91Ó°ÊÓ

Given in the question that,

Normal distribution with mean is $4.5$minutes

Standard deviation is$0.9$minutes

The given data is

$\mathrm{μ}=4.5$

$\mathrm{σ}=0.9$

$25th$percentile $Q_1$

The $X^{th}$percentile is the data value that has $x\%$of all data values below it, suggesting that $25\%$of all data values are below it.

In the normal probability table of the appendix, find the $z$-score that corresponds to a probability of $0.25$. The closest probability is $0.2514$, which is found in the row $-0.6$and column of the normal probability table, and so the equivalent $z$-score is

$-0.6+.07=-0.67$

$Q_1=-0.67$

$75th$percentile

The $X^{th}$percentile is the data value that has $x\%$of all data values below it, implying that $75\%$of all data values are below the ${75}^{th}$percentile.

In the normal probability table of the appendix, find the $z$-score that corresponds to a probability of $0.75$. The closest probability is $0.7486$, which is found in the normal probability table's row $0.6$and column $0.07$, and so the equivalent $z$-score is

$0.6+.07=0.67$

$Q_3=0.67$

Outliers

The difference between the third and first quartile is the interquartile range (IQR):

localid="1650687556358" $$\begin{gathered} IQR=Q_3-Q_1 \\ =0.67-(-0.67) \\ =1.34 \end{gathered}$$

Outliers are observations that are more than $1.5$times the IQR above $Q_3$or below $Q_1$

localid="1650687571854" $$\begin{gathered} Q_3+1.5IQR=0.67+1.5(1.34) \\ =2.68 \end{gathered}$$

localid="1650687586729" $$\begin{gathered} Q_1-1.5IQR=-0.67-1.5(1.34) \\ =-2.68 \end{gathered}$$

Outliers are defined as $z$-scores that are less than $-2.68$ or greater than $2.68$. The $z$-score is the difference between the mean and the standard deviation:

localid="1650687600637" $$\begin{gathered} z=\frac{7-4.5}{0.9} \\ ≈2.78 \end{gathered}$$

Since $2.78$is above $2.68$, the -minute shower would be classified as an outlier.

Normal distribution has

Mean is$4.5$minutes

Standard deviation is$0.9$minutes

According to the information

$\mathrm{μ}=4.5$

$\mathrm{σ}=0.9$

The$z$-score is the difference between the mean and the standard deviation:

localid="1650687713162" $$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{7-4.5}{0.9} \\ ≈2.78 \end{gathered}$$

Using table A, calculate the corresponding probability:

localid="1650687761801" $$\begin{gathered} P(X>7)=P(Z>2.78) \\ =P(Z<-2.78) \\ =0.0027 \end{gathered}$$

localid="1650687773704" $$\begin{gathered} P(X≤7)=P(Z<2.78) \\ =0.9973 \end{gathered}$$

Multiplication and complement rule

$P(AandB)=P\left(A\right)P\left(B\right)$

$P(notA)=1-P\left(A\right)$

Then we get,

$P(Atleast2of10days>7)=1-P(0of10days>7)-P(1of10days>7)$

$=1-0.{9973}^{10}-10×0.{9973}^9×0.0027=0.000323.$

Normal distribution has a

Mean is$4.5$minutes

Standard deviation is$0.9$minutes.

From the given values

With a mean of $\mathrm{μ}$and a standard deviation of $\mathrm{σ}/\sqrt{n}$the sample mean follows a normal distribution.

The$z$-score is calculated by dividing the sample mean by the standard deviation:

localid="1650688607451" $$\begin{gathered} z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{\mathrm{σ}/\sqrt{n}} \\ =\frac{5-4.5}{0.9/\sqrt{10}} \\ ≈1.76 \end{gathered}$$

Using table A, calculate the corresponding probability:

localid="1650688598819" $$\begin{gathered} P(X>5)=P(Z>1.76) \\ =P(Z<-1.76) \\ =0.0392 \end{gathered}$$.