Question

$X~N(60,9)$. Suppose that you form random samples of 25 from this distribution. Let X be the random variable of averages. Let $\mathrm{\Sigma }X$be the random variable of sums. For parts, c through f, sketch the graph, shade the region, label and scale the horizontal axis for $\stackrel{-}{X}$, and find the probability.

a. Sketch the distributions of X and $\stackrel{-}{X}$on the same graph.

b. $\stackrel{-}{X}~$

c. $P(\stackrel{-}{X}<60)=$

d. Find the 30th percentile for the mean.

e. $P(56<\stackrel{-}{X}<62)$

f. $P(18<\stackrel{-}{X}<58)=$

g. $\mathrm{\Sigma X}~$

h. Find the minimum value for the upper quartile for the sum.

i. $$P(1,400<\mathrm{\Sigma X}<1,550)=$

Short answer

a. The graph has been shown

b.$\overline{X}~N(60,1.8)$

c.$P(\overline{X}<60)=0.50$

d.$59.06$

e.role="math" localid="1652455429651" $P(56<\stackrel{-}{X}<60)=0.853$

f. $P(18<\stackrel{-}{X}<58)=0.133$

g.$\sum X~N(1500,45)$

h.$1530.35$

i.$P\left(1400<\sum X<1550\right)=0.853$

Step-by-step solution

Given Information

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 $\mathrm{\Sigma }X$

Explanation Part (a)

The graph is represented,

Explanation Part (b)

Given,

The mean is${\mu }_x=60$ and the standard deviation is${\sigma }_x=9$

sample size n = $25$

Using,

$\overline{X}~N\left({\mu }_x,\frac{{\sigma }_x}{\sqrt{n}}\right)$we get $\overline{X}~N\left(60,\frac{9}{\sqrt{25}}\right)$

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

Explanation Part (c)

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

Using a calculator we get,

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

$=0.50$

Explanation Part (d)

We know,

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

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

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

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

Explanation Part (e)

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

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

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

Explanation Part (f)

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

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

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

Explanation Part (g)

Sample size n = $25$

standard deviation ${\sigma }_x=9$

mean ${\mu }_x=60$

Calculating the mean of the sums $\mu \sum x$,

$$\begin{gathered} =n{\mu }_x=25\times 60 \\ =1500 \end{gathered}$$

Calculating the standard deviation of sums $\sigma \sum x$,

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

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

Explanation Part (h)

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

$\sum X~N(1500,45)$

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

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

Explanation Part (i)

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

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