NM Shah (2017) Solutions for Class 11 Science Economics Chapter 7 Measures Of Dispersion are provided here with simple step-by-step explanations. These solutions for Measures Of Dispersion are extremely popular among class 11 Science students for Economics Measures Of Dispersion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NM Shah (2017) Book of class 11 Science Economics Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NM Shah (2017) Solutions. All NM Shah (2017) Solutions for class 11 Science Economics are prepared by experts and are 100% accurate.
Page No 273:
Question 1:
The daily wages of ten workers are given below. Find out range and its coefficient.
No. of Workers | : | A | B | C | D | E | F | G | H | I | J |
Wages in (â¹) | : | 175 | 50 | 50 | 55 | 100 | 90 | 125 | 145 | 70 | 60 |
Answer:
Range = Largest item (L) − Smallest item (S)
=175 − 50 = 125
Page No 273:
Question 2:
Following are the marks obtained by students in Sec . A and Sec . B . Compare the range of marks of students in two sections:
Marks ( Section A) | : | 20 | 25 | 28 | 45 | 15 | 30 |
Marks ( Section B) | : | 45 | 52 | 36 | 42 | 28 | 25 |
Answer:
For Section A
Range = Largest item (L) − Smallest item (S)
= 45 − 15 = 30
For Section B
Range = Largest item (L) − Smallest item (S)
= 52 − 25 = 27
Thus, range of marks of students in section A is greater than range of marks of students in section B.
Page No 273:
Question 3:
Find range and coefficient of range of the following:
(a) Per day earning of seven agricultural labourers in â¹:
60, | 72, | 36, | 85, | 35, | 52, | 72 |
(b) Mean temperature deviation from normal (2002).
Jan. | Feb. | Mar. | Apr. | May | June |
+1.5 | +2.4 | +3.1 | −1.5 | −0.4 | +3.3 |
July | Aug. | Sep. | Oct. | Nov. | Dec. |
−0.1 | −0.6 | −1.5 | −0.6 | −1.9 | −6.1 |
Answer:
(a) Range = Largest item (L) − Smallest item (S)
(b) Range = Largest item (L) − Smallest item (S)
Page No 273:
Question 4:
Calculate range and coefficient of range for the following data:
Income (â¹) | : | 50 | 70 | 80 | 90 | 100 | 120 | 130 | 150 |
No of workers | : | 2 | 8 | 12 | 7 | 4 | 3 | 8 | 6 |
Answer:
Range = Largest item (L) − Smallest item (S)
= 150 − 50 = 100
Page No 273:
Question 5:
Calculate range and coefficient of range of the following data:
X | : | 10 | 15 | 20 | 30 | 40 | 50 |
f | : | 4 | 12 | 7 | 3 | 5 | 2 |
Answer:
Range = Largest Value (L) − Smallest Value (S)
= 50 − 10 = 40
Page No 273:
Question 6:
Find the range and coefficient of range of the following:
Age in Years | : | 5-10 | 10-15 | 15-20 | 20-25 |
Frequency | : | 10 | 15 | 20 | 5 |
Answer:
Range = Largest item (L) − Smallest item (S)
= 25 − 5 = 20
Page No 273:
Question 7:
Calculate range and coefficient of range for the following distribution:
Class | : | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 |
Frequency | : | 2 | 8 | 15 | 35 | 20 | 10 |
Answer:
The given series is an inclusive series. For the calculation of range the given series must first be converted into exclusive series as follows:
Inclusive class | Exclusive class | Frequency |
1−5 | 0.5−5.5 | 2 |
6−10 | 5.5−10.5 | 8 |
11−15 | 10.5−15.5 | 15 |
16−20 | 15.5−20.5 | 35 |
21−25 | 20.5−25.5 | 20 |
26−30 | 25.5−30.5 | 10 |
Range = Largest item (L) − Smallest item (S)
= 30.5 − 0.5 = 30
Page No 274:
Question 8:
The following table gives the height of 100 persons . Calculate dispersion by Range Method.
Height (in centimetres) |
No. of Persons |
Below 162 | 2 |
Below 163 | 8 |
Below 164 | 19 |
Below 165 | 32 |
Below 166 | 45 |
Below 167 | 58 |
Below 168 | 85 |
Below 169 | 93 |
Below 170 | 100 |
Answer:
Height | Cumulative Frequency | Frequency |
161−162 | 2 | 2 |
162−163 | 8 | 8−2 = 6 |
163−164 | 19 | 19−8 = 11 |
164−165 | 32 | 32−19 = 13 |
165−166 | 45 | 45−32 = 13 |
166−167 | 58 | 58−45 = 13 |
167−168 | 85 | 85−58 = 27 |
168−169 | 93 | 93−85 = 8 |
169−170 | 100 | 100−93 = 7 |
For the calculation of range, the given data must be converted in continuous series as done above.
Range = Largest item (L) − Smallest item (S)
= 170 − 161 = 9
Page No 274:
Question 9:
Calculate Quartile Deviation and its Coefficient of Rajesh's daily income.
Months | : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
Income (â¹) | : | 239 | 250 | 251 | 251 | 257 | 258 | 260 | 261 | 262 | 262 | 273 |
Answer:
Month | Income (Rs) |
Month | Income (Rs) |
1 | 239 | 7 | 260 |
2 | 250 | 8 | 261 |
3 | 251 | 9 | 262 |
4 | 251 | 10 | 262 |
5 | 257 | 11 | 273 |
6 | 258 |
N = 11
Thus, the quartile deviation and coefficient of quartile deviation are 5.5 and 0.021 respectively.
Page No 274:
Question 10:
Find the Quartile Deviation and its Coefficient from the following data relating to the daily wages of seven workers:
Daily Wages ( in â¹) | : | 50 | 90 | 70 | 40 | 80 | 65 | 60 |
Answer:
S. No. | Daily Wages |
1 | 40 |
2 | 50 |
3 | 60 |
4 | 65 |
5 | 70 |
6 | 80 |
7 | 90 |
For the calculation of quartile deviation, we first arrange the given data in ascending order as above. The quartile deviation is then calculated in the following manner.
Thus, the quartile deviation and coefficient of quartile deviation are 15 and 0.23 respectively.
Page No 274:
Question 11:
Find out Quartile Deviation and Coefficient of Quartile Deviation of the following items:
145 | 130 | 200 | 210 | 198 |
234 | 159 | 160 | 178 | 257 |
260 | 300 | 345 | 360 | 390 |
Answer:
S. No. | Item |
1 | 130 |
2 | 145 |
3 | 159 |
4 | 160 |
5 | 178 |
6 | 198 |
7 8 9 10 11 12 13 14 15 |
200 210 234 257 260 300 345 360 390 |
We first arrange the given data in ascending order as above.
Thus, the quartile deviation and coefficient of quartile deviation are 70 and 0.304 respectively.
Page No 274:
Question 13:
Find out Quartile Deviation , Coefficient of Quartile Deviation of the following data:
X | : | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
f | : | 6 | 17 | 29 | 38 | 25 | 14 | 9 | 1 |
Answer:
X | f | Cumulative Frequency |
10 | 6 | 6 |
15 | 17 | 23 |
20 | 29 | 52 |
25 | 38 | 90 |
30 | 25 | 115 |
35 | 14 | 129 |
40 | 9 | 138 |
45 | 1 | 139 |
N = 139 |
The 35th item corresponds to 52 in the cumulative frequency.
The 105th item corresponds to 115 in the cumulative frequency.
Thus, coefficient of quartile deviation is 0.2.
Page No 274:
Question 14:
Calculate the Quartile Deviation and Coefficient of Quartile Deviation of the following data:
Marks | : | 5-9 | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 |
Students | : | 1 | 3 | 8 | 5 | 4 | 2 | 2 |
Answer:
Marks | Students (f) |
Cumulative Frequency |
5−9 | 1 | 1 |
10−14 | 3 | 1 + 3 = 4 |
15−19 | 8 | 4 + 8 = 12 |
20−24 | 5 | 12 + 5 = 17 |
25−29 | 4 | 17 + 4 = 21 |
30−34 | 2 | 21 + 2 = 23 |
35−39 | 2 | 23 + 2 = 25 |
N = Æ©f = 25 |
This corresponds to the class interval (15−19).
This corresponds to the class interval (20−24).
This corresponds to the class interval (25−29).
Page No 275:
Question 16:
Calculate the Semi-interquartile Range and its Coefficient of the following data:
Marks | : | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
No. of Students | : | 4 | 8 | 10 | 14 | 10 | 9 | 5 |
Answer:
Marks | No. of Students (f) |
c.f |
0−10 | 4 | 4 |
10−20 | 8 | 12 |
20−30 | 10 | 22 |
30−40 | 14 | 36 |
40−50 | 10 | 46 |
50−60 | 9 | 55 |
60−70 | 5 | 60 |
N = 60 |
This corresponds to the class interval 20−30, so this is the first quartile class.
The third quartile class is given by the size of the 3item, i.e. the 3item, which is the 45th item.
This corresponds to the class interval 40−50, so this is the third quartile class.
Page No 275:
Question 18:
Calculate the Interquartile Range for the data given below:
Class | : | 35-40 | 30-35 | 25-30 | 20-25 | 15-20 | 10-15 | 5-10 | 0-5 |
Frequency | : | 1 | 4 | 9 | 11 | 10 | 6 | 5 | 4 |
Answer:
Class | Frequency (f) |
c.f. |
0−5 | 4 | 4 |
5−10 | 5 | 9 |
10−15 | 6 | 15 |
15−20 | 10 | 25 |
20−25 | 11 | 36 |
25−30 | 9 | 45 |
30−35 | 4 | 49 |
35−40 | 1 | 50 |
N = 50 |
The first quartile class is given by the size of the item, i.e. the item, which is the 12.5th item.
This corresponds to the class interval 10−15, so this is the first quartile class.
The third quartile class is given by the size of the 3item, i.e. the 3item, which is the 37.5th item.
This corresponds to the class interval 25−30, so this is the third quartile class.
Page No 275:
Question 19:
Calculate the Mean deviation from median of the following data:
Marks | : | 41 | 66 | 59 | 38 | 54 | 21 | 32 | 49 | 68 |
Answer:
Marks (X) |
Deviation from Median D= |X−M| (M = 49) |
21 | 28 |
32 | 17 |
38 | 11 |
41 | 8 |
49 | 0 |
54 | 5 |
59 | 10 |
66 | 17 |
68 | 19 |
ΣD = 115 |
For the calculation of mean deviation, the given data is first arranged in the ascending order as done above. Then median is calculated in the following manner.
Thus, mean deviation from median is 12.77.
Page No 275:
Question 20:
Find Coefficient of Mean deviation from median :
Price (Rs) | : | 25 | 28 | 32 | 32 | 36 | 48 | 44 | 45 | 50 | 50 |
Answer:
Price | Deviation from Median D = |X−M| M = 40 |
25 | 15 |
28 | 12 |
32 | 8 |
32 | 8 |
36 | 4 |
44 | 4 |
45 | 5 |
48 | 8 |
50 | 10 |
50 | 10 |
ΣD = 84 |
Thus, coefficient of mean deviation from median is 0.2.
Page No 275:
Question 21:
Calculate the Mean Deviation from mean and median of the following data:
X | : | 54 | 71 | 57 | 52 | 49 | 45 | 72 | 57 | 47 |
Answer:
X | Deviation from Median DM =|X−M| M = 54 |
Deviation from Mean |
45 | 9 | 11 |
47 | 7 | 9 |
49 | 5 | 7 |
52 | 2 | 4 |
54 | 0 | 2 |
57 | 3 | 1 |
57 | 3 | 1 |
71 | 17 | 15 |
72 | 18 | 16 |
ΣX = 504 | Σ|X−Md| = 64 |
Page No 275:
Question 22:
Calculate Mean deviation from (i) arithmetic mean, (ii) Median:
Marks | : | 7 | 4 | 10 | 9 | 15 | 12 | 7 | 9 | 7 |
Answer:
Marks (X) |
Deviation from Mean |
Deviation from Median DM =|X−M| |
4 | 4.89 | 5 |
7 | 1.89 | 2 |
7 | 1.89 | 2 |
7 | 1.89 | 2 |
9 | 0.11 | 0 |
9 | 0.11 | 0 |
10 | 1.11 | 1 |
12 | 3.11 | 3 |
15 | 6.11 | 6 |
ΣX = 80 | Σ|X−Md| = 21 |
Page No 275:
Question 23:
Find out the average deviation from median for the following distribution:
Values | : | 6 | 12 | 18 | 24 | 30 | 36 | 42 |
Frequency | : | 4 | 7 | 9 | 18 | 15 | 10 | 5 |
Answer:
Value (X) |
Frequency (f) |
c.f | Deviation from Median |DM|=|X−M| (M = 24) |
f|DM| |
6 | 4 | 4 | 18 | 72 |
12 | 7 | 11 | 12 | 84 |
18 | 9 | 20 | 6 | 54 |
24 | 18 | 38 | 0 | 0 |
30 | 15 | 53 | 6 | 90 |
36 | 10 | 63 | 12 | 120 |
42 | 5 | 68 | 18 | 90 |
Σf = N = 68 | Σf|dm| = 510 |
This corresponds to value 24. Thus, median is 24.
Page No 275:
Question 24:
Calculate Mean Deviation from median:
No. of tomatoes per plant | : | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
No. of Plants | : | 2 | 5 | 7 | 11 | 18 | 24 | 12 | 8 | 6 | 4 | 3 |
Answer:
Tomatoes | Frequency (f) |
c.f | Deviation from Median |DM|=|X−M| (M = 5) |
f|DM| |
0 | 2 | 2 | 5 | 10 |
1 | 5 | 7 | 4 | 20 |
2 | 7 | 14 | 3 | 21 |
3 | 11 | 25 | 2 | 22 |
4 | 18 | 43 | 1 | 18 |
5 | 24 | 67 | 0 | 0 |
6 | 12 | 79 | 1 | 12 |
7 | 8 | 87 | 2 | 16 |
8 | 6 | 93 | 3 | 18 |
9 | 4 | 97 | 4 | 16 |
10 | 3 | 100 | 5 | 15 |
Σf = N = 100 | Σf|dm|=168 |
This corresponds to 5. Thus, median is 5.
Page No 275:
Question 25:
Calculate (a) Median Coefficient of dispersion from the following data:
Size of item | : | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
Frequency | : | 2 | 4 | 5 | 3 | 2 | 1 | 4 |
Answer:
Size of Item (X) |
Frequency (f) |
c.f | Deviation from Median |DM|=|X−M| (M = 8) |
f|DM| |
4 | 2 | 2 | 4 | 8 |
6 | 4 | 6 | 2 | 8 |
8 | 5 | 11 | 0 | 0 |
10 | 3 | 14 | 2 | 6 |
12 | 2 | 16 | 4 | 8 |
14 | 1 | 17 | 6 | 6 |
16 | 4 | 21 | 8 | 32 |
Σf = N = 21 | Σf|dm| = 68 |
This corresponds to size 8. Thus, median is 8.
Size (X) |
Frequency (f) |
fX | Deviation from Mean |
|
4 | 2 | 8 | 5.71 | 11.42 |
6 | 4 | 24 | 3.71 | 14.84 |
8 | 5 | 40 | 1.71 | 8.55 |
10 | 3 | 30 | 0 .29 | 0.87 |
12 | 2 | 24 | 2.29 | 4.58 |
14 | 1 | 14 | 4.29 | 4.29 |
16 | 4 | 64 | 6.29 | 25.16 |
Σf = 21 | ΣfX = 204 |
Page No 276:
Question 28:
Calculate Mean and Mean Deviation and coefficient of M.D. for the following distribution:
Weekly wages | : | 20-40 | 40-60 | 60-80 | 80-100 |
Workers | : | 20 | 40 | 30 | 10 |
Answer:
Weekly wages (X) |
Mid value m |
f | fm | ||
20−40 | 30 | 20 | 600 | 26 | 520 |
40−60 | 50 | 40 | 2000 | 6 | 240 |
60−80 | 70 | 30 | 2100 | 14 | 420 |
80−100 | 90 | 10 | 900 | 34 | 340 |
Σf = 100 | Σfm = 5600 |
Page No 276:
Question 30:
Find Mean Deviation from median of the marks secured by 100 students in a class-test as given below:
Marks | : | 60-63 | 63-66 | 66-69 | 69-72 | 72-75 |
No. of Students | : | 5 | 18 | 42 | 27 | 8 |
Answer:
Marks | Mid Value | f | c.f | Deviation from Median DM = |m−M| (M= 67.93) |
f|DM| |
60−63 | 61.5 | 5 | 5 | 6.43 | 32.15 |
63−66 | 64.5 | 18 | 23 | 3.43 | 61.74 |
66−69 | 67.5 | 42 | 65 | 0 .43 | 18.06 |
69−72 | 70.5 | 27 | 92 | 2.57 | 69.39 |
72−75 | 73.5 | 8 | 100 | 5.57 | 44.56 |
Σf = 100 | Σf|DM| = 225.9 |
This corresponds to the class interval 66−69, so this is the median class.
Page No 276:
Question 31:
Using Mean Deviation from median of the income group of 5 and 7 members given below, compare which of the group has more variability?
Group A | : | 4000 | 4200 | 4400 | 4600 | 4800 | ||
Group B | : | 3000 | 4000 | 4200 | 4400 | 4600 | 4800 | 5800 |
Answer:
For Group A
Income (X) |
Deviation from median DM=|X−M| (M=4400) |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
Σ|DM|=1200 |
For Group B
Income |
Deviation from median DM=|X−M| (M=4400) |
3000 | 1400 |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
5800 | 1400 |
Σ|DM|=4000 |
In case of group B, as the coefficient of MD is more, so group B has greater variation.
Page No 276:
Question 32:
Calculate the standard Deviation of wage earner's daily earnings:
Week | : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Earnings (in â¹) | : | 54 | 62 | 63 | 65 | 68 | 71 | 73 | 78 | 82 | 84 |
Answer:
S.No. (Week) |
Farming (X) |
Deviation |
Square of deviation |
1 2 3 4 5 6 7 8 9 10 |
54 62 63 65 68 71 73 78 82 84 |
−16 −8 −7 −5 −2 1 3 8 12 14 |
256 64 49 25 4 1 9 64 144 196 |
ΣX = 700 |
The mean of the given series can be calculated in the following manner.
To calculate standard deviation, we use the following formula.
Thus, standard deviation of wage earner's daily earnings is 9.01.
Page No 276:
Question 33:
Calculate Standard Deviation of the following two series . Which series has more variability:
A: | 58 | 59 | 60 | 65 | 66 | 52 | 75 | 31 | 46 | 48 |
B: | 56 | 87 | 89 | 46 | 93 | 65 | 44 | 54 | 78 | 68 |
Answer:
For A
S. No. | (X) | Deviation |
x2 |
1 2 3 4 5 6 7 8 9 10 |
58 59 60 65 66 52 75 31 46 48 |
2 3 4 9 10 −4 19 −25 −10 −8 |
4 9 16 81 100 16 361 625 100 64 |
ΣX = 560 | Σx2 = 1376 |
For B
S. No. | Y | Deviation |
y2 |
1 2 3 4 5 6 7 8 9 10 |
56 87 89 46 93 65 44 54 78 68 |
−12 19 21 −22 25 −3 −24 −14 10 0 |
144 361 441 484 625 9 576 196 100 0 |
ΣY = 680 | Σy2 = 2936 |
Since the coefficient of variation of series B is more, it is said to have more variability.
Page No 277:
Question 35:
Calculate Mean and Standard Deviation of income of 8 employees of a firm:
Income (â¹): | 100 | 120 | 140 | 120 | 180 | 140 | 120 | 150 |
Answer:
S. No. (X) |
Deviation |
x2 |
100 120 140 120 180 140 120 150 |
−33.75 −13.75 6.25 −13.75 46.25 6.25 −13.75 16.25 |
1139.06 189.06 39.06 189.06 2139.06 39.06 189.06 264.06 |
ΣX= 1070 | Σx2 = 4187.48 |
Thus, mean standard deviation is 22.87.
Page No 277:
Question 36:
Calculate Mean and Standard Deviation from the following data:
Marks ( Above) | : | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
No. of Students | : | 100 | 90 | 75 | 50 | 25 | 15 | 5 | 0 |
Answer:
As we are given marks (above) series, we can convert it into classes and frame the following table.
Marks | Mid Value (m) |
cf | Frequency (f) |
fm | Deviation from mean |
x2 | fx2 |
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
5 15 25 35 45 55 65 75 |
100 90 75 50 25 15 5 0 |
10 15 25 25 10 10 5 0 |
50 225 625 875 450 550 325 0 |
-26 -16 -6 4 14 24 34 44 |
676 256 36 16 196 576 1156 1936 |
6760 3840 900 400 1960 5760 5780 0 |
Σf = 100 | Σfm = 3100 | Σfx2=25400 |
Page No 277:
Question 37:
Calculate Mean and Variance for the following frequency distribution:
X | : | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
f | : | 2 | 5 | 8 | 16 | 21 | 18 | 13 | 10 | 4 | 3 |
Answer:
X | f | fX | Deviation |
x2 | fx2 |
1 2 3 4 5 6 7 8 9 10 |
2 5 8 16 21 18 13 10 4 3 |
2 10 24 64 105 108 91 80 36 30 |
−4.5 −3.5 −2.5 −1.5 −.5 .5 1.5 2.5 3.5 4.5 |
20.25 12.25 6.25 2.25 .25 .25 2.25 6.25 12.25 20.25 |
40.5 61.25 50 36 5.25 4.5 29.25 62.5 49 60.75 |
Σf = 100 | ΣfX = 550 | Σfx2 = 399 |
Thus, mean and variance are 5.5 and 3.99 respectively.
Page No 277:
Question 38:
Calculate Mean, Standard Deviation and Variance:
Variable | : | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequency | : | 2 | 5 | 7 | 13 | 21 | 16 | 8 | 3 |
Answer:
Variable | Mid Value (m) | Frequency (f) |
fm | Deviation Mean |
x2 | fx2 |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 |
2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 |
2 5 7 13 21 16 8 3 |
5 37.5 87.5 227.5 472.5 440 260 112.5 |
19.4 14.4 9.4 4.4 0.6 5.6 10.6 15.6 |
376.36 207.36 88.36 19.36 0.36 31.36 112.36 243.36 |
752.72 1036.8 618.52 251.68 7.56 501.76 898.88 730.08 |
Σf = 75 | Σfm = 1642.5 | Σfx2 = 4798 |
Page No 277:
Question 39:
Calculate the Coefficient of Variation for the following distribution of the wages of 200 workers in a factory:
Wages ( in â¹) | : | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 | 100-109 | 110-119 |
No. of workers | : | 11 | 23 | 40 | 60 | 35 | 16 | 9 | 6 |
Answer:
In order to calculate coefficient of variation, first we will convert the given inclusive series into exclusive series as follows:
Wages | Mid Point (m) |
Frequency (f) |
fm | Deviation from |
x2 | fx2 |
39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 69.5 − 79.5 79.5 − 89.5 89.5 − 99.5 99.5 − 109.5 109.5 − 119.5 |
44.5 54.5 64.5 74.5 84.5 94.5 104.5 114.5 |
11 23 40 60 35 16 9 6 |
489.5 1253.5 2580 4470 2957.5 1512 940.5 687 |
29.95 19.95 9.95 .05 10.05 20.05 30.05 40.05 |
897 398 99 .002 101 402 903 1604 |
9867 9154 3960 0.12 3535 6432 8127 9624 |
Σf = 200 | Σfm = 14890 | Σfx2 = 50699.12 |
Page No 277:
Question 40:
The following are the scores made by two batsmen A and B in a series of innings:
A | : | 12 | 115 | 6 | 73 | 7 | 19 | 119 | 36 | 84 | 29 |
B | : | 47 | 12 | 76 | 42 | 4 | 51 | 37 | 48 | 13 | 0 |
Who is better as a run-getter ? Who is more consistent?
Answer:
For Batsman A
S. No. | Score (XA) |
Deviation from mean |
|
1 2 3 4 5 6 7 8 9 10 |
12 115 6 73 7 19 119 36 84 29 |
−38 65 −44 23 −43 −31 69 −14 34 −21 |
144 4225 1936 529 1849 961 4761 196 1156 441 |
N = 10 | ΣxA = 500 |
For Batsman B
S. No. | Score XB |
Deviation from Mean |
|
1 2 3 4 5 6 7 8 9 10 |
47 12 76 42 4 51 37 48 13 0 |
14 −21 43 9 −29 18 4 15 −20 −33 |
196 441 1849 81 841 324 16 225 400 1089 |
N = 10 | ΣxB = 330 |
Since average score of A is more than B, he is better as run-getter but since, coefficient of variance is less for B, therefore B is more consistent.
Page No 277:
Question 41:
The index number of prices of cotton and coal shares in 1998 were as under:
Index Number of Prices:
Cotton | : | 188 | 178 | 173 | 164 | 172 | 183 | 184 | 185 | 211 | 217 | 232 | 240 |
Coal | : | 131 | 130 | 130 | 129 | 129 | 129 | 127 | 127 | 130 | 137 | 140 | 142 |
Which of these two shares do you consider more variable in prices:
Answer:
S. No. | Price X |
Deviation from mean |
x2 |
1 2 3 4 5 6 7 8 9 10 11 12 |
188 178 173 164 172 183 184 185 211 217 232 240 |
5.9 15.9 20.9 29.9 21.9 10.9 9.9 8.9 17.1 23.1 38.1 46.1 |
34.81 252.81 436.81 894.01 479.61 118.81 98.01 79.21 292.41 533.61 1451.61 2125.61 |
ΣX = 2327 | Σx2 = 6796.92 |
S. No. | Price X |
Deviation from mean |
x2 |
1 2 3 4 5 6 7 8 9 10 11 12 |
131 130 130 129 129 129 127 127 130 137 140 142 |
−.75 −1.75 −1.75 −2.75 −2.75 −2.75 −4.75 −4.75 −1.75 5.25 8.25 10.25 |
.56 3.06 3.06 7.56 7.56 7.56 22.56 22.56 3.06 27.56 68.06 105.06 |
ΣX = 1581 | Σx2= 278.22 |
Since coefficient of variance is more of cotton than that of coal. So, prices of cotton shares are more variable.
Page No 278:
Question 42:
Calculate the arithmetic mean, standard deviation and variance from the following distribution:
Class | : | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequency | : | 2 | 5 | 7 | 13 | 21 | 16 | 8 | 3 |
Answer:
Size | Mid Value (m) |
(f) | fm | Deviation from mean x |
x2 | fx2 |
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40 |
2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 |
2 5 7 13 21 16 8 3 |
5 37.5 87.5 227.5 472.5 440 260 112.5 |
19.4 14.4 9.4 4.4 .6 5.6 10.6 15.6 |
376.36 207.36 88.36 19.36 .36 31.36 112.36 243.36 |
752.72 1036.8 618.52 251.68 7.56 501.76 898.88 730.08 |
Σf = 75 | Σfm = 1642.5 | Σfx2 = 4798 |
Page No 278:
Question 43:
Calculate arithmetic mean, standard deviation and variance from the following series:
Marks | : | 70-80 | 60-70 | 50-60 | 40-50 | 30-40 | 20-30 |
No of students | : | 7 | 11 | 22 | 0 | 15 | 5 |
Answer:
Marks | Mid Value (m) |
f | fm | Deviation from mean |
x2 | fx2 |
20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 |
25 35 45 55 65 75 |
5 15 0 22 11 7 |
125 525 0 1210 715 525 |
26.67 16.67 6.67 3.33 13.33 23.33 |
711.29 277.89 44.49 11.09 177.69 544.29 |
3556.45 4168.35 0 243.98 1954.59 3810.03 |
Σf = 60 | Σfm = 3100 | Σfx2 = 13733.4 |
Page No 278:
Question 45:
You are given the following data about height of boys and girls:
Boys | Girls | |
Number | 72 | 38 |
Average height( in inches) | 68 | 61 |
Variance of distribution (in inches) | 9 | 4 |
(a) Calculate Coefficient of Variation.
(b) Decide whose height is more variable.
Answer:
(a)
For Boys
For Girls
(b) Since coefficient of variance of boys is more than that of girls, thus the height of boys is more variable.
Page No 278:
Question 46:
Lives of two models of refrigerators in a recent survey are as under:
Life No fo Years | : | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 |
No of Refrigerators | : | ||||||
Model A | : | 5 | 16 | 13 | 7 | 5 | 4 |
Model B | : | 2 | 7 | 12 | 19 | 9 | 1 |
What is the average life of each model of refrigerators? Which model has more uniformity?
Answer:
Model A:
Years | Mid Value (m) |
f | fm | Deviation from mean |
||
0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 |
1 3 5 7 9 11 |
5 16 13 7 5 4 |
5 48 65 49 45 44 |
−4.12 −2.12 −.12 1.88 3.88 5.88 |
16.97 4.49 0.01 3.53 15.05 34.57 |
84.85 71.84 0.13 24.71 75.25 138.28 |
Σf = 50 | Σfm = 256 |
Average life of model A is 5.12 year.
Model B:
Years | Mid Value (m) |
f | fm | Deviation from mean |
||
0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 |
1 3 5 7 9 11 |
2 7 12 19 9 1 |
2 21 60 133 81 11 |
−5.16 −3.16 −1.16 0.84 2.84 4.84 |
26.62 9.98 1.34 0.70 8.06 23,42 |
53.24 69.86 16.08 13.3 72.54 23.42 |
Σf = 50 | Σfm = 308 |
Average life of model B is 6.16 year.
Since the coefficient of variation of model B is less than that of A, thus model B has more uniformity.
Page No 279:
Question 48:
Two brands of Tyres are tested with the following results:
Life ( '000 miles) | : | 20-25 | 25-30 | 30-35 | 35-40 | 45-50 | 50-55 |
Brand X | 8 | 15 | 12 | 18 | 13 | 9 | |
Brand Y | 6 | 20 | 32 | 30 | 12 | 0 |
Answer:
For Brand 'X'
Life |
Mid value m |
f | fm | Deviation from mean |
x2 | fx2 |
20−25 | 22.5 | 8 | 180 | −12.67 | 160.53 | 1284.24 |
25−30 | 27.5 | 15 | 412.5 | −7.67 | 58.83 | 882.45 |
30−35 | 32.5 | 12 | 390 | −2.67 | 7.13 | 85.56 |
35−40 | 37.5 | 18 | 675 | 2.33 | 5.43 | 97.74 |
40−45 | 42.5 | 13 | 552.5 | 7.33 | 53.73 | 698.49 |
45−50 | 47.5 | 9 | 427.5 | 12.33 | 152.03 | 1368.27 |
Σf =75 | Σfm =2637.5 | Σfx2= 4416.75 |
For Brand 'Y'
Life |
Mid value m |
f | fm | Deviation from mean |
x2 | fx |
20−25 | 22.5 | 6 | 135 | −11.1 | 123.21 | 739.26 |
25−30 | 27.5 | 20 | 550 | −6.1 | 37.21 | 744.2 |
30−35 | 32.5 | 32 | 1040 | −1.1 | 1.21 | 38.72 |
35−40 | 37.5 | 30 | 1125 | 3.9 | 15.21 | 456.3 |
40−45 | 42.5 | 12 | 510 | 8.9 | 79.21 | 950.52 |
45−50 | 47.5 | 0 | 0 | 13.9 | 193.21 | 0 |
Σf=100 | Σfm=3360 |
As the coefficient of variation is more for brand X, it is said to be more variable than brand Y.
Page No 279:
Question 49:
From the data given below state which series is more consistent:
Variable | : | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Series A | : | 10 | 18 | 32 | 40 | 22 | 18 |
Series B | : | 18 | 22 | 40 | 32 | 18 | 10 |
Answer:
For Series A
Variable | Mid value (m) |
f | fm | Deviation from mean |
||
10−20 | 15 | 10 | 150 | −27.14 | 736.58 | 7365.8 |
20−30 | 25 | 18 | 450 | −17.14 | 293.78 | 5288.04 |
30−40 | 35 | 32 | 1120 | −7.14 | 50.98 | 1631.36 |
40−50 | 45 | 40 | 1800 | 2.86 | 8.18 | 327.2 |
50−60 | 55 | 22 | 1210 | 12.86 | 165.38 | 3638.36 |
60−70 | 65 | 18 | 1170 | 22.86 | 522.58 | 9406.44 |
Σf=140 | Σfm=5900 |
For Series B
Variable |
Mid value (m) |
f | fm | Deviation from mean |
||
10−20 | 15 | 18 | 270 | −22.86 | 522.58 | 9406.44 |
20−30 | 25 | 22 | 550 | −12.86 | 165.38 | 3638.36 |
30−40 | 35 | 40 | 1400 | −2.86 | 8.18 | 327.2 |
40−50 | 45 | 32 | 1440 | 7.14 | 50.98 | 1631.36 |
50−60 | 55 | 18 | 990 | 17.14 | 293.78 | 5288.04 |
60−70 | 65 | 10 | 650 | 27.14 | 736.58 | 7365.8 |
Σf=140 | Σfm=5300 |
As coefficient of variation for series 'A' is less than series 'B', therefore, series A is more consistent.
Page No 279:
Question 50:
The following table gives the distribution of wages in the two branches of a factory:
Monthly wages ( in â¹) | : | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 | Total |
Number of workers | : | ||||||
Branch A | : | 167 | 207 | 253 | 205 | 168 | 1000 |
Branch B | : | 63 | 93 | 157 | 105 | 82 | 500 |
(a) Which branch pays higher average wages?
(b) Which branch has greater variability in wages in relation to the average wages?
(c) What is the average monthly wage for the factory as a whole?
(d) What is the variance of wages of all the workers in the two branchesA and B taken together?
Answer:
For Branch A
Monthly Wages | Mid Value (m) |
f | fm | Deviation from mean |
||
100 − 150 150 − 200 200 − 250 250 − 300 300 − 350 |
125 175 225 275 325 |
167 207 253 205 168 |
20875 36225 56925 56375 54600 |
−100 −50 0 50 100 |
10000 2500 0 2500 10000 |
1670000 517500 0 512500 1680000 |
Σf = 1000 | Σfm = 225000 |
For Branch B
Monthly wage | Mid value (m) |
f | fm | Deviation from mean |
||
100 − 150 150 − 200 200 − 250 250 − 300 300 − 350 |
125 175 225 275 325 |
63 93 157 105 82 |
7875 16275 35325 28875 26650 |
−105 −55 −5 45 95 |
11025 3025 25 2025 9025 |
694575 281325 3925 212625 740050 |
Σf = 500 | Σfm = 115000 |
(a) Branch B pays higher average monthly wages i.e. Rs 230.
(b) Branch A has greater variability since its coefficient of variation is higher than that of B.
(c)
Thus, the average monthly wage for the factory as a whole is Rs 226.67.
(d)
Thus, the variance of wages of all the workers in the two branches- A and B taken together is Rs 4217.95.
Page No 279:
Question 52:
Goals scored by two teams A and B in football matches were as follows:
No. of Goals in a match | : | 0 | 1 | 2 | 3 | 4 |
No.of Matches: A | : | 27 | 9 | 8 | 5 | 4 |
B | : | 17 | 9 | 6 | 5 | 3 |
Answer:
X | f | fX | Deviation from mean |
||
0 1 2 3 4 |
27 9 8 5 4 |
0 9 16 15 16 |
−1.06 −0.06 0.94 1.94 2.94 |
1.12 0.0036 0.88 3.76 8.64 |
30.24 0.03 7.04 18.8 34.56 |
Σf = 53 | ΣfX = 56 |
X | f | fX | Deviation from mean |
||
0 1 2 3 4 |
17 9 6 5 3 |
0 9 12 15 12 |
−1.2 −0.2 0.8 1.8 2.8 |
1.44 0.04 0.64 3.24 7.84 |
24.48 0.36 3.84 16.2 23.52 |
Σf = 40 | ΣfX= 48 |
Since, coefficient of variation is lesser for team B, it is more consistent.
Page No 279:
Question 53:
Find mean and the standard deviation of the following two groups taken together:
Group | Number | Mean | S.D. |
A | 113 | 159 | 22.4 |
B | 121 | 149 | 20.0 |
Answer:
Now, to calculate combined standard deviation, we need to find,
Combined Standard Deviation
Thus, mean and standard deviation of the two groups-A and B taken together are 23.33 and 21.8 respectively.
Page No 280:
Question 54:
The number examined , the mean weight and standard deviations in each group of examination by two medical examiners are given below . Calculate mean and standard deviation of both the groups taken together.
Medical Examiner | Number Examined | Mean Weight (lbs) |
Standard deviation (lbs) |
A | 50 | 113 | 6.5 |
B | 60 | 120 | 8.2 |
Answer:
Combined Standard Deviation
Thus, mean and standard deviation of both the groups taken together are 116.82 and 8.25 respectively.
Page No 280:
Question 55:
The following data gives arithmetic means, standard deviations of three sub-groups . Calculate arithmetic mean and standard deviation of the whole group.
Sub-group | No. of Men | Average Wage (in â¹) |
Standard deviation (in â¹) |
A | 50 | 61.0 | 8.0 |
B | 100 | 70.0 | 9.0 |
C | 120 | 80.5 | 10.0 |
Answer:
Combined Standard Deviation
Thus, arithmetic mean and standard deviation of the whole group are 73 and 11.9 respectively.
Page No 280:
Question 56:
A sample of 35 values has mean 80 and Standard Deviation 4 . A second sample of 65 values from same population has mean 70 and standard deviation 3. Find the mean and standard deviation of combined sample of 100 values.
Answer:
Given,
Combined Standard Deviation
Thus, mean and standard deviation of combined sample of 100 values are 73.5 and 5.84 respectively.
Page No 280:
Question 57:
For a group of 50 male workers , the mean and standard deviation of their weekly wages are â¹ 63 and â¹ 9 respectively. For a group of 40 female workers , these are â¹ 54 and â¹ 6 respectively. Find mean and standard deviation for a combined group of 90 workers.
Answer:
Given,
Combined Standard Deviation
Thus, mean and standard deviation for a combined group of 90 workers are 59 and 9 respectively.
Page No 280:
Question 58:
Coefficient of variation of two series are 58% and 69% and their standard deviation are 21.2 and 15.6 . What are their means?
Answer:
Let the first series be A.
Now,
Coefficient of variation of A = 58%
σA = 21.2
Let the second series B.
Now,
Coefficient of variation of B= 69%
σB = 21.2
Thus, means of series A and series B are 36.55 and 22.608 respectively.
Page No 280:
Question 59:
If the coefficient of variation of X series is 14.6% and that of Y series is 36.9% and their means are 101.2 and 101.25 respectively. Find their standard deviation.
Answer:
For first series,
Coefficient of variation of A = 14.6%
For second series,
Coefficient of variation of B= 36.9%
Thus, standard deviations of first and second series are 14.775 and 37.36 respectively.
Page No 280:
Question 60:
Draw a Lorenz curve from the following:
Income (â¹ '000) | : | 20 | 40 | 60 | 100 | 160 |
No of Persons ('000) | ||||||
Class A | : | 10 | 20 | 40 | 50 | 80 |
Class B | : | 16 | 14 | 10 | 6 | 4 |
Answer:
Income | Cumulative Income |
Cumulative Percentage |
Person of Class A |
Cumulative |
Cumulative Percentage |
Person of Class B |
Cumulative Persons |
Cumulative Percentage |
20 40 60 100 160 |
20 60 120 220 380 |
5.26 15.78 31.57 57.89 100 |
10 20 40 50 80 |
10 30 70 120 200 |
5 15 35 60 100 |
16 14 10 6 4 |
16 30 40 46 50 |
32 60 80 92 100 |
Page No 280:
Question 61:
Following is the frequency distribution of marks obtained by students in Economics and Statistics . Analyse the data by drawing a Lorenz Curve.
Marks (mid-values) | : | 5 | 15 | 25 | 35 | 45 | 55 | 65 | 75 | 85 | 95 |
No.of students (Economics) | : | 10 | 12 | 13 | 14 | 22 | 27 | 20 | 12 | 11 | 9 |
No. of students (Statistics) | : | 1 | 2 | 26 | 50 | 59 | 40 | 10 | 8 | 3 | 1 |
Answer:
Marks | Cumulative | Cumulative Percentage |
Students in Economics |
Cumulative students | Cumulative Percentage |
Students in Statistics | Cumulative Students |
Cumulative Percentage |
5 15 25 35 45 55 65 75 85 95 |
5 20 45 80 125 180 245 320 405 500 |
1 4 9 16 25 36 49 64 81 100 |
10 12 13 14 22 27 20 12 11 9 |
10 22 35 49 71 98 118 130 141 150 |
6.67 14.67 23.33 32.67 47.33 65.33 78.67 86.67 94 100 |
1 2 26 50 59 40 10 8 3 1 |
1 3 29 79 138 178 188 196 199 200 |
0.5 1.5 14.5 39.5 69 89 94 98 99.5 100 |
View NCERT Solutions for all chapters of Class 11