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

$$P(X < 20) = \sum_{X = 14}^{19}\mspace{2mu} P(X) = \frac{11}{15}$$

Create a free account to view solution

View Solution Free
Topic: Probability·Practice all Probability questions

More Probability Questions