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A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
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Solution
Let X denote the age of the selected student.
∴ Possible values of X are 14, 15, 16, 17, 18, 19, 20, 21.
There are 2 students of age 14, 1 student of age 15, 2 students of age 16, 3 students of age 17, 1 student of age 18, 2 students of age 19, 3 students of age 20, 1 student of age 21.
∴ P(X = 14) = `(2)/(15), "P"("X" = 15) = (1)/(15),"P"("X" = 16) = (2)/(15),"P"("X" = 17) = (3)/(15),"P"("X" = 18) = (1)/(15),"P"("X" = 19) = (2)/(15),"P"("X" = 20) = (3)/(15),"P"("X" = 21) = (1)/(15),`
∴ Mean of X
= E(X)
= \[\sum\limits_{i=1}^{8} x_i.\text{P}(x_i)\]
=`14 xx (2)/(15) + 15 xx (1)/(15) + 16 xx (2)/(15) + 17 xx (3)/(15) + 18 xx (1)/(15) + 19 xx (2)/(15) + 20 xx (3)/(15) + 21 xx (1)/(15)`
= `(1)/(15)(28 + 15 + 32 + 51 + 18 + 38 + 60 + 21)`
= `(263)/(15)`
= 17.53
E(X^{2}) = \[\sum\limits_{i=1}^{8} x_i^2\text{P}(x_i)\]
= `(1)/(15)(14^2 xx 2 + 15^2 + 16^2 xx 2 + 17^2 xx 3 + 18^2 + 19^2 xx 2 + 20^2 xx +21^2)`
= `(4683)/(15)`
= `(1561)/(5)`
Variance of X = Var(X)
= E(X^{2}) – [E(X)]^{2}
= `(1561)/(5)  (263/15)^2`
= `(70245  69169)/(225)`
= `(1076)/(225)`
= 4.8
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