 15 −19 20 −24 25 −29 30− 34 |
4.5 − 9.5 9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 |
4 6 3 2 6 4 |
Page No 10.84:
Question 7:
Calculate interquartile range, quartile deviation and coefficient of quartile deviation from the following data:
Family | A | B | C | D | E | F | G |
Income (in â¹) | 1,200 | 1,400 | 1,500 | 1,700 | 2,000 | 2,100 | 2,200 |
Answer:
Family | S. No. | Income Rs |
A | 1 | 1200 |
B | 2 | 1400 |
C | 3 | 1500 |
D | 4 | 1700 |
E | 5 | 2000 |
F | 6 | 2100 |
G | 7 | 2200 |
N = 7 |
Page No 10.84:
Question 8:
find out the value of quartile deviation and its coefficient from the following data:
Roll No. | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
Marks | 20 | 28 | 40 | 12 | 30 | 15 | 50 |
Answer:
Roll No. | Marks in Ascending Order |
1 | 12 |
2 | 15 |
3 | 20 |
4 | 28 |
5 | 30 |
6 | 40 |
7 | 50 |
N = 7 |
Page No 10.84:
Question 9:
Form the following figures, find the quartile deviation and its coefficient:
Height (cm) | 150 | 151 | 152 | 153 | 154 | 155 | 156 | 157 | 158 |
No. of students | 15 | 20 | 32 | 35 | 33 | 22 | 20 | 12 | 10 |
Answer:
Height (c.m) |
No. of Students | Cumulative Frequency |
150 | 15 | 15 |
151 | 20 | 15 + 20 = 35 |
152 | 32 | 35 + 32 = 67 |
153 | 35 | 67 + 35 = 102 |
154 | 33 | 102 + 33 = 135 |
155 | 22 | 135 + 22 = 157 |
156 | 20 | 157 + 20 = 177 |
157 | 12 | 177 + 12 = 189 |
158 | 10 | 189 + 10 = 199 |
N = 199 |
Page No 10.84:
Question 10:
Compute coefficient of Quartile deviation form the following data:
Marks | 10 | 20 | 30 | 40 | 50 | 60 |
No. of students | 4 | 7 | 15 | 8 | 7 | 2 |
Answer:
Marks | No. of Students | Cumulative Frequency |
10 | 4 | 4 |
20 | 7 | 7 + 4 = 11 |
30 | 15 | 11 + 15 = 26 |
40 | 8 | 26 + 8 = 34 |
50 | 7 | 34 + 7 = 41 |
60 | 2 | 41 + 2 = 43 |
N = 43 |
Page No 10.84:
Question 11:
Calculate Quartile Deviation and its Coefficient from the following data:
Size | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 | 40−45 | 45−50 |
Frequency | 6 | 10 | 18 | 30 | 15 | 12 | 10 | 6 | 4 |
Answer:
Size | Frequency | Cumulative Frequency |
5−10 | 6 | 6 |
10−15 | 10 | 16 |
15−20 | 18 | 34 |
20−25 | 30 | 64 |
25−30 | 15 | 79 |
30−35 | 12 | 91 |
35−40 | 10 | 101 |
40−45 | 6 | 107 |
45−50 | 4 | 111 |
Σf=N = 111 |
Q1 = Size of item
= Size of item
= Size of 27.75th item
27.75 item lies in group 15−20 and falls within 34thc.f. of the series
Like wise,
Q3 = Size of item
= Size of 3
= Size of 83.25th item
83.25th item lies in the group 30-35 within the 33rd c.f. of the series.
Page No 10.85:
Question 12:
Calculate coefficient of quartile deviation from the following data:
X (Less than) | 200 | 300 | 400 | 500 | 600 |
Frequency | 8 | 20 | 40 | 46 | 50 |
Answer:
Converting less than cumulative frequency into normal frequency distribution:
X | c.f. | Frequency |
100−200 | 8 | 8 |
200−300 | 20 | 20−8=12 |
300−400 | 40 | 40−20=20 |
400−500 | 46 | 46−40=6 |
500−600 | 50 | 50−46=4 |
N=50 |
Q1 = Size of item
= Size of item
= Size of 12.5th item
12.5th item lies in group 200−300 and falls within 20thc.f. of the series.
Here,
lâ=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Likewise,
Q3 = Size of item
= Size of item
= Size of 37.5th item
37.5th item lies in group 300−400 and fall within the 40th c.f. of the series
Page No 10.85:
Question 13:
Estimate an appropriate measure of the following data:
Wages (â¹) | Less than 25 | 25−30 | 30−35 | 35−40 | Above 40 |
No. of workers | 2 | 10 | 26 | 16 | 7 |
Answer:
Wages | No. of worker (Frequency) |
Cumulative Frequency |
Less than 25 | 2 | 2 |
25−30 | 10 | 12 |
30−35 | 26 | 38 |
35−40 | 16 | 54 |
Above 40 | 7 | 61 |
N=61 |
Q1 = Size of item
= Size of item
= Size of 15.25th item
15.25th item lies in group 30−35 and falls within 38thc.f. of the series.
Here,
lâ=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval
Likewise,
Q3 = Size of item
= Size of item
= Size of 45.75th item
47.75th item lies in group 35−40 and fall within the 54thc.f. of the series
Page No 10.85:
Question 14:
Calculate the mean deviaiton from median and its coefficient from the data: 100, 150, 80, 90, 160, 200, 140
Answer:
Sr. No | Values (X) |
Deviation from Median M= 140 |
1 2 3 4 5 6 7 |
80 90 100 140 150 160 200 |
60 50 40 0 10 20 60 |
N=7 |
(M) Median = Size of
= Size of
= Size of 4th item
= 140
Page No 10.85:
Question 15:
Compute Mean deviation and its coefficient by mean from the data given below:
X | 210 | 220 | 225 | 225 | 225 | 240 | 250 | 270 | 280 |
Answer:
Sr. No | Values (X) |
Deviation from Mean |
1 | 210 | 28 |
2 | 220 | 18 |
3 | 225 | 13 |
4 | 225 | 13 |
5 | 225 | 13 |
6 | 235 | 3 |
7 | 240 | 2 |
8 | 250 | 12 |
9 | 270 | 32 |
10 | 280 | 42 |
N=10 | ΣX = 2380 | Σ = 176 |
Page No 10.85:
Question 16:
Following are the marks of the students. Find mean deviation and coefficient mean deviation from mean.
Marks | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
No. of students | 16 | 32 | 36 | 44 | 28 | 18 | 12 | 14 |
Answer:
Marks (X) |
Frequency (f) |
fX | Deviation from Mean |
|
5 | 16 | 80 | 15.2 | 243.2 |
10 | 32 | 320 | 10.2 | 326.4 |
15 | 36 | 540 | 5.2 | 187.2 |
20 | 44 | 880 | 0.2 | 8.8 |
25 | 28 | 700 | 4.8 | 134.4 |
30 | 18 | 540 | 9.8 | 176.4 |
35 | 12 | 420 | 14.8 | 177.6 |
40 | 14 | 560 | 19.8 | 277.2 |
ΣX = 180 | Σf = 200 | ΣfX=4040 | Σ=1531.2 |
Page No 10.85:
Question 17:
Find out the mean deviation from the median and its coefficient.
Marks | 10 | 11 | 12 | 13 | 14 |
No. of students | 3 | 12 | 18 | 12 | 3 |
Answer:
Marks (X) |
Frequency (f) |
Cumulative frequency (c.f.) |
Deviation from Median |
|
10 | 3 | 3 | 2 | 6 |
11 | 12 | 15 | 1 | 12 |
12 | 18 | 33 | 0 | 0 |
13 | 12 | 45 | 1 | 12 |
14 | 3 | 48 | 2 | 6 |
N=Σf = 48 | Σf|dM| =36 |
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 24.5 th item is 33 (in the c.f. column), which is corresponding to 12.
Hence, median is 12.
Thus we calculate the deviation of the values from 12.
Page No 10.85:
Question 18:
Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students:
Scores | 140−150 | 150−160 | 160−170 | 170−180 | 180−190 | 190−200 |
Frequency | 4 | 6 | 10 | 18 | 9 | 3 |
Answer:
Scores (X) |
Frequency (f) |
Cumulative Frequency (c.f.) |
Mid -Values (m) |
Deviation from Median |
|
140−150 | 4 | 4 | 145 | 27.78 | 111.12 |
150−160 | 6 | 10 | 155 | 17.78 | 106.68 |
160−170 | 10 | 165 | 7.78 | 77.8 | |
−180 | 38 | 175 | 2.22 | 39.96 | |
180−190 | 9 | 47 | 185 | 12.22 | 109.98 |
190−200 | 3 | 50 | 195 | 22.22 | 66.66 |
N=Σf = 50 | Σf|dM|=512.2 |
Median class is given by the size of item, i.e. item, which is 25th item.
This corresponds to the class interval of 170-180, so this is the median class.
Scores (X) |
Frequency (f) |
Mid -Values (m) |
fm | Deviation from Mean |
|
140−150 | 4 | 145 | 580 | 26.2 | 104.8 |
150−160 | 6 | 155 | 930 | 16.2 | 97.2 |
160−170 | 10 | 165 | 1650 | 6.2 | 62 |
170−180 | 18 | 175 | 3150 | 3.8 | 68.4 |
180−190 | 9 | 185 | 1665 | 13.8 | 124.2 |
190−200 | 3 | 195 | 585 | 23.8 | 71.4 |
Σf = 50 | Σfm=8560 | Σ=528 |
Page No 10.86:
Question 19:
Find the mean deviation from mean and its coefficient for the given data:
X | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 |
F | 3 | 5 | 7 | 2 | 9 | 4 |
Answer:
Values (X) |
Frequency (f) |
Mid-Values (m) |
fm | ||
0−10 | 3 | 5 | 15 | 27 | 81 |
10−20 | 5 | 15 | 75 | 17 | 85 |
20−30 | 7 | 25 | 175 | 7 | 49 |
30−40 | 2 | 35 | 70 | 3 | 6 |
40−50 | 9 | 45 | 405 | 13 | 117 |
50−60 | 4 | 55 | 220 | 23 | 92 |
Σf = 30 | Σfm=960 | Σ=430 |
Page No 10.86:
Question 20:
Calculate mean deviation and its coefficient from the following figures:
Class-Interval | Frequency |
Less than 10 | 5 |
Less than 20 | 12 |
Less than 30 | 20 |
Less than 40 | 35 |
Less than 50 | 54 |
Less than 60 | 60 |
Answer:
Converting less than cumulative frequency distribution into simple frequency distribution:
Class Interval | Cumulative Frequency (c.f.) |
Frequency (f) |
Mid-Values (m) |
||
0−10 | 5 | 5 | 5 | 31.67 | 158.35 |
10−20 | 12 | 7 | 15 | 21.67 | 151.69 |
20−30 | 20 | 8 | 25 | 11.67 | 93.36 |
30−40 | 35 | 15 | 35 | 1.67 | 25.65 |
40−50 | 54 | 19 | 45 | 8.33 | 158.27 |
50−60 | 60 | 6 | 55 | 18.33 | 109.98 |
Σf=N=60 | Σf|dM|=696.7 |
Median class is given by the size of , i.e. , which is the 30th item.
This corresponds to the class interval of 30-40, so this is the median class.
Page No 10.86:
Question 21:
Calculate the Standard deviation from the following values:8 , 9, 15, 23, 5, 11, 19, 8, 10, 12.
Answer:
Values (X) |
x2 | |
8 | −4 | 16 |
9 | −3 | 9 |
15 | 3 | 9 |
23 | 11 | 121 |
5 | −7 | 49 |
11 | −1 | 1 |
19 | 7 | 49 |
8 | −4 | 16 |
10 | −2 | 4 |
12 | 0 | 0 |
ΣX=120 | Σx2=274 |
Hence, the standard deviation of the above series is 5.23
Page No 10.86:
Question 22:
Find the standard deviation for the following data: 3, 5, 6, 7, 10, 12, 15, 18.
Answer:
Values (X) |
x2 | |
3 | −6.5 | 42.25 |
5 | −4.5 | 20.25 |
6 | −3.5 | 12.25 |
7 | −2.5 | 6.25 |
10 | .5 | .25 |
12 | 2.5 | 6.25 |
15 | 5.5 | 30.25 |
18 | 8.5 | 72.25 |
ΣX=76 | Σx2=190 |
Hence, standard deviation of the above series is 4.87
Page No 10.86:
Question 23:
Find out the standard deviation of the height of 10 men given below: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.
Answer:
Values (X) |
x2 | |
160 | −3 | 9 |
160 | −3 | 9 |
161 | −2 | 4 |
162 | −1 | 1 |
163 | 0 | 0 |
163 | 0 | 0 |
163 | 0 | 0 |
164 | 1 | 1 |
164 | 1 | 1 |
170 | 7 | 49 |
ΣX=1630 | Σx2=74 |
Hence, standard deviation of the above series is 2.72
Page No 10.86:
Question 24:
Calculate standard deviation of the given below:
Size | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Frequency | 3 | 7 | 22 | 60 | 85 | 32 | 8 |
Answer:
Size (X) |
Frequency (f) |
fX | x2 | fx2 | |
3 | 3 | 9 | 3.59 | 12.89 | 38.67 |
4 | 7 | 28 | 2.59 | 6.71 | 46.97 |
5 | 22 | 110 | 1.59 | 2.53 | 55.66 |
6 | 60 | 360 | 0.59 | 0.35 | 21 |
7 | 85 | 595 | 0.41 | 0 .17 | 14.45 |
8 | 32 | 256 | 1.41 | 1.99 | 63.68 |
9 | 8 | 72 | 2.41 | 5.81 | 46.48 |
N=Σf=217 | ΣfX=1430 | Σfx2=286.91 |
Hence, standard deviation of the above series is 1.149
Page No 10.86:
Question 25:
Find the value of standard deviation and coefficient of variation from the following:
Variables | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
Frequency | 6 | 8 | 16 | 15 | 32 | 11 | 12 |
Answer:
Size (X) |
Frequency (f) |
fX | x2 | fx2 | |
10 | 6 | 60 | −34 | 1156 | 6936 |
20 | 8 | 160 | −24 | 576 | 4608 |
30 | 16 | 480 | −14 | 196 | 3136 |
40 | 15 | 600 | −4 | 16 | 240 |
50 | 32 | 1600 | +6 | 36 | 1152 |
60 | 11 | 660 | 16 | 256 | 2816 |
70 | 12 | 840 | 26 | 676 | 8112 |
Σf=100 | ΣfX=4400 | Σfx2=27000 |
Page No 10.86:
Question 26:
Measurements are made to the nearest cm. of the heights of 10 children. Calculate mean and standard deviation.
Heights (cms) | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
No. of children | 2 | 0 | 15 | 29 | 25 | 12 | 10 | 4 | 3 |
Answer:
Height (X) |
Frequency (f) |
fX | x2 | fx2 | |
60 | 2 | 120 | −3.89 | 15.13 | 30.26 |
61 | 0 | 0 | −2.89 | 8.35 | 0 |
62 | 15 | 930 | −1.89 | 3.57 | 53.55 |
63 | 29 | 1827 | −0.89 | 0.79 | 22.91 |
64 | 25 | 1600 | 0.11 | 0.01 | 0.25 |
65 | 12 | 780 | 1.11 | 1.23 | 14.76 |
66 | 10 | 660 | 2.11 | 4.45 | 44.5 |
67 | 4 | 268 | 3.11 | 9.67 | 38.68 |
68 | 3 | 204 | 4.11 | 16.89 | 50.67 |
Σf=100 | ΣfX=6389 | Σfx2=255.58 |
Hence mean of the above series is 63.89 cms and standard deviation is 1.6 cms.
Page No 10.87:
Question 27:
Calculate standard deviation for the given data:
Age (in yrs) | 20−25 | 25−30 | 30−35 | 35−40 | 40−45 | 45−50 | 50−55 |
No. of workers | 17 | 11 | 8 | 5 | 4 | 3 | 2 |
Answer:
Age (X) |
No. of Workers (f) |
Mid-Values (m) |
fm | m2 | |
20−25 | 17 | 22.5 | 382.5 | 506.25 | 8606.25 |
25−30 | 11 | 27.5 | 302.5 | 756.25 | 8318.75 |
30−35 | 8 | 32.5 | 260 | 1056.25 | 8450 |
35−40 | 5 | 37.5 | 187.5 | 1406.25 | 7031.25 |
40−45 | 4 | 42.5 | 170 | 1806.25 | 7225 |
45−50 | 3 | 47.5 | 142.5 | 2256.25 | 6768.75 |
50−55 | 2 | 52.5 | 105 | 2756.25 | 5512.5 |
Σf=50 | Σfm=1550 | Σfm2=51912.5 |
Hence, standard deviation of the above series is 8.79 years.
Page No 10.87:
Question 28:
Calculate Standard deviation from the following series:
Class | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 2 | 4 | 6 | 8 | 6 | 4 | 2 |
Answer:
Class Interval (X) |
Frequency (f) |
Mid-Values (m) |
fm | m2 | |
0−10 | 2 | 5 | 10 | 25 | 50 |
10−20 | 4 | 15 | 60 | 225 | 900 |
20−30 | 6 | 25 | 150 | 625 | 3750 |
30−40 | 8 | 35 | 280 | 1225 | 9800 |
40−50 | 6 | 45 | 270 | 2025 | 12150 |
50−60 | 4 | 55 | 220 | 3025 | 12100 |
60−70 | 2 | 65 | 130 | 4225 | 8450 |
Σf=32 | Σfm=1120 | Σfm2=47200 |
Hence, standard deviation of the above series is 15.81
Page No 10.87:
Question 29:
From the following figures, find the standard deviation and the coefficient of variation:
Marks | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
No. of persons | 5 | 10 | 20 | 40 | 30 | 20 | 10 | 4 |
Answer:
Marks (X) |
No. of Persons (f) |
Mid-Values (m) |
fm | m2 | fm2 |
0−10 | 5 | 5 | 25 | 25 | 125 |
10−20 | 10 | 15 | 150 | 225 | 2250 |
20−30 | 20 | 25 | 500 | 625 | 12500 |
30−40 | 40 | 35 | 1400 | 1225 | 49000 |
40−50 | 30 | 45 | 1350 | 2025 | 60750 |
50−60 | 20 | 55 | 1100 | 3025 | 60500 |
60−70 | 10 | 65 | 650 | 4225 | 42250 |
70−80 | 4 | 75 | 300 | 5625 | 22500 |
Σf=139 | Σfm=5475 | Σfm2=249875 |
Page No 10.87:
Question 31:
The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3 and the standard deviations are 8 and 7. Find the mean and the standard deviation of the sample of size 150 obtained by combining the two samples.
Answer:
Calculating combined standard deviation,
To calculate combined standard deviation, we need to find the combined mean of the observations.
Thus, Combined mean is
Hence, the combined standard deviation is 7.56
Page No 10.87:
Question 32:
For a group of 100 males, mean and standard deviation of their daily wages are â¹ 36 and â¹ 9 respectively. For a group of 50 females, it is â¹ 45 and â¹ 6. Find the standard deviation for the whole group.
Answer:
Calculating combined standard deviation,
To calculate combined standard deviation, we need to find the combined mean of the observations.
Thus, Combined mean is
Hence, the combined standard deviation is Rs 9.16
Page No 10.87:
Question 33:
The profit of two business concerns for 5 years are as given below. Draw Lorenz Curves to show the distribution.
Year | 2001 | 2002 | 2003 | 2004 | 2005 |
Firm A | 15 | 30 | 45 | 60 | 50 |
Firm B | 20 | 30 | 45 | 60 | 45 |
Answer:
Years | Firm A | Cumulative Profits of Firm A | Cumulative Percentage Profits of Firm A (x) |
Firm B | Cumulative Profits of Firm B | Cumulative Percentage Profits of Firm B (y) |
Coordinates of Lorenz Curve (x,y) |
2001 | 15 | 15 | 20 | 20 | (7.5,10) | ||
2002 | 30 | 45 | 30 | 50 | (22.5, 25) | ||
2003 | 45 | 90 | 45 | 95 | (45, 47.5) | ||
2004 | 60 | 150 | 60 | 155 | (75,77.5) | ||
2005 | 50 | 200 | 45 | 200 | (100,100) | ||
=200 | =200 |
Page No 10.87:
Question 34:
The given table shows the daily income of workers of two factories. Draw the Lorenz Curves for both the factories.
Daily Income (â¹) | 0−100 | 100−200 | 200−300 | 300−400 | 400−500 |
Factory A | 8 | 7 | 5 | 3 | 2 |
Factory B | 15 | 6 | 2 | 1 | 1 |
Answer:
Daily Income | Mid Value | Cumulative Mid Value |
% Cumulative Mid Value |
No. of worker (f) |
(c.f.) | % Cumulative Frequency |
0−100 | 50 | 50 | 4 | 8 | 8 | 32 |
100−200 | 150 | 200 | 16 | 7 | 15 | 60 |
200−300 | 250 | 450 | 36 | 5 | 20 | 80 |
300−400 | 350 | 800 | 64 | 3 | 23 | 92 |
400−500 | 450 | 1250 | 100 | 2 | 25 | 100 |
Daily Income | Mid Value | Cumulative Mid Value |
% Cumulative Mid Value |
No. of worker (f) |
(c.f.) | % Cumulative Frequency |
0−100 | 50 | 50 | 4 | 15 | 15 | 60 |
100−200 | 150 | 200 | 16 | 6 | 21 | 84 |
200−300 | 250 | 450 | 36 | 2 | 23 | 92 |
300−400 | 350 | 800 | 64 | 1 | 24 | 96 |
400−500 | 450 | 1250 | 100 | 1 | 25 | 100 |
Page No 10.87:
Question 35:
Find the mean deviation from mean and its coefficient for the given data:
Marks (more than) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
No. of students | 200 | 180 | 150 | 100 | 40 | 15 | 5 |
Answer:
Marks (X) |
Frequency (f) |
Mid-Values (m) |
fm | ||
0−10 | 20 | 5 | 100 | 24.5 | 490 |
10−20 | 30 | 15 | 450 | 14.5 | 435 |
20−30 | 50 | 25 | 1250 | 4.5 | 225 |
30−40 | 60 | 35 | 2100 | 5.5 | 330 |
40−50 | 25 | 45 | 1125 | 15.5 | 387.5 |
50−60 | 10 | 55 | 550 | 25.5 | 255 |
60−70 | 5 | 65 | 325 | 35.5 | 177.5 |
Σf=200 | Σfm=5900 | Σ=2300 |
Page No 10.88:
Question 36:
Calculate the Mean and Standard Deviation from the following distribution.
Age (years) | 15−19 | 20−24 | 25−29 | 30−34 | 35−39 | 40−44 |
No. of Persons | 4 | 20 | 38 | 24 | 10 | 4 |
Answer:
In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below:
Age (X) |
Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
14.5−19.5 | 4 | 17 | 68 | 289 | 1156 |
19.5−24.5 | 20 | 22 | 440 | 484 | 9680 |
24.5−29.5 | 38 | 27 | 1026 | 729 | 27702 |
29.5−34.5 | 24 | 32 | 768 | 1024 | 24576 |
34.5−39.5 | 10 | 37 | 370 | 1369 | 13690 |
39.5−44.5 | 4 | 42 | 168 | 1764 | 7056 |
Σf=100 | Σfm=2840 | Σfm2=83860 |
Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.
Page No 10.88:
Question 37:
The following table gives the weights of one hundred persons. Copute the coefficient of dispersion by the method of limits.
Class-interval | 40−45 | 45−50 | 50−55 | 55−60 | 60−65 | 65−70 | 70−75 | 75−80 | 80−85 | 85−90 |
No. of persons | 4 | 13 | 8 | 14 | 9 | 16 | 17 | 9 | 8 | 2 |
Answer:
Given:
Upper limit of the Highest Class Interval (H) = 90
Lower limit of the Lowest Class Interval (L) = 40
Range = Highest Value − Lowest Value
i.e R= H− L
Substituting the given values in the formula.
R= 90 − 40= 50
Hence, range of the above series is 50 kg and coefficient of range is 0.384
Page No 10.88:
Question 38:
Calculate standard deviation of the following data:
Age in years (below) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
No. of persons | 15 | 30 | 53 | 75 | 100 | 110 | 115 | 125 |
Answer:
Age (X) |
Cumulative Frequency (c.f.) |
Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
0−10 | 15 | 15-0=15 | 5 | 75 | 25 | 375 |
10−20 | 30 | 30-15=15 | 15 | 225 | 225 | 3375 |
20−30 | 53 | 53-15=23 | 25 | 575 | 625 | 14375 |
30−40 | 75 | 75-23=22 | 35 | 770 | 1225 | 26950 |
40−50 | 100 | 100-75=25 | 45 | 1125 | 2025 | 50625 |
50−60 | 110 | 110-100=10 | 55 | 550 | 3025 | 30250 |
60−70 | 115 | 115-110=5 | 65 | 325 | 4225 | 21125 |
70−80 | 125 | 125-115=10 | 75 | 750 | 5625 | 56250 |
Σf=125 | Σfm=4395 | Σfm2=203325 |
Hence, standard deviation of the above series is 19.76 years.
Page No 10.88:
Question 39:
Price of a particular item in 10 years in two cities are given below, which city has more stable prices?
City A | 55 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |
City B | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |
Answer:
City A | City B | ||||
XA | XB | ||||
55 | 2 | 4 | 108 | 3 | 9 |
54 | 1 | 1 | 107 | 2 | 4 |
52 | −1 | 1 | 105 | 0 | 0 |
53 | 0 | 0 | 105 | 0 | 0 |
56 | 3 | 9 | 106 | 1 | 1 |
58 | 5 | 25 | 107 | 2 | 4 |
52 | −1 | 1 | 104 | −1 | 1 |
50 | −3 | 9 | 103 | −2 | 4 |
51 | −2 | 4 | 104 | −1 | 1 |
49 | −4 | 16 | 101 | −4 | 16 |
ΣXA=530 | ΣXB=1050 |
Since C.V. of city B is less. Therefore, prices are more stable in city B.
Page No 10.88:
Question 40:
For a distribution, the coefficient of variation is 22.5% and mean is 7.5. Calculate standard deviation.
Answer:
Given,
Coefficient of variation= 22.5%
Mean, = 7.5
Hence, standard deviation is 1.69
Page No 10.88:
Question 41:
Find out the arithmetic mean and standard deviation from the following data:
Variable | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 |
Frequency | 2 | 9 | 29 | 54 | 11 | 5 |
Answer:
Variable (X) |
Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
5−10 | 2 | 7.5 | 15 | 56.25 | 112.5 |
10−15 | 9 | 12.5 | 112.5 | 156.25 | 1406.25 |
15−20 | 29 | 17.5 | 507.5 | 306.25 | 8881.25 |
20−25 | 54 | 22.5 | 1215.0 | 506.25 | 27337.5 |
25−30 | 11 | 27.5 | 302.5 | 756.25 | 8318.75 |
30−35 | 5 | 32.5 | 162.5 | 1056.25 | 5281.25 |
Σf=110 | Σfm=2315 | Σfm2=51337.5 |
Page No 10.88:
Question 42:
The mean and standard deviation of a series of 20 items are 20 and 5 respectively. While calculating these measures, an item of 13 was wrongly read as 30. Find out the correct mean and standard deviation.
Answer:
Calculating correct Mean, by using the following observations:
Squaring both sides
425 × 20 = Σx2
= 8500
= − (Incorrect item)2 + (Correct item)2
= 8500 − (30)2 + (13)2
= 8500 − 900 + 169 = 7769
Hence, correct mean and standard deviation are 19.15 and 4.66 respectively.
Page No 10.88:
Question 43:
Following are the marks obtained by two students: Mollie and Isha, in 10 sets of examinations:
Marks of obtained by Mollie | 44 | 80 | 76 | 48 | 52 | 72 | 68 | 56 | 60 | 54 |
Marks of obtained by Isha | 48 | 75 | 54 | 60 | 63 | 69 | 72 | 51 | 57 | 66 |
Answer:
Mollie | Isha | ||||
XM | XI | ||||
44 | −17 | 289 | 48 | −13.5 | 182.25 |
80 | 19 | 361 | 75 | 13.5 | 182.25 |
76 | 15 | 225 | 54 | −7.5 | 56.25 |
48 | −13 | 169 | 60 | −1.5 | 2.25 |
52 | −9 | 81 | 63 | 1.5 | 2.25 |
72 | 11 | 121 | 69 | 7.5 | 56.25 |
68 | 7 | 49 | 72 | 10.5 | 110.25 |
56 | −5 | 25 | 51 | −10.5 | 110.25 |
60 | −1 | 1 | 57 | −4.5 | 20.25 |
54 | −7 | 49 | 66 | 4.5 | 20.25 |
ΣXM=610 | ΣXI=615 |
Isha is more consistent in securing makes as her C.V. (14.01)% is less than that of Mollie's C.V. (19.18)%
Page No 10.89:
Question 44:
Calculate coefficient of variation from the following data:
Marks (more than) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of students | 100 | 90 | 75 | 50 | 20 | 10 | 5 | 0 |
Answer:
Converting more than cumulative frequency distribution into simple frequency distribution:
Marks (X) |
No. of Students (f) |
Mid-Values (m) |
fm | m2 | fm2 |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 −70 70 −80 |
100-90=10 90-75=15 75-50=25 50-20=30 20-10=10 10-5=5 5-0=5 0-0=0 |
5 15 25 35 45 55 65 75 |
50 225 625 1050 450 275 325 0 |
25 225 625 1225 2025 3025 4225 5625 |
250 3375 15625 36750 20250 15125 21125 0 |
Σf =100 | Σfm = 3000 | Σfm2 =112500 |
Page No 10.89:
Question 45:
In two Towns A and B, daily pocket money and the standard deviation are given below:
Town | Average Daily Pocket | Standard Deviation | No. of Teenagers |
A | 34.5 | 5.0 | 476 |
B | 28.5 | 4.5 | 524 |
(i) Which town, A or B, pays out the larger amount of daily pocket money?
(ii) What is the average daily pocket money of all teenagers taken together?
(iii) Calculate coefficient of variation of each town. Which town is more variable in terms of pocket money?
Answer:
(i) Town A
Daily pocket money = Average Daily Pocket × No. of teenagers
= 34.5 × 476
= Rs 16422
Town B
Daily Pocket Money = 28.5 × 524
= Rs 14934
Town 'A' pays larger amount of daily pocket money.
(ii) Solution:
In order to find out average daily pocket money of all teenagers, we need find out the combined mean of their pocket money.
Thus,
(iii)
Town B is more variable in terms of pocket money as its C.V. is higher than that of C.V. of Town A.
Page No 10.89:
Question 46:
The prices of share of Company X and Company Y are given below. State, which company is more stable?
Company X | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 34 | 60 |
Company Y | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |
Answer:
Company X | Company Y | ||||
X | x2 | Y | y2 | ||
25 50 45 30 70 42 36 48 34 60 |
−19 6 1 −14 26 −2 −8 4 −10 16 |
361 36 1 196 676 4 64 16 100 256 |
10 70 50 20 95 55 42 60 48 80 |
−43 17 −3 −33 42 2 −11 7 −5 27 |
1849 289 9 1089 1764 4 121 49 25 729 |
ΣX = 440 | Σx2 = 1710 | ΣY = 530 | Σy2 = 5928 |
As C.V of prices of shares of Co. X is less than that of the prices of shares of Co. Y therefore, price of share of Co. X is more stable.
Page No 10.89:
Question 47:
A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50 instead of 40. Find the correct mean and standard deviation.
Answer:
Calculating correct Mean, by using the following observations:
Squaring both sides
Hence, correct mean and standard deviation are 39.9 and 5 respectively.
Page No 10.89:
Question 48:
From the following data, calculate standard deviation of the two groups A and B. Which group is more consistent?
Class Interval | Group A | Group B |
5−10 | 2 | 9 |
10−15 | 9 | 11 |
15−20 | 29 | 18 |
20−25 | 54 | 32 |
25−30 | 11 | 27 |
30−35 | 5 | 13 |
Answer:
For group A
Class Interval | Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 |
2 9 29 54 11 5 |
7.5 12.5 17.5 22.5 27.5 32.5 |
15 112.5 507.5 1215 302.5 162.5 |
56.25 156.25 306.25 506.25 756.25 1056.25 |
112.5 1406.25 8881.25 27337.5 8318.75 5281.25 |
Σf = 110 | Σfm = 2315 | Σfm2 = 51337.5 |
For group B
Class Interval | Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 |
9 11 18 32 27 13 |
7.5 12.5 17.5 22.5 27.5 32.5 |
67.5 137.5 315 720 742.5 422.5 |
56.25 156.25 306.25 506.25 756.25 1056.25 |
506.25 1718.75 5512.5 16200 20418.75 13731.25 |
Σf = 110 | Σfm = 2405 | Σfm2 = 58087.5 |
Group A is more consistent as C.V. of group A is less than C.V. of group B.
Page No 10.89:
Question 49:
From the following data of marks, calculate standard deviation. What will be the value of standard deviation, if marks obtained by each student is increased by one?
Marks Obtained | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
No. of students | 32 | 41 | 57 | 98 | 123 | 83 | 46 | 17 | 3 |
Answer:
Marks (X) |
Frequency (f) |
fX | x2 | fx2 | |
1 2 3 4 5 6 7 8 9 |
32 41 57 98 123 83 46 17 3 |
32 82 171 392 615 498 322 136 27 |
−3.55 −2.55 −1.55 −0.55 0.45 1.45 2.45 3.45 4.45 |
12.60 6.50 2.40 0.30 0.20 2.10 6.00 11.90 19.80 |
403.2 266.5 136.8 29.4 24.6 174.3 276 202.3 59.4 |
Σf = 500 | Σfx = 2275 | Σfx2 = 1572.5 |
As, standard deviation is independent of the change in origin, i.e. it will not get affected if the value of the series is increased or decreased by a constant quantity. Therefore it will remain the same.
Page No 10.90:
Question 50:
From the following data of two workers, identify who is a more consistent worker?
Worker | ||
A | B | |
Average Time in completing a job | 40 | 42 |
Standard Deviation | 8 | 6 |
Answer:
Worker B is more consistent as his C.V. (14.29)% is less than that of C.V of worker A(20%).
Page No 10.90:
Question 51:
Find the standard deviation and coefficient of standard deviation:
X (less than) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
Frequency | 12 | 30 | 65 | 107 | 157 | 202 | 222 | 230 |
Answer:
Converting less than frequency distribution into simple frequency distribution:
X | Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
12-0=12 30-12=18 65-30=35 107-65=42 157-107=50 202-157=45 222-202=20 230-222=8 |
5 15 25 35 45 55 65 75 |
60 270 875 1470 2250 2475 1300 600 |
25 225 625 1225 2025 3025 4225 5625 |
300 4050 21875 51450 101250 136125 84500 45000 |
Σf = 230 | Σfm = 9300 | Σfm2 = 444550 |
Page No 10.90:
Question 52:
Find mean, standard deviation and coefficient of variation.
Class-interval | 0−4 | 4−8 | 8−12 | 12−16 | 16−20 | 20−24 |
Frequency | 10 | 15 | 20 | 25 | 20 | 10 |
Answer:
Class Interval | Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
0 − 4 4 − 8 8 − 12 12 − 16 16 − 18 20 − 24 |
10 15 20 25 20 10 |
2 6 10 14 18 22 |
20 90 200 350 360 220 |
4 36 100 196 324 484 |
40 540 2000 4900 6480 4840 |
Σf = 100 | Σfm = 1240 | Σfm2 = 18800 |
Page No 10.90:
Question 53:
The following table shows the marks obtained by 60 students. Calculate mean and standard deviation.
Marks (more than) | 70 | 60 | 50 | 40 | 30 | 20 |
No. of students | 7 | 18 | 40 | 40 | 55 | 60 |
Answer:
Converting more than cumulative frequency into ordinary continuos series:
Marks X |
Frequency (f) |
Mid-Values (m) |
fm | m2 | fm2 |
20 −30 30 −40 40 −50 50 −60 60 −70 70 −80 |
60-55=5 55-40=15 40-40=0 40-18=22 18-7=11 7-0=7 |
25 35 45 55 65 75 |
125 525 0 1210 715 525 |
625 1225 2025 3025 4225 5625 |
3125 18375 0 66550 46475 39375 |
Σf = 60 | Σfm = 3100 | Σfm2 = 173900 |
Hence, mean and standard deviation of the above series are 51.67 marks and 15.11 marks respectively.
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