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GATE 2025 CE (CE1) – Question 14

Engineering Mathematics · Probability and Statistics: Probability, conditional probability, descriptive statistics · 1 mark · Multiple choice

$X$ is a random variable that can take any one of the values 0, 1, 7, 11, and 12. The probability mass function for $X$ is

$P(X = 0) = 0.4$; $P(X = 1) = 0.3$; $P(X = 7) = 0.1$; $P(X = 11) = 0.1$; $P(X = 12) = 0.1$

Then, the variance of $X$ is

  1. 20.81
  2. 28.40
  3. 31.70
  4. 10.89

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Correct answer: (A) 20.81

Explanation

The mean is $E[X] = 0.3 + 0.7 + 1.1 + 1.2 = 3.3$. Then $E[X^2] = 0.3(1) + 0.1(49) + 0.1(121) + 0.1(144) = 0.3 + 4.9 + 12.1 + 14.4 = 31.7$. The variance is $31.7 - 3.3^2 = 31.7 - 10.89 = 20.81$.