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Questions › JNU › JNU SIS 2011 › Q20
4 marks, −1Statistics › Random variables and standard distributions › Random variables, PMFs, PDFs and CDFsModerateabout 4 min

JNU SIS 2011 Question 20: Random variables and standard distributions

The probability density function of a random variable yyy is αy(1−y)2\alpha y(1-y)^2αy(1−y)2 for yyy lying between 0 and 1 (α\alphaα is a constant). Find the probability of yyy lying between 0⋅250\cdot 250⋅25 and 0⋅750\cdot 750⋅75.

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The probability density function of a random variable y is α y(1-y)² for y lying between 0 and 1 (α is a constant). Find the probability of y lying between 0· 25 and 0· 75. (A) 0· 2234 (B) 0· 0000 (C) 0· 9876 (D) 0· 6875

Also written as: The probability density function of a random variable y is alpha y(1-y)^2 for y lying between 0 and 1 (alpha is a constant). Find the probability of y lying between 0* 25 and 0* 75.

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

What this question tests

This question is on Random variables, PMFs, PDFs and CDFs, part of Random variables and standard distributions in Statistics.

Describing a random variable by its mass, density or distribution function.

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Question from the JNU SIS 2011 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.