ISRO EC 2023 Q.17: solution and answer

ISRO 2023 · ECQ.17 · 1 mark · MCQ

Random variable x is with uniform distribution in the span [-4, 1], mean and variance of x is

(A)
-1.5, 25/12
(B)
-1.5, 5/12
(C)
-2.5, 5/12
(D)
-2.5, 1/4

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Answer and solution

ANSWER

The mean is $- 1.5$ and the variance is $\frac{25}{12}$ , which is option (A).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

For a uniform distribution on $[a, b]$ , the density is $f (x) = \frac{1}{b - a}$ on the interval, and the mean is the midpoint: $μ = \frac{a + b}{2}$ .
Here $a = - 4$ and $b = 1$ , so $μ = \frac{- 4 + 1}{2} = - 1.5$ .
The variance of a uniform distribution is $σ^{2} = \frac{( b - a ) ^{2}}{12}$ . This follows from $E [x^{2}] = \frac{b ^{3} - a ^{3}}{3 ( b - a )} = \frac{a ^{2} + ab + b ^{2}}{3}$ , so $σ^{2} = \frac{a ^{2} + ab + b ^{2}}{3} - \frac{( a + b ) ^{2}}{4} = \frac{( b - a ) ^{2}}{12}$ .
The width is $b - a = 1 - (- 4) = 5$ , so $σ^{2} = \frac{5 ^{2}}{12} = \frac{25}{12}$ .
Mean $- 1.5$ and variance $\frac{25}{12}$ match option (A). Options (C) and (D) have the wrong mean, $- 2.5$ . Option (B) has the right mean but uses $\frac{5}{12}$ , which is the width not squared.

KEY IDEA

For a uniform distribution on $[a, b]$ , the mean is the midpoint $\frac{a + b}{2}$ and the variance is $\frac{( b - a ) ^{2}}{12}$ .

WATCH OUT FOR

Using the width $b - a$ instead of its square in the variance, which gives $\frac{5}{12}$ . Another slip is taking the mean as $\frac{b - a}{2}$ or adding the endpoints wrongly, which gives $- 2.5$ .

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO Probability & statistics questions, with notes

ISRO EC 2023 Q.17: solution and answer

Answer and solution

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