Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Probability Distributions

Question:

Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).

Assertion (A): If a random variable X has the following probability distribution

$X$

1

2

3

4

$P(X)$

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{5}$

$\frac{3}{10}$

then the mean of the probability distribution is $\frac{37}{15}$.

Reason (R): Variance of the probability distribution = $∑p_i{x_i}^2-(∑p_ix_i)^2$ where x's represent the values of a random variable and ${p_i}'s$ represent the probabilities of ${x_i}'s$.

Select the correct answer from the options given below:

Options:

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Assertion (A) is true, but Reason (R) is false.

Assertion (A) is false, but Reason (R) is true.

Correct Answer:

Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Explanation:

The correct answer is Option (2) → Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Mean (μ) = $∑p_ix_i = p_1x_1 + p_2x_2 + p_3x_3 + p_4x_4$

$=\frac{1}{3}×1+\frac{1}{6}×2+\frac{1}{5}×3+\frac{3}{10}×4$

$=\frac{1}{3}+\frac{1}{3}+\frac{3}{5}+\frac{6}{5}$

$=\frac{5+5+9+18}{15}=\frac{37}{15}$

∴ Assertion is true.

Also, Reason is true.

Reason is not the correct explanation of Assertion.