Question

Suppose x is a normally distributed random variable with μ=48 and σ=5. Find a value x0 of the random variable x that satisfies the following equations or statements.

a.​ 10% of the values of x are less than x0.

b.​ 1% of the values of x are greater than x0.

Answer 1

$\mu=48$

$\sigma=5$

Use the distribution function of the standard normal distribution

$\displaystyle\Phi(x)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^x e^{-\frac{t^2}{2}}dt$

Note that

$\Phi(-\infty)=0$

$\Phi(+\infty)=1$

a. $P(X<x_0)=0.1$

$P(-\infty<X<x_0)=0.1$

$\Phi\left(\dfrac{x_0-\mu}{\sigma}\right)-\Phi(-\infty)=0.1$

$\Phi\left(\dfrac{x_0-48}{5}\right)-0=0.1$

$\Phi\left(\dfrac{x_0-48}{5}\right)=0.1$

$\dfrac{x_0-48}{5}=\Phi^{-1}(0.1)$

$x_0-48=5\cdot \Phi^{-1}(0.1)$

$x_0=5\cdot \Phi^{-1}(0.1)+48$

$x_0\approx5\cdot (-1.28)+48=-6.41+48=41.59$

b. $P(X>x_0)=0.01$

$P(x_0<X<+\infty)=0.01$

$\Phi(+\infty)-\Phi\left(\dfrac{x_0-\mu}{\sigma}\right)=0.01$

$1-\Phi\left(\dfrac{x_0-48}{5}\right)=0.01$

$-\Phi\left(\dfrac{x_0-48}{5}\right)=0.01-1$

$\Phi\left(\dfrac{x_0-48}{5}\right)=1-0.01$

$\Phi\left(\dfrac{x_0-48}{5}\right)=0.99$

$\dfrac{x_0-48}{5}=\Phi^{-1}(0.99)$

$x_0-48=5\cdot\Phi^{-1}(0.99)$

$x_0=5\cdot\Phi^{-1}(0.99)+48$

$x_0\approx5\cdot2.33+48=11.63+48=59.63$