Exercise Miscellaneous - Statistics NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: 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.
Ans: 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.

Q2: 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.
Ans: Let the remaining two observation be x and y
The observation 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 observation are 6 and 8

Q3: 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.
Ans:


We know that,



Q4: 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_x and a² σ², respectively (a ≠ 0).
Ans: The given n observations are x₁, x₂ … x_n.
Mean =
Variance = σ²

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

Therefore, mean of the observations, ax₁, ax₂ … ax_n, is
Substituting the values of x_iand in (1), we obtain

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

Q5: 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.
Ans: (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


= √4.09
= 2.02
(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



Q6: 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.
Ans: Number of observations (n) = 100
Incorrect mean  = 20Incorrect standard deviation (σ) = 3

Incorrect sum of observations  = 2000
Correct sum of observations  = 2000 - 21 - 21 - 18 = 2000 - 60 = 1940
∴ Correct mean = (correct sum)/(100 - 3) = 1940/97 = 20