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Question

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19, and 20yr. One student is selected in such a manner that each has the same chance of being of chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and Standard Deviation (SD) of X.

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Solution

Here, total students=15

The ages of students in ascending order are 14, 14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 20, 20, 20, 21

Now, P(X=14)=215,P(X=15)=115,P(X=16)=215,P(X=17)=315

P(X=18)=115,P(X=19)=215,P(X=20)=315,P(X=21)=115

Therefore, the probability distribution of random variable X is as follows:

X 14 15 16 17 18 19 20 21Number of students21231231P(X)215 115 215315115215315115

The third row gives the probability distribution of X.

Mean X=XP(X)=14×2+15×1+16×2+17×3+18×1+19×2+20×3+21×115

=28+15+32+51+18+38+60+2115=26315=17.53

Variance X=X2P(X)Mean2


[(14)2×2+(15)2×1+(16)2×2+(17)2×3+(18)2×1+(19)2×2+(20)2×3+(21)2×1]15(26315)2=392+225+512+867+324+722+1200+44115(26315)2=468315(26315)2=312.2307.4=4.8

SD of X=Variance=4.8=2.19


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