ProbabilityHard
Question
The probability distribution of a random variable X is given below:
| $$X$$ | 4 k | $$\frac{30}{7}k$$ | $$\frac{32}{7}k$$ | $$\frac{34}{7}k$$ | $$\frac{36}{7}k$$ | $$\frac{38}{7}k$$ | $$\frac{40}{7}k$$ | 6 k |
|---|---|---|---|---|---|---|---|---|
| $$P(X)$$ | $$\frac{2}{15}$$ | $$\frac{1}{15}$$ | $$\frac{2}{15}$$ | $$\frac{1}{5}$$ | $$\frac{1}{15}$$ | $$\frac{2}{15}$$ | $$\frac{1}{5}$$ | $$\frac{1}{15}$$ |
If $E(X) = \frac{263}{15}$, then $P(X < 20)$ is equal to :
Options
A.$\frac{3}{5}$
B.$\frac{8}{15}$
C.$\frac{11}{15}$
D.$\frac{14}{15}$
Solution
$E(X) = \sum X_{i}P\left( X_{i} \right) = \frac{526k}{15 \times 7} = \frac{263}{15} \Rightarrow k = \frac{7}{2}$
| X | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 |
|---|---|---|---|---|---|---|---|---|
| $$P(X)$$ | $$\frac{2}{15}$$ | $$\frac{1}{15}$$ | $$\frac{2}{15}$$ | $$\frac{1}{5}$$ | $$\frac{1}{15}$$ | $$\frac{2}{15}$$ | $$\frac{1}{5}$$ | $$\frac{1}{15}$$ |
$$P(X < 20) = \sum_{X = 14}^{19}\mspace{2mu} P(X) = \frac{11}{15}$$
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