The probability distribution of a random variable X is given below: X, 4k, 30k7, 32k7, 34k…

Question

The probability distribution of a random variable $X$ is given below:
 $X$, $4k$, $\dfrac{30k}{7}$, $\dfrac{32k}{7}$, $\dfrac{34k}{7}$, $\dfrac{36k}{7}$, $\dfrac{38k}{7}$, $\dfrac{40k}{7}$, $6k$ |
---|---|---|---|---|---|---|---|---|
 $P(X)$, $\dfrac{2}{15}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$, $\dfrac{2}{15}$, $\dfrac{1}{5}$, $\dfrac{1}{15}$ |

If $E(X) = \dfrac{263}{15}$, then $P(X<20)$ is equal to:

Accepted Answer

$$E(X)=\frac{k}{15}\left[8+\frac{30}{7}+\frac{64}{7}+\frac{102}{7}+\frac{36}{7}+\frac{76}{7}+\frac{120}{7}+6\right]=\frac{k}{15}\left[14+\frac{428}{7}\right]=\frac{526k}{105}$$

Setting $\dfrac{526k}{105}=\dfrac{263}{15}$ gives $k=\dfrac{7}{2}=3.5$.

Substituting $k=3.5$ into each value of $X$:

$$X_1=14,\ X_2=15,\ X_3=16,\ X_4=17,\ X_5=18,\ X_6=19,\ X_7=20,\ X_8=21$$

So $X<20$ corresponds to $X_1,\dots,X_6$ (i.e. values 14 through 19):

$$P(X<20) = \frac{2}{15}+\frac{1}{15}+\frac{2}{15}+\frac{1}{5}+\frac{1}{15}+\frac{2}{15} = \frac{11}{15}$$

Solutions · 1