Question

Marcella takes a shower every morning when she gets up. Her time in the shower varies according to a Normal distribution with mean $4.5$minutes and standard deviation $0.9$minutes.

(a) If Marcella took a $7$-minute shower, would it be classified as an outlier? Justify your answer.

(b) Suppose we choose $10$days at random and record the length of Marcella’s shower each day. What’s the probability that her shower time is $7$minutes or higher on at least $2$of the days? Show your work.

(c) Find the probability that the mean length of her shower times on these 10 days exceeds$5$ minutes. Show your work

Short answer

(a) If Marcella took a $7$-minute shower, yes it is classified as an outlier.

(b) The probability is $0.000323.$

(c) The probability that the mean length of her shower times is$0.0392$.

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that,

Normal distribution with mean is $4.5$minutes

Standard deviation is$0.9$minutes

Part (a) Step 2: Explanation

The given data is

$\mu =4.5$

$\sigma =0.9$

$25th$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" $$\begin{gathered} z=\frac{7-4.5}{0.9} \\ \approx 2.78 \end{gathered}$$

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

Part (b) Step 1: Given information

Normal distribution has

Mean is$4.5$minutes

Standard deviation is$0.9$minutes

Part (b) Step 2: Explanation

According to the information

$\mu =4.5$

$\sigma =0.9$

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

localid="1650687713162" $$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{7-4.5}{0.9} \\ \approx 2.78 \end{gathered}$$

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

Multiplication and complement rule

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

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

Then we get,

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

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

Part (c) Step 1: Given information

Normal distribution has a

Mean is$4.5$minutes

Standard deviation is$0.9$minutes.

Part (c) Step 2: Explanation

From the given values

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

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

localid="1650688607451" $$\begin{gathered} z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}} \\ =\frac{5-4.5}{0.9/\sqrt{10}} \\ \approx 1.76 \end{gathered}$$

Using table A, calculate the corresponding probability:

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