91Ó°ÊÓ

We have,

$X~N(60,9)$

sample size n =$25$

The random variable of averages is$\stackrel{-}{X}$and the Random variable of sums is $\mathrm{Î\pounds }X$

The graph is represented,

Given,

The mean is${\mathrm{μ}}_x=60$ and the standard deviation is${\mathrm{σ}}_x=9$

sample size n = $25$

Using,

$\stackrel{¯}{X}~N\left({\mathrm{μ}}_x,\frac{{\mathrm{σ}}_x}{\sqrt{n}}\right)$we get $\stackrel{¯}{X}~N\left(60,\frac{9}{\sqrt{25}}\right)$

$=\stackrel{-}{X}~N(60,1.8)$

The probability of the distribution $\stackrel{-}{X}~N(60,1.8)$be $P(\stackrel{-}{X}<60)$

Using a calculator we get,

$P(\stackrel{¯}{X}<60)=\text{Normalcdf}(60,60,1.8)$

$=0.50$

We know,

sample size n = $25$and $X~N(60,9)$

For $\stackrel{-}{X}$the $30th$percentile is,

$P(\stackrel{-}{X}<k)=0.30$

$$\begin{gathered} \stackrel{-}{X}=\text{invnorm}(0.30,60,1.8) \\ =59.06 \end{gathered}$$

Sample size n = $25$for distribution $X~N(60,9)$

Calculating role="math" localid="1652457644091" $P(56<\stackrel{-}{X}<62)$using a calculator,

$$\begin{gathered} =P(56<\stackrel{¯}{X}<60)=\text{Normalcdf}(56,60,60,1.8) \\ =0.853 \end{gathered}$$

Sample size n = $25$for distribution $\stackrel{-}{X}~N(60,1.8)$

Calculating $P(18<\stackrel{-}{X}<58)$using a calculator,

$$\begin{gathered} P(18<\stackrel{¯}{X}<58)=\text{Normalcdf}(18,58,60,1.8) \\ =0.133 \end{gathered}$$

Sample size n = $25$

standard deviation ${\mathrm{σ}}_x=9$

mean ${\mathrm{μ}}_x=60$

Calculating the mean of the sums $\mathrm{μ}∑x$,

$$\begin{gathered} =n{\mathrm{μ}}_x=25×60 \\ =1500 \end{gathered}$$

Calculating the standard deviation of sums $\mathrm{σ}∑x$,

role="math" localid="1652458063627" $={\mathrm{σ}}_x\left(\sqrt{n}\right)=9×\sqrt{25}=45$

Hence the distribution is$∑X~N(1500,45)$

The minimum value for the upper quartile for the sum is,

$∑X~N(1500,45)$

The upper quartile is $75th$percentile, Using calculator we get,

$$\begin{gathered} =\text{invnorm}(0.75,1500,45) \\ =1530.35 \end{gathered}$$

Using a calculator we find $P\left(1400<∑X<1550\right)$,

$$\begin{gathered} =\text{normalcdf}(1400,1550,1500,45) \\ =0.853 \end{gathered}$$