JEE Main 2021MathematicsProbabilityEasyNumerical

JEE Main 2021 — Probability Question with Solution

JEE Main 2021 (01 Sep Shift 2)

Question

Let X be a random variable with distribution.

x -2 -1 3 4 6
P(X=x) 15 a 13 15 b

If the mean of X is 2.3 and variance of X is σ2, then 100σ2 is equal to :

Enter your answer

Show full solutionCorrect answer: 781
Correct answer
781

Step-by-step explanation

Given mean is μ=∑pixi=2.3

⇒-25-a+1+45+6b=2.3

⇒6b-a=0.9   ...(1)

Also ∑pi=1

⇒15+a+13+15+b=1

⇒a+b=415   ...(2)

Solving, we get a=110; b=16

Variance is σ2=∑pixi2-μ2

⇒σ2=15(4)+110(1)+13(9)+15(16)+16(36)-5.29

⇒σ2=7.81

⇒100σ2=781

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About this question

This is a previous-year question from JEE Main 2021, covering the Probability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.