Question ID: #945
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:
- (1) $\frac{3}{5}$
- (2) $\frac{8}{15}$
- (3) $\frac{11}{15}$
- (4) $\frac{14}{15}$
Solution:
The expected value $E(X)$ is given by $\sum X_i P(X_i)$.
$$E(X) = 4k\left(\frac{2}{15}\right) + \frac{30k}{7}\left(\frac{1}{15}\right) + \frac{32k}{7}\left(\frac{2}{15}\right) + \frac{34k}{7}\left(\frac{3}{15}\right) + \frac{36k}{7}\left(\frac{1}{15}\right) + \frac{38k}{7}\left(\frac{2}{15}\right) + \frac{40k}{7}\left(\frac{3}{15}\right) + 6k\left(\frac{1}{15}\right)$$
We simplify the expression. Factor out $\frac{k}{105}$ (where $105 = 15 \times 7$):
$$E(X) = \frac{k}{105} [ 4(7)(2) + 30(1) + 32(2) + 34(3) + 36(1) + 38(2) + 40(3) + 6(7)(1) ]$$
$$E(X) = \frac{k}{105} [ 56 + 30 + 64 + 102 + 36 + 76 + 120 + 42 ]$$
$$E(X) = \frac{k}{105} [ 526 ]$$
Given $E(X) = \frac{263}{15}$:
$$\frac{526k}{105} = \frac{263}{15}$$
$$\frac{526k}{7 \times 15} = \frac{263}{15}$$
$$\frac{526k}{7} = 263$$
$$2k = 7 \Rightarrow k = \frac{7}{2} = 3.5$$
Now we calculate the values of $X$:
1. $X_1 = 4(3.5) = 14$
2. $X_2 = \frac{30}{7}(3.5) = 15$
3. $X_3 = \frac{32}{7}(3.5) = 16$
4. $X_4 = \frac{34}{7}(3.5) = 17$
5. $X_5 = \frac{36}{7}(3.5) = 18$
6. $X_6 = \frac{38}{7}(3.5) = 19$
7. $X_7 = \frac{40}{7}(3.5) = 20$
8. $X_8 = 6(3.5) = 21$
We need to find $P(X < 20)$. This corresponds to $X \in \{14, 15, 16, 17, 18, 19\}$.
Summing the corresponding probabilities:
$$P(X<20) = P(X=14) + P(X=15) + P(X=16) + P(X=17) + P(X=18) + P(X=19)$$ $$P(X<20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{2}{15}$$ $$P(X<20) = \frac{11}{15}$$
Ans. (3)
Was this solution helpful?
YesNo