The probability distribution of a random variable X is given below : X4k(30)/(7)k[(32)/(7)…

Question

The probability distribution of a random variable X is given below :
X4k$\frac{30}{7}k$$\frac{32}{7}k$$\frac{34}{7}k$$\frac{36}{7}k$$\frac{38}{7}k$$\frac{40}{7}k$6kP(X)$\frac{2}{15}$$\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 :

Accepted Answer

The expected value of a discrete random variable is given by the formula :

$E(X)=\sum {X}_{i}\cdot P({X}_{i})$

Using the data from the table :

$E(X)=(4k\cdot \frac{2}{15})+(\frac{30k}{7}\cdot \frac{1}{15})+(\frac{32k}{7}\cdot \frac{2}{15})+(\frac{34k}{7}\cdot \frac{1}{5})+(\frac{36k}{7}\cdot \frac{1}{15})+(\frac{40k}{7}\cdot \frac{1}{5})+(6k\cdot \frac{1}{15})$

Take $\frac{k}{15}$ factor common

$\begin{gathered} E(X)=\frac{k}{15}[8+\frac{30}{7}+\frac{64}{7}+\frac{102}{7}+\frac{36}{7}+\frac{76}{7}+\frac{120}{7}+6] \ E(X)=\frac{k}{15}[14+\frac{30+64+102+36+76+120}{7}] \ E(X)=\frac{k}{15}[14+\frac{428}{7}] \end{gathered}$

$E(X)=\frac{k}{15}[\frac{98+428}{7}]=\frac{k}{15}\cdot \frac{526}{7}=\frac{526k}{105}$

Given that $E(X)=\frac{263}{15}$ :

$\begin{gathered} \frac{263}{15}=\frac{526k}{105} \ 263=\frac{526k}{7} \ k=\frac{263\times 7}{526}=\frac{1\times 7}{2}=\frac{7}{2}=3.5 \end{gathered}$

Identify $X$ values where $XSubstitute $k=\frac{7}{2}$ into each $X$ :

$ $${X}_{1}=4\times \frac{7}{2}=14(Calculate $P(X$$

Sum the probabilities for $X=14,15,16,17,18$, and 19:

$P(X<20)=P(X=14)+P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)$

$\Rightarrow$ $P(X<20)=\frac{2}{15}+\frac{1}{15}+\frac{2}{15}+\frac{3}{15}+\frac{1}{15}+\frac{2}{15}$

$\Rightarrow$ $P(X<20)=\frac{11}{15}

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