The probability distribution of a random variable X is given below :15 2/151 If E(X) then P(X < 20) is eq [2026]

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Solution

Q.
The probability distribution of a random variable X is given below :

X 4k $\frac{30}{7}$ k $\frac{32}{7}$ k $\frac{34}{7}$ k $\frac{36}{7}$ k $\frac{38}{7}$ k $\frac{40}{7}$ k 6k

P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$

If E(X) $= \frac{263}{15}$ then P(X < 20) is equal to : [2026]

1 $\frac{8}{15}$

2 $\frac{11}{15}$

3 $\frac{3}{5}$

4 $\frac{14}{15}$

Ans.
(2)

$E (X) = \sum x_{i} P (X_{i}) = \frac{526 k}{15 \times 7} = \frac{263}{15} \Rightarrow k = \frac{7}{2}$

X 14 15 16 17 18 19 20 21

P(X) $\frac{2}{15}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$ $\frac{2}{15}$ $\frac{1}{5}$ $\frac{1}{15}$

$P (X < 20) = \sum_{x = 14}^{19} P (X) = \frac{11}{15}$

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