Question

Let X be a normal random variable with mean $12$and variance $4$. Find the value of $c$such that localid="1646649699736" $P\left\{X>c\right\}=0.10$.

Short answer

The value of $c$ is $14.56$

Step-by-step solution

Step 1. Given information.

$X$is a normal random variable with meanlocalid="1646649744005" $\left(\mu \right)=12$and variance${\left(\sigma \right)}^2=4$

Step 2. Find value of c. 

As given in the question,

Mean$\left(\mu \right)=12$and Variance ${\sigma }^2=4$.

Hence, Standard Deviation $\left(\sigma \right)=2$.

As required,

$$\begin{gathered} P\left[X>c\right]=0.10 \\ 1-P\left[X\le c\right]=1-0.10 \\ P\left[X\le c\right]=0.90 \\ P\left[\frac{X-M}{\sigma }\le \frac{c-M}{\sigma }\right]=0.90 \\ P\left[Z\le \frac{c-12}{2}\right]=0.90 \\ \frac{c-12}{2}=1.28 \\ c=\left(1.28\times 2\right)+12 \\ c=14.56 \end{gathered}$$

Hence, the required value of $c$is $14.56$.