Question

NCERT Solutions Class 11 Maths Chapter 15 Statistics is designed by the experts in the subject. NCERT Solutions are made intuitive with the help of pictograms, images, equations, diagrams, and graphs to make it easy for students to comprehend and revise every concept.

In NCERT Solutions Class 11 given here, we have discussed all the topics in in-depth detail. Topics discussed in the chapter are:

• Statistics – statistics is the part of mathematics that particularly deals with the collection, organization, analysis, interpretation, and presentation of data. While applying statistics to any scientific, industrial, or social problem, it is the convention to start with a statistical population or a statistical model. Statistics is mainly categorized into two types:
  • Descriptive statistics – in descriptive statistics the given data is summarised through given observation. It is done with the help of man or standard deviation which help in the summarization of the sample of the population using some parameters.
  • Inferential statistics – in inferential statistics the meaning of descriptive statistics is determined. It helps us conclude the given data of any random variation.

• Measures of dispersion – a measure of the dispersion of data describes the range in which the data varies. It defines the disparity between one kind of data from another. Dispersion is the range in which the average value of the distribution is calculated. It helps us determine the extent to which individual items vary from one another and their mean value.

• Range – it is the measurement of the range of highest and lowest value of any given data. If A max and A min is the two extreme value given data then the Range = A_max – A_min.

• Mean deviation – mean deviation is the arithmetic deviation of a range of data from its mean value.

• Variance – the variance is the measure of deviation meaning how far a set of numbers ranges from its average value.

• Standard deviation of ungrouped/grouped data – standard deviation is the measure of the deviation of data. It helps us determine how much the given data ranges around the mean or the average of the given data.

• Analysis of frequency distributions with equal means but different variances.

Our NCERT Solutions Class 11 Maths is a great way to learn and revise for tough competitive exams like JEE Mains, JEE Advance, etc.

Answer 1

NCERT Solutions Class 11 Maths Chapter 15 Statistics

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 \(\bar x\), i.e., \(x_i-\bar x,\) are -6, -3, -2, -1, 0, 2, 3, 7

The absolute values of the deviations, i.e \(|x_i-\bar x|,\) are 6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

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 \(\bar x,\) i.e, x_i - \(\bar x\), are

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

The absolute values of the deviations, i.e. |x_i - \(\bar x\)|, are

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

The required mean deviation about the mean is

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.x_i  - M 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, |x_i - M|, 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

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, |x_i - M| 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

5. Find the mean deviation about the mean for the data.

Answer :

6. Find the mean deviation about the mean for the data

Answer :

7. Find the mean deviation about the median for the data.

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.

Here, N = 26, which is even.

Median is the mean of 13th and 14th 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. |x_i - M|, are

Answer 2

8. Find the mean deviation about the median for the data.

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.

Here, N = 29, which is odd

\(\therefore\) Median = \((\frac{29+1}2)^{th}\) 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. |x_i - M|, are

9. Find the mean deviation about the mean for the data.

Answer :

The following table is formed.

Here,

10. Find the mean deviation about the mean for the data

Answer :

The following table is formed.

Here,

11. Find the mean deviation about median for the following data:

Answer :

The following table is formed.

The class interval containing the \((\frac N2)^{th}\) or 25^th item is 20 - 30.

Therefore,  20 - 30 is the median class.

It is known that,

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

∴ Median = 20 + \(\frac{25-14}{14}\times10\) = 20 + \(\frac{110}{14}\) = 20 + 7.85

= 27.85

Thus, mean deviation about the median is given by,

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

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.

The class interval containing the \(\frac N2^{th}\) or 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

\(\therefore\) Median

 = 35.5 + \(\frac{50-37}{26}\times5\) = 35.5 + \(\frac{13\times5}{26}\) = 35.5 + 2.5 = 38

Thus, mean deviation about the median is given by,

13. 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.

Variance(\(\sigma^2\)) = 

14. Find the mean and variance for the first n natural numbers

Answer :

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

Answer 3

15. 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

Mean

The following table is obtained.

Variance \((\sigma^2)\)

16. Find the mean and variance for the data.

Answer :

The data is obtained in tabular form as follows.

Here,

17. Find the mean and variance for the data

Answer :

The data is obtained in tabular form as follows.

Here,

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

Answer :

The data is obtained in tabular form as follows.

Mean,

Answer 4

19. Find the mean and variance for the following frequency distribution. 

Answer :

Mean,

20. Find the mean and variance for the following frequency distribution.

Answer :

Mean,

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

Answer :

Mean,

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

Answer :

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Mean,

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

Answer :

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

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

Mean,

The standard deviation of group B is calculated as follows.

Mean,

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.

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

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.

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.

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.

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

(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 \((\sigma_1^2)\) = 100

∴ Standard deviation of the distribution of wages in firm

A \((\sigma_1)\) = \(\sqrt{100}\) = 10

Variance of the distribution of wages in firm B \((\sigma_2^2)\) = 121

∴ Standard deviation of the distribution of wages in firm B(\(\sigma_2^2\)) = \(\sqrt{121}\) = 11

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.

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

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.

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.

Answer 5

27. 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 :

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

28. 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

x² + y² + 2xy = 144 …(3)

From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy = 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.

Answer 6

29. 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

x² + y² + 2xy = 196 … (3)

From (2) and (3), we obtain

2xy = 196 – 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

x² + y² – 2xy

= 100 – 96

⇒ (x – y)² = 4

⇒ x – y = ± 2 … (5)

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

 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.

30. 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 x₁, x₂, x₃, x₄, x₅, and x₆.

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

If each observation is multiplied by 3 and the resulting observations are yi, then

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

Substituting the values of x_i and \(\bar x\)i n (2), we obtain

Therefore, variance of new observations = \((\frac16\times864)=144\)

Hence, the standard deviation of new observations is \(\sqrt{144}=12\)

31. Given that is the mean and σ² is the variance of n observations x₁, x₂ … x_n.

Prove that the mean and variance of the observations ax₁, ax₂, ax₃ …ax_n are \(a\bar x\) and a² σ² , respectively (a ≠ 0).

Answer :

The given n observations are x₁, x₂ … x_n.

Mean  = \(\bar x\)

Variance = \(\sigma^2\)

If each observation is multiplied by a and the new observations are y_i, then

Therefore, mean of the observations, ax₁, ax₂ … ax_n, is  \(a\bar x\).

Substituting the values of x_i and \(\bar x\) in (1), we obtain

Thus, the variance of the observations, ax₁, ax₂ … ax_n, is a² \(\sigma^2\).

32. 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 = \(\frac{correct\,sum}{19}\) = \(\frac{192}{19}=10.1\) 

Standard deviation

(ii) When 8 is replaced by 12, Incorrect sum of observations = 200

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

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

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 \(\frac{standard\,deviation}{Mean}\times100\) 

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.

34. 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 \((\bar x)\)= 20

Incorrect standard deviation (\(\sigma\)) = 3

∴ Incorrect sum of observations = 2000

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