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. One student is selected in such a manner that each has the same chance of being 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 of X.

Given: The class of 15 students with their ages.

Form the given information we can draw a table:

X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |

f | 2 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |

P(X = 14)

P(X = 15)

P(X = 16)

P(X = 17)

P(X = 18)

P(X = 19)

P(X = 20)

P(X = 21)

Hence, the required probability distribution is,

X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |

P(X) |

Therefore E(X) is:

⇒ E(X) = 17.53

And E(X^{2}) is:

⇒ E(X^{2}) = 312.2

Then Variance, Var(X) = E(X^{2}) – (E(X))^{2}

= 312.2 – (17.53)^{2}

= 312.2 – 307.417 ≈ 4.78

And Standard deviation

⇒ Standard deviation ≈ 2.19

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