91Ó°ÊÓ

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

Then probability density operation is determined by:

$f\left(x\right)=\frac{1}{\sqrt{2\mathrm{π}}\mathrm{σ}}{e}^{-\frac{1}{2}\frac{(x-\mathrm{μ}{)}^2}{{\mathrm{σ}}^2}},x∈R.$

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 $0.25$.

We have the following:

$0.25=P(X<x)$

$={∫}_{-∞}^{x}f\left(t\right)dt$

Substitute the given expression,

$={∫}_{-∞}^x\frac{1}{\sqrt{2\mathrm{π}}5}{e}^{-\frac{1}{2}\frac{(t-4{)}^2}{5^2}}dt$

$={∫}_{-∞}^x\frac{1}{\sqrt{2\mathrm{π}}5}{e}^{-\frac{1}{2}{\left(\frac{t-4}{5}\right)}^2}dt$

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

$={∫}_{-∞}^{\frac{x-4}{5}}\frac{1}{\sqrt{2\mathrm{π}}}{e}^{-\frac{u^2}{2}}du$

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

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

$=0.84$

Adding the given expression,

$x=0.8â‹\dots 5+4$

We get,

$=0.63$.