NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 15 – Statistics Miscellaneous Exercise

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

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 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

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 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 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 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 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  and a² σ², respectively (a ≠ 0).

Answer:

The given n observations are x₁, x₂ … x_n.

Mean = 

Variance = σ²

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 .

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

Thus, the variance of the observations, ax₁, ax₂ … ax_n, is a² σ².

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 = 2080

Now, 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