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Questions › JNU › JNU SIS 2020 › Q31
0 marksStatistics › Random variables and standard distributions › The Poisson distributionModerateabout 1 min

JNU SIS 2020 Question 31: Random variables and standard distributions

The probability density function of a random variable is given as f(x)=λxe−λx!f(x) = \dfrac{\lambda^{x} e^{-\lambda}}{x!}f(x)=x!λxe−λ​ where x = 0, 1, 2, . . . . then the expected value of x2x^{2}x2 is :

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The probability density function of a random variable is given as f(x) = (λ^x e^(-λ))/x! where x = 0, 1, 2,.... then the expected value of x² is: (A) λ (B) λ² (C) λ(λ + 1) (D) 1 - λ²

Also written as: The probability density function of a random variable is given as f(x) = (lambda^x e^(-lambda))/x! where x = 0, 1, 2,.... then the expected value of x^2 is:

Random variables and standard distributions has come up 173 times in 7 exams since 2005.

What this question tests

This question is on The Poisson distribution, part of Random variables and standard distributions in Statistics.

Counts of rare events, where mean equals variance.

TestStop has 9 past-paper questions on this subtopic.

Practise Random variables and standard distributionsMore on The Poisson distribution

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Question from the JNU SIS 2020 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.