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Question 75 of 88
Q.

The probability distribution of a random variable X, the number of defects per 10 meters of a fabric is given by

x01234
P(X = x)0.450.350.150.030.02

Find the variance of X.

OR

For the following probability density function (p.d.f.) of X, find: (i) P(X < 1), (ii) P(|X| < 1)

if f(x)=x218f(x) = \dfrac{x^2}{18}, −3<x<3-3 < x < 3

=0= 0, otherwise

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2019Subjective· 3mImportance★★★★★
85% · 75/88 Questions
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Part 1: Var(X)=E(X2)−[E(X)]2\mathrm{Var}(X)=E(X^2)-[E(X)]^2 from the table. Part 2 (OR): integrate the pdf over the given ranges.

Part 1 — Variance from the given table:

x01234
P(X=x)0.450.350.150.030.02

E(X)=∑xP(x)=0(.45)+1(.35)+2(.15)+3(.03)+4(.02)=0+.35+.30+.09+.08=0.82E(X) = \sum xP(x) = 0(.45)+1(.35)+2(.15)+3(.03)+4(.02) = 0+.35+.30+.09+.08 = 0.82

E(X2)=∑x2P(x)=0(.45)+1(.35)+4(.15)+9(.03)+16(.02)=0+.35+.60+.27+.32=1.54E(X^2) = \sum x^2P(x) = 0(.45)+1(.35)+4(.15)+9(.03)+16(.02) = 0+.35+.60+.27+.32 = 1.54

Var(X)=E(X2)−[E(X)]2=1.54−(0.82)2=1.54−0.6724=0.8676\mathrm{Var}(X) = E(X^2)-[E(X)]^2 = 1.54 - (0.82)^2 = 1.54-0.6724 = 0.8676


Part 2 (OR) — pdf f(x)=x218f(x)=\dfrac{x^2}{18}, −3<x<3-3<x<3:

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