NCERT Solutions Class 11 Mathematics Chapter 15 Statistics

Exercise 15.1

Question. 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, 

Question. 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,

Question. 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

Question. 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

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

Answer :

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

Answer :

Question. 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 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, |x_i – M|, are  

Question. 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. 
∴ Median = [(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  

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

Answer :
The following table is formed. 

Income per day	Number of person fi	Mid – point xi	fi xi
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

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

Answer :
The following table is formed. 

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

Answer :
The following table is formed. 

Question. 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 (N/2)^th or  50^th item is  35.5 - 40.5 . 
Therefore, 35.5 - 40.5 is the median class.
It is known that, 

Exercise 15.2

Question. 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

The following table is obtained. 

Question. Find the mean and variance for the first n natural numbers 
Answer :
The mean of first n natural numbers is calculated as follows. 
Mean = (Sum of all observations)/(Number of observations) 

Question. 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. 

Question. Find the mean and variance for the data

Answer :
The data is obtained in tabular form as follows. 

Question. Find the mean and variance for the data

Answer :
The data is obtained in tabular form as follows.

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

Answer :
The data is obtained in tabular form as follows.

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

Answer :

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

Answer :

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

Answer :

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

Answer :

Exercise 15.3

Question. 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. 

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. 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 

Question. 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 (σ₁²) = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ₁) = √100 = 10
Variance of the distribution of wages in firm B (σ₂²) = 121
∴ Standard deviation of the distribution of wages in firm B (σ₂²) = √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.

Question. 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. 

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. 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.

Miscellaneous Solutions

Question. 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.

Question. 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 observations be x₁, x₂, x₃, x₄, x₅, and x₆.
It is given that mean is 8 and standard deviation is 4.

Question. Given that x 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 ax and a² σ², respectively (a ≠ 0).
Answer :

Question. 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 = (Correct sum )/19 = 192/19 = 10.1 

(ii) When 8 is replaced by 12, 
Incorrect sum of observations = 200
Correct sum of observations = 200 - 8 + 12 = 204 
∴ Correct mean = (Correct sum)/20 = 204/20 = 10.2 

Question. 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 (Standard deviation )/Mean  × 100 

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. 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 x = 20 
Incorrect standard deviation (σ) = 3