The variable $X$ is normally distributed the mean is $\mu(X)=60.0$ and the std deviation is $\omega(X)=40.0$, find $X_0$ s. t. $P(X<X_0)=0.95$

Question

The variable $X$ is normally distributed the mean of $X$ is $\mu(X) = 60.0$ and the standard deviation of $X$ is $\omega(X) = 4.0$, find $X_0$ such that $P(X < X_0) = 0.95$

I know we need to use the $z$ score formula $$z = \frac{X - \mu}\omega\implies \frac{X - 60.0}{40.0}$$

Correct answer is $66.58$ please explain

Answer 1

We have

$P(X\leq x)=\Phi\left( \frac{x - 60.0}{4.0} \right)=0.95$

$\Phi\left( z \right)$ is the cdf of the standard normal distribution. Next step is to take the inverse function of $\Phi(z)$

$\frac{x - 60.0}{4.0}=\Phi^{-1}\left( 0.95\right)$

Now you use a table for that function. At row $z=1.6$ in combination with the columns $.04$ and $.05$ we can read off that

$\Phi{(1.64)}=0.94950$ and $\Phi{(1.65)}=0.95053$

Now we apply a linear approximation. The arithmetic mean of $ 0.94950$ and $0.95053$ is $0.950015\approx 0.95$. Since we apply linear approximation we calculate the arethmetic mean of $1.64$ and $1.65$ as well. This is $1.645$. Thus the equation is

$$\frac{x - 60.0}{4.0}=1.645$$

All that remains to do is to solve the equation for $x$.