Free download NCERT Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1, Ex 15.2, Ex 15.3, and Miscellaneous Exercise PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2020-21.

Topics and Sub Topics in Class 11 Maths Chapter 15 Statistics:

Section Name | Topic Name |

15 | Statistics |

15.1 | Introduction |

15.2 | Measures of Dispersion |

15.3 | Range |

15.4 | Mean Deviation |

15.5 | Variance and Standard Deviation |

15 .6 | Analysis of Frequency Distributions |

**Contents**show

**NCERT Solutions for Class 11 Maths** Chapter 15

**Question 1:**

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

**Answer:**

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e., are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

**Question 2:**

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

**Answer:**

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

**Question 3:**

Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

**Answer:**

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e.are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations,, are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

**Question 4:**

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

**Answer:**

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e.are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,, are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

**Question 5:**

Find the mean deviation about the mean for the data.

x_{i} | 5 | 10 | 15 | 20 | 25 |

f_{i} | 7 | 4 | 6 | 3 | 5 |

**Answer:**

x_{i} | f_{i} | f_{i} x_{i} | ||

5 | 7 | 35 | 9 | 63 |

10 | 4 | 40 | 4 | 16 |

15 | 6 | 90 | 1 | 6 |

20 | 3 | 60 | 6 | 18 |

25 | 5 | 125 | 11 | 55 |

25 | 350 | 158 |

**Question 6:**

Find the mean deviation about the mean for the data

x_{i} | 10 | 30 | 50 | 70 | 90 |

f_{i} | 4 | 24 | 28 | 16 | 8 |

**Answer:**

x_{i} | f_{i} | f_{i} x_{i} | ||

10 | 4 | 40 | 40 | 160 |

30 | 24 | 720 | 20 | 480 |

50 | 28 | 1400 | 0 | 0 |

70 | 16 | 1120 | 20 | 320 |

90 | 8 | 720 | 40 | 320 |

80 | 4000 | 1280 |

**Question 7:**

Find the mean deviation about the median for the data.

x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |

f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |

**Answer:**

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

x_{i} | f_{i} | c.f. |

5 | 8 | 8 |

7 | 6 | 14 |

9 | 2 | 16 |

10 | 2 | 18 |

12 | 2 | 20 |

15 | 6 | 26 |

Here, N = 26, which is even.

Median is the mean of 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e.are

|x_{i} – M| | 2 | 0 | 2 | 3 | 5 | 8 |

f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |

f_{i} |x_{i} – M| | 16 | 0 | 4 | 6 | 10 | 48 |

and

**Question 8:**

Find the mean deviation about the median for the data

x_{i} | 15 | 21 | 27 | 30 | 35 |

f_{i} | 3 | 5 | 6 | 7 | 8 |

**Answer:**

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

x_{i} | f_{i} | c.f. |

15 | 3 | 3 |

21 | 5 | 8 |

27 | 6 | 14 |

30 | 7 | 21 |

35 | 8 | 29 |

Here, N = 29, which is odd.

observation = 15^{th} observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e.are

|x_{i} – M| | 15 | 9 | 3 | 0 | 5 |

f_{i} | 3 | 5 | 6 | 7 | 8 |

f_{i} |x_{i} – M| | 45 | 45 | 18 | 0 | 40 |

∴

**Page No 361:**

**Question 9:**

Find the mean deviation about the mean for the data.

Income per day | Number of persons |

0-100 | 4 |

100-200 | 8 |

200-300 | 9 |

300-400 | 10 |

400-500 | 7 |

500-600 | 5 |

600-700 | 4 |

700-800 | 3 |

**Answer:**

The following table is formed.

Income per day | Number of persons f_{i} | Mid-point x_{i} | f_{i} x_{i} | ||

0 – 100 | 4 | 50 | 200 | 308 | 1232 |

100 – 200 | 8 | 150 | 1200 | 208 | 1664 |

200 – 300 | 9 | 250 | 2250 | 108 | 972 |

300 – 400 | 10 | 350 | 3500 | 8 | 80 |

400 – 500 | 7 | 450 | 3150 | 92 | 644 |

500 – 600 | 5 | 550 | 2750 | 192 | 960 |

600 – 700 | 4 | 650 | 2600 | 292 | 1168 |

700 – 800 | 3 | 750 | 2250 | 392 | 1176 |

50 | 17900 | 7896 |

Here,

**Question 10:**

Find the mean deviation about the mean for the data

Height in cms | Number of boys |

95-105 | 9 |

105-115 | 13 |

115-125 | 26 |

125-135 | 30 |

135-145 | 12 |

145-155 | 10 |

**Answer:**

The following table is formed.

Height in cms | Number of boys f_{i} | Mid-point x_{i} | f_{i} x_{i} | ||

95-105 | 9 | 100 | 900 | 25.3 | 227.7 |

105-115 | 13 | 110 | 1430 | 15.3 | 198.9 |

115-125 | 26 | 120 | 3120 | 5.3 | 137.8 |

125-135 | 30 | 130 | 3900 | 4.7 | 141 |

135-145 | 12 | 140 | 1680 | 14.7 | 176.4 |

145-155 | 10 | 150 | 1500 | 24.7 | 247 |

Here,

**Question 11:**

Find the mean deviation about median for the following data:

Marks | Number of girls |

0-10 | 6 |

10-20 | 8 |

20-30 | 14 |

30-40 | 16 |

40-50 | 4 |

50-60 | 2 |

**Answer:**

The following table is formed.

Marks | Number of girls f_{i} | Cumulative frequency (c.f.) | Mid-point x_{i} | |x_{i} â€“ Med.| | |f_{i} x_{i} â€“ Med.| |

0-10 | 6 | 6 | 5 | 22.85 | 137.1 |

10-20 | 8 | 14 | 15 | 12.85 | 102.8 |

20-30 | 14 | 28 | 25 | 2.85 | 39.9 |

30-40 | 16 | 44 | 35 | 7.15 | 114.4 |

40-50 | 4 | 48 | 45 | 17.15 | 68.6 |

50-60 | 2 | 50 | 55 | 27.15 | 54.3 |

50 | 517.1 |

The class interval containing theor 25^{th} item is 20 â€“ 30.

Therefore, 20 â€“ 30 is the median class.

It is known that,

Here,* l *= 20, C = 14, *f* = 14, *h* = 10, and N = 50

âˆ´ Median =

Thus, mean deviation about the median is given by,

**Question 12:**

Calculate the mean deviation about median age for the age distribution of 100 persons given below:

Age | Number |

16-20 | 5 |

21-25 | 6 |

26-30 | 12 |

31-35 | 14 |

36-40 | 26 |

41-45 | 12 |

46-50 | 16 |

51-55 | 9 |

**Answer:**

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

Age | Number f_{i} | Cumulative frequency (c.f.) | Mid-point x_{i} | |x_{i} – Med.| | |f_{i} x_{i} – Med.| |

15.5-20.5 | 5 | 5 | 18 | 20 | 100 |

20.5-25.5 | 6 | 11 | 23 | 15 | 90 |

25.5-30.5 | 12 | 23 | 28 | 10 | 120 |

30.5-35.5 | 14 | 37 | 33 | 5 | 70 |

35.5-40.5 | 26 | 63 | 38 | 0 | 0 |

40.5-45.5 | 12 | 75 | 43 | 5 | 60 |

45.5-50.5 | 16 | 91 | 48 | 10 | 160 |

50.5-55.5 | 9 | 100 | 53 | 15 | 135 |

100 | 735 |

The class interval containing theor 50^{th} item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here,* l *= 35.5, C = 37, *f *= 26, *h* = 5, and N = 100

Thus, mean deviation about the median is given by,

**Page No 371:**

**Question 1:**

Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

**Answer:**

6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

x_{i} | ||

6 | –3 | 9 |

7 | –2 | 4 |

10 | –1 | 1 |

12 | 3 | 9 |

13 | 4 | 16 |

4 | –5 | 25 |

8 | –1 | 1 |

12 | 3 | 9 |

74 |

**Question 2:**

Find the mean and variance for the first *n *natural numbers

**Answer:**

The mean of first *n* natural numbers is calculated as follows.

**Question 3:**

Find the mean and variance for the first 10 multiples of 3

**Answer:**

The first 10 multiples of 3 are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, *n *= 10

The following table is obtained.

x_{i} | ||

3 | –13.5 | 182.25 |

6 | –10.5 | 110.25 |

9 | –7.5 | 56.25 |

12 | –4.5 | 20.25 |

15 | –1.5 | 2.25 |

18 | 1.5 | 2.25 |

21 | 4.5 | 20.25 |

24 | 7.5 | 56.25 |

27 | 10.5 | 110.25 |

30 | 13.5 | 182.25 |

742.5 |

**Question 4:**

Find the mean and variance for the data

xi | 6 | 10 | 14 | 18 | 24 | 28 | 30 |

f i | 2 | 4 | 7 | 12 | 8 | 4 | 3 |

**Answer:**

The data is obtained in tabular form as follows.

x_{i} | f i | f_{i}x_{i} | |||

6 | 2 | 12 | –13 | 169 | 338 |

10 | 4 | 40 | –9 | 81 | 324 |

14 | 7 | 98 | –5 | 25 | 175 |

18 | 12 | 216 | –1 | 1 | 12 |

24 | 8 | 192 | 5 | 25 | 200 |

28 | 4 | 112 | 9 | 81 | 324 |

30 | 3 | 90 | 11 | 121 | 363 |

40 | 760 | 1736 |

Here, N = 40,

**Question 5:**

Find the mean and variance for the data

xi | 92 | 93 | 97 | 98 | 102 | 104 | 109 |

f i | 3 | 2 | 3 | 2 | 6 | 3 | 3 |

**Answer:**

The data is obtained in tabular form as follows.

x_{i} | f i | f_{i}x_{i} | |||

92 | 3 | 276 | –8 | 64 | 192 |

93 | 2 | 186 | –7 | 49 | 98 |

97 | 3 | 291 | –3 | 9 | 27 |

98 | 2 | 196 | –2 | 4 | 8 |

102 | 6 | 612 | 2 | 4 | 24 |

104 | 3 | 312 | 4 | 16 | 48 |

109 | 3 | 327 | 9 | 81 | 243 |

22 | 2200 | 640 |

Here, N = 22,

**Question 6:**

Find the mean and standard deviation using short-cut method.

x_{i} | 60 | 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 |

f_{i} | 2 | 1 | 12 | 29 | 25 | 12 | 10 | 4 | 5 |

**Answer:**

The data is obtained in tabular form as follows.

x_{i} | f_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

60 | 2 | –4 | 16 | –8 | 32 |

61 | 1 | –3 | 9 | –3 | 9 |

62 | 12 | –2 | 4 | –24 | 48 |

63 | 29 | –1 | 1 | –29 | 29 |

64 | 25 | 0 | 0 | 0 | 0 |

65 | 12 | 1 | 1 | 12 | 12 |

66 | 10 | 2 | 4 | 20 | 40 |

67 | 4 | 3 | 9 | 12 | 36 |

68 | 5 | 4 | 16 | 20 | 80 |

100 | 220 | 0 | 286 |

Mean,

**Question 7:**

Find the mean and variance for the following frequency distribution.

Classes | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |

Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |

**Answer:**

Class | Frequency f_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

0-30 | 2 | 15 | –3 | 9 | –6 | 18 |

30-60 | 3 | 45 | –2 | 4 | –6 | 12 |

60-90 | 5 | 75 | –1 | 1 | –5 | 5 |

90-120 | 10 | 105 | 0 | 0 | 0 | 0 |

120-150 | 3 | 135 | 1 | 1 | 3 | 3 |

150-180 | 5 | 165 | 2 | 4 | 10 | 20 |

180-210 | 2 | 195 | 3 | 9 | 6 | 18 |

30 | 2 | 76 |

Mean,

**Page No 372:**

**Question 8:**

Find the mean and variance for the following frequency distribution.

Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

Frequencies | 5 | 8 | 15 | 16 | 6 |

**Answer:**

Class | Frequencyf_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

0-10 | 5 | 5 | –2 | 4 | –10 | 20 |

10-20 | 8 | 15 | –1 | 1 | –8 | 8 |

20-30 | 15 | 25 | 0 | 0 | 0 | 0 |

30-40 | 16 | 35 | 1 | 1 | 16 | 16 |

40-50 | 6 | 45 | 2 | 4 | 12 | 24 |

50 | 10 | 68 |

Mean,

**Question 9:**

Find the mean, variance and standard deviation using short-cut method

Heightin cms | No. of children |

70-75 | 3 |

75-80 | 4 |

80-85 | 7 |

85-90 | 7 |

90-95 | 15 |

95-100 | 9 |

100-105 | 6 |

105-110 | 6 |

110-115 | 3 |

**Answer:**

Class Interval | Frequency f_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

70-75 | 3 | 72.5 | –4 | 16 | –12 | 48 |

75-80 | 4 | 77.5 | –3 | 9 | –12 | 36 |

80-85 | 7 | 82.5 | –2 | 4 | –14 | 28 |

85-90 | 7 | 87.5 | –1 | 1 | –7 | 7 |

90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |

95-100 | 9 | 97.5 | 1 | 1 | 9 | 9 |

100-105 | 6 | 102.5 | 2 | 4 | 12 | 24 |

105-110 | 6 | 107.5 | 3 | 9 | 18 | 54 |

110-115 | 3 | 112.5 | 4 | 16 | 12 | 48 |

60 | 6 | 254 |

Mean,

**Question 10:**

The diameters of circles (in mm) drawn in a design are given below:

Diameters | No. of children |

33-36 | 15 |

37-40 | 17 |

41-44 | 21 |

45-48 | 22 |

49-52 | 25 |

**Answer:**

Class Interval | Frequency f_{i} | Mid-point x_{i} | f_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

32.5-36.5 | 15 | 34.5 | –2 | 4 | –30 | 60 |

36.5-40.5 | 17 | 38.5 | –1 | 1 | –17 | 17 |

40.5-44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |

44.5-48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |

48.5-52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |

100 | 25 | 199 |

Here, N = 100, *h* = 4

Let the assumed mean, A, be 42.5.

Mean,

**Page No 375:**

**Question 1:**

From the data given below state which group is more variable, A or B?

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |

Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |

**Answer:**

Firstly, the standard deviation of group A is calculated as follows.

Marks | Group A f_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

10-20 | 9 | 15 | –3 | 9 | –27 | 81 |

20-30 | 17 | 25 | –2 | 4 | –34 | 68 |

30-40 | 32 | 35 | –1 | 1 | –32 | 32 |

40-50 | 33 | 45 | 0 | 0 | 0 | 0 |

50-60 | 40 | 55 | 1 | 1 | 40 | 40 |

60-70 | 10 | 65 | 2 | 4 | 20 | 40 |

70-80 | 9 | 75 | 3 | 9 | 27 | 81 |

150 | –6 | 342 |

Here, *h *= 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks | Group Bf_{i} | Mid-pointx_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

10-20 | 10 | 15 | –3 | 9 | –30 | 90 |

20-30 | 20 | 25 | –2 | 4 | –40 | 80 |

30-40 | 30 | 35 | –1 | 1 | –30 | 30 |

40-50 | 25 | 45 | 0 | 0 | 0 | 0 |

50-60 | 43 | 55 | 1 | 1 | 43 | 43 |

60-70 | 15 | 65 | 2 | 4 | 30 | 60 |

70-80 | 7 | 75 | 3 | 9 | 21 | 63 |

150 | –6 | 366 |

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

**Question 1:**

From the data given below state which group is more variable, A or B?

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |

Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |

**Answer:**

Firstly, the standard deviation of group A is calculated as follows.

Marks | Group Af_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

10-20 | 9 | 15 | â€“3 | 9 | â€“27 | 81 |

20-30 | 17 | 25 | â€“2 | 4 | â€“34 | 68 |

30-40 | 32 | 35 | â€“1 | 1 | â€“32 | 32 |

40-50 | 33 | 45 | 0 | 0 | 0 | 0 |

50-60 | 40 | 55 | 1 | 1 | 40 | 40 |

60-70 | 10 | 65 | 2 | 4 | 20 | 40 |

70-80 | 9 | 75 | 3 | 9 | 27 | 81 |

Î£ | 150 | â€“6 | 342 |

Here, *h *= 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks | Group Bf_{i} | Mid-point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |

10-20 | 10 | 15 | â€“3 | 9 | â€“30 | 90 |

20-30 | 20 | 25 | â€“2 | 4 | â€“40 | 80 |

30-40 | 30 | 35 | â€“1 | 1 | â€“30 | 30 |

40-50 | 25 | 45 | 0 | 0 | 0 | 0 |

50-60 | 43 | 55 | 1 | 1 | 43 | 43 |

60-70 | 15 | 65 | 2 | 4 | 30 | 60 |

70-80 | 7 | 75 | 3 | 9 | 21 | 63 |

Î£ | 150 | â€“6 | 366 |

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

**Question 2:**

From the prices of shares X and Y below, find out which is more stable in value:

X | 35 | 54 | 52 | 53 | 56 | 58 | 52 | 50 | 51 | 49 |

Y | 108 | 107 | 105 | 105 | 106 | 107 | 104 | 103 | 104 | 101 |

**Answer:**

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

x_{i} | ||

35 | –16 | 256 |

54 | 3 | 9 |

52 | 1 | 1 |

53 | 2 | 4 |

56 | 5 | 25 |

58 | 7 | 49 |

52 | 1 | 1 |

50 | –1 | 1 |

51 | 0 | 0 |

49 | –2 | 4 |

350 |

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

y_{i} | ||

108 | 3 | 9 |

107 | 2 | 4 |

105 | 0 | 0 |

105 | 0 | 0 |

106 | 1 | 1 |

107 | 2 | 4 |

104 | –1 | 1 |

103 | –2 | 4 |

104 | –1 | 1 |

101 | –4 | 16 |

40 |

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

**Question 3:**

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A | Firm B | |

No. of wage earners | 586 | 648 |

Mean of monthly wages | Rs 5253 | Rs 5253 |

Variance of the distribution of wages | 100 | 121 |

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

**Answer:**

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ_{1}) =

Variance of the distribution of wages in firm = 121

∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

**Page No 376:**

**Question 4:**

The following is the record of goals scored by team A in a football session:

No. of goals scored | 0 | 1 | 2 | 3 | 4 |

No. of matches | 1 | 9 | 7 | 5 | 3 |

For the team B, mean number of goals scored per match was 2 with a standard

deviation 1.25 goals. Find which team may be considered more consistent?

**Answer:**

The mean and the standard deviation of goals scored by team A are calculated as follows.

No. of goals scored | No. of matches | f_{i}x_{i} | x_{i}^{2} | f_{i}x_{i}^{2} |

0 | 1 | 0 | 0 | 0 |

1 | 9 | 9 | 1 | 9 |

2 | 7 | 14 | 4 | 28 |

3 | 5 | 15 | 9 | 45 |

4 | 3 | 12 | 16 | 48 |

25 | 50 | 130 |

Thus, the mean of both the teams is same.

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

**Question 5:**

The sum and sum of squares corresponding to length *x *(in cm) and weight *y*

(in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

**Answer:**

Here, N = 50

∴ Mean,

Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

**Page No 380:**

**Question 1:**

The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

**Answer:**

Let the remaining two observations be *x* and* y*.

Therefore, the observations are 6, 7, 10, 12, 12, 13, *x*, *y*.

From (1), we obtain

*x*^{2} + *y*^{2} + 2*xy* = 144 …(3)

From (2) and (3), we obtain

2*xy *= 64 … (4)

Subtracting (4) from (2), we obtain

*x*^{2} + *y*^{2 }– 2*xy* = 80 – 64 = 16

⇒ *x* – *y* = ± 4 … (5)

Therefore, from (1) and (5), we obtain

*x* = 8 and *y* = 4, when *x* – *y* = 4

*x* = 4 and *y* = 8, when *x* – *y* = –4

Thus, the remaining observations are 4 and 8.

**Question 2:**

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.

**Answer:**

Let the remaining two observations be *x* and* y*.

The observations are 2, 4, 10, 12, 14, *x*, *y*.

From (1), we obtain

*x*^{2} + *y*^{2} + 2*xy* = 196 … (3)

From (2) and (3), we obtain

2*xy *= 196 – 100

⇒ 2*xy* = 96 … (4)

Subtracting (4) from (2), we obtain

*x*^{2} + *y*^{2 }– 2*xy* = 100 – 96

⇒ (*x* – *y*)^{2} = 4

⇒ *x* – *y* = ± 2 … (5)

Therefore, from (1) and (5), we obtain

*x* = 8 and *y* = 6 when *x* – *y* = 2

*x* = 6 and *y* = 8 when *x* – *y* = – 2

Thus, the remaining observations are 6 and 8.

**Question 3:**

The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

**Answer:**

Let the observations be *x*_{1}, *x*_{2}, *x*_{3}, *x*_{4}, *x*_{5}, and *x*_{6}.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are *y*_{i}, then

From (1) and (2), it can be observed that,

Substituting the values of *x*_{i} and in (2), we obtain

Therefore, variance of new observations =

Hence, the standard deviation of new observations is

**Question 4:**

Given that is the mean and σ^{2} is the variance of *n *observations *x*_{1}, *x*_{2} … *x*_{n}. Prove that the mean and variance of the observations *ax*_{1}, *ax*_{2}, *ax*_{3} …*ax*_{n }are and *a*^{2} σ^{2}, respectively (*a* ≠ 0).

**Answer:**

The given *n* observations are *x*_{1}, *x*_{2} … *x*_{n}.

Mean =

Variance = σ^{2}

If each observation is multiplied by *a* and the new observations are *y*_{i}, then

Therefore, mean of the observations, *ax*_{1}, *ax*_{2} … *ax*_{n}, is .

Substituting the values of *x*_{i}and in (1), we obtain

Thus, the variance of the observations, *ax*_{1}, *ax*_{2} … *ax*_{n}, is *a*^{2} σ^{2}.

**Question 5:**

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

**(i)** If wrong item is omitted.

**(ii)** If it is replaced by 12.

**Answer:**

**(i)** Number of observations (*n*) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean

Standard deviation, σ = 1n∑i=1nxi2 – 1n2∑i=1nxi2⇒2 = 1n∑i=1nxi2 – 1n∑i=1nxi2⇒2 = 1n∑i=1nxi2 – x¯2 as, 1n∑i=1nx = x ⇒2 = 120×Incorrect∑i=1nxi2 – 102⇒4 = 120×Incorrect∑i=1nxi2 – 100⇒120×Incorrect∑i=1nxi2 = 104⇒Incorrect∑i=1nxi2 = 2080Now, correct∑i=1nxi2 = Incorrect∑i=1nxi2 – 82⇒correct∑i=1nxi2 = 2080 – 64 = 2016∴ correct Standard Deviation = 1ncorrect∑i=1nxi2 – correct mean2⇒correct Standard Deviation = 119×2016 – 192192⇒correct Standard Deviation = 201619-192192⇒correct Standard Deviation = 144019 = 121019⇒correct Standard Deviation = 12 × 3.16219 = 1.997**(ii)** When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

**Question 6:**

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject | Mathematics | Physics | Chemistry |

Mean | 42 | 32 | 40.9 |

Standard deviation | 12 | 15 | 20 |

Which of the three subjects shows the highest variability in marks and which shows the lowest?

**Answer:**

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by .

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

**Question 7:**

The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

**Answer:**

Number of observations (*n*) = 100

Incorrect mean

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

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