# 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. $P(X<X_0)=0.95$

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

• Are you sure the answer is supposed to be $66.58$? – Math1000 May 25 '18 at 11:16
• edited, changed 40.0 to 4.0 – gtx May 25 '18 at 11:43
• Using Mathematica: $$\texttt{InverseCDF[NormalDistribution[60, 4], 0.95]} = 66.5794$$ – Math1000 May 25 '18 at 12:05
• You have SD 4 in one place and 40 in another. That is probably a source of confusion. (Maybe you changed it in only one of two parts of your question.) Mathematica result is accurate; result in Answer by @callculus (+1) turns out to be about as good as you can get from printed normal tables. In R statistical software, I used qnorm(.95, 50, 4) to get 56.57941. – BruceET May 26 '18 at 19:11

## 1 Answer

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 $x$.