Question

$X~N(4,5)$

Find the maximum of $x$in the bottom quartile.

Short answer

The maximum value in the bottom quartile is $0.63$.

Step-by-step solution

Given Information

Let variable $\mathrm{X}$has normal allocation with mean $\mu$and variance ${\sigma }^2$.

Then probability density operation is determined by:

$f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}\frac{(x-\mu {)}^2}{{\sigma }^2}},x\in R.$

Explanation

Let $\mathrm{X}$has normal distribution with mean $4$and standard deviation $5,X~N\left(4,5^2\right)$.

Now we need to find the maximum value in the bottom quartile.

In the bottom quartile, the minimum value is the minimum value of the distribution, and the maximum value is the first quartile.

The first quartile is${25}^{th}$percentile of this distribution. It means that we need to find the value $x$such that the probability that variable $X$is less than $x$is $0.25$.

Calculation

We have the following:

$0.25=P(X<x)$

$={\int }_{-\infty }^{x}f\left(t\right)dt$

Substitute the given expression,

$={\int }_{-\infty }^x\frac{1}{\sqrt{2\pi }5}{e}^{-\frac{1}{2}\frac{(t-4{)}^2}{5^2}}dt$

$={\int }_{-\infty }^x\frac{1}{\sqrt{2\pi }5}{e}^{-\frac{1}{2}{\left(\frac{t-4}{5}\right)}^2}dt$

$\left|u=\frac{t-4}{5},du=\frac{1}{5}dt\right|$

$={\int }_{-\infty }^{\frac{x-4}{5}}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{u^2}{2}}du$

$=\Phi \left(\frac{x-4}{5}\right)$

$\frac{x-4}{5}=NORMSINV(0.25)$

$=0.84$

Adding the given expression,

$x=0.8\cdot 5+4$

We get,

$=0.63$.