RBSE Solutions for Class 11 Maths Chapter 15 Statistics Miscellaneous Exercise

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 15 Statistics Miscellaneous Exercise Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 15 Statistics Miscellaneous Exercise

Question 1.
Mean and variance of 8 observations are 9 and 9.25 respectively. 1f out of these, 6 observations are 6, 7, 10, 12, 12 and 13. Find remaining two observations.
Answer:
Let remaining two observations are x₇ and x₈
We have, mean of 8 observations = x̄ = 9
and variance σ² = 925

Thus, 6² 7² + 10² + 12² + 12² + 13² + x²₇ + x²₈ = 722
= x²₇ + x²₈ = 722 - (6² + 10² + 12² + 12² + 13²)
= x²₇ + x²₈ = 722 - (36 + 49 + 100 + 144 + 144 + 169)
= x²₇ + x²₈ = 722 - 642
= x²₇ + x²₈ = 80 .................. (ii)
Squaring equation (1) and then subtracting equation

(ii) (x₇ + x₈)² = 12²

Now, subtracting equation (iii) from equation (ii)
x²₇ + x²₈ - 2x₇x₈ = 80 - 64
⇒ (x₇ - x₈)² = 16
⇒ x₇ - x₈ = ±4 ........ (iv)
Now, adding equation (j) and (iv),
x₇ + x₈ = 12
x₇ - x₈ = ±4
On adding 2x₇ = 16 and 2x₇ = 8
∴ x₇ = 8 and x₇ = 4
Then, x₈ = 4 and x₈ = 8
Thus, remaining two observations are 8 and 4 or 4 and 8.

Question 2.
Mean and variance of seven observations are 8 and 16 respectively. 1f out of these five observations are 2, 4, 10, 12, 14, then find remaining two observations.
Answer:
Let remaining two observations are x₆ and x₇.
We have, mean of 7 observations x̄ = 8
and Variance σ² = 16

and 7 × 8 = 42 + x₆ + x₇ = 56 = 42 + x₆ + x₇
⇒ x₆ + x₇ = 56 - 42
⇒ x₆ + x₇ = 14 .................... (iii)

Thus, 2² + 4² + 10² + 12² + 14² + x²₆ + x²₇ = 560
⇒ x²₆ + x²₇ = 560 - (2² + 4² + 10² + 12² + 14²)
⇒ x²₆ + x²₇ = 560 - (4 + 16 + 100 + 144 + 196)
⇒ x²₆ + x²₇ = 560 - 460
⇒ x₆² + x²₇ = 100 ................. (iii)
Squaring equation (ii) and then subtracting equation

(iii)

Again, subtracting equation (iv) from equation (iii)
⇒ x²₆ + x²₇ - 2x₆x₇ = 100 - 96
⇒ x²₆ + x²₇ - 2x₆x₇ = 4
⇒ (x₆ - x₇)² = 4
x₆ - x₇ = ±2 ................ (v)
Now, adding equation (ii) and (v),

⇒ x₆ = 8 and x₆ = 6
Then, x₇ = 6 and x₇ = 8
Thus, remaining two observations are 8 and 6 or 6 and 8.

Question 3.
Mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, then find new mean and standard deviation of resulting observations.
Answer:
We have, no. of observations n = 6
Mean of observation x̄ = 8
Then, standard deviation of observation σ = 4

Thus, resulting mean = 24
and resulting standard deviation = 12

Question 4.
Given that x̄ is the mean an σ² is the variance of n observations x₁, x₂, x₃, ....... x_n, then prove that mean and variance of observations ax₁, ax₂, ax₃, ................... ax_n are ax̄ and a²σ² (a ≠ 0) respectively.
Answer:

Question 5.
Mean and standard deviation of 20 observations are found to be 10 and 2 respectively. After checking it is found that observation 8 was incorrect. Find mean and standard deviation of each of the following. If
(i) If wrong item is omitted.
(ii) If It Is replaced by 12.
Answer:
We have, no. of observations n = 20, mean x̄ = 10
and standard deviation σ = 2

Question 6.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects Mathematics, Physics and Chemistry:

Which of the three subject shows the highest variability in marks and which shows the lowest?
Answer:
We have:

Clearly, there are more variation in Chemistry and less in Mathematics.

Question 7.
Mean and standard deviations 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:
We have, mean of 100 observations, x̄ = 20
∴ Sum of 100 observations = 200 × 100 = 2000
Out of these, three observations 21, 21 are 18 were incorrect, then
Sum of remaining 97 observations