Chapter 12.2, Problem 49E

(a)

To determine

To Explain: that it would be categorize as an outlier, justify the answer.

Answer to Problem 49E

Yes

Explanation of Solution

Given:

  $\begin{array}{l}\mu =\text{ 4}\text{.5}\\ \sigma =\text{0}\text{.9}\end{array}$

Formula used:

  $\begin{array}{l}IQR=Q_3-Q_1\\ \text{for}\text{outiliers}\text{observations}\\ Q_3+1.5IQR\\ Q_3-1.5IQR\\ z=\frac{x-\mu }{\sigma }\end{array}$

Calculation:

Outliers

The interquartile range IQR

  $IQR=Q_3-Q_1=0.67-\left(-0.67\right)=1.34$

Outliers are observations

  $\begin{array}{l}{Q}_3+1.5IQR=0.67+1.5\left(1.34\right)=2.68\\ Q_3-1.5IQR=-0.67-1.5\left(1.34\right)=-2.68\end{array}$

It is observed that outliers are all :-scores that are below -2.68 or above 2.68.

The z-score

  $z=\frac{7-4.5}{0.9}=2.78$

Since 2.78 is beyond 2.68 it will consider the 7-minute shower as an outlier.

25th percentile $Q_1$

The percentile reflects the data value with below it x percent of all data values, which means that 25 percent of all data values are below the 25th percentile.

Calculating the z-score usual probability row, which corresponds to a probability of 0.25. it is noticed that the nearest probability is 0.2514, which is in the row-0.6 and in the normal probability table column.07 and that the resulting z-score is -0.6+.07=-0.67.

  $Q_1=-0.67$

75^th percentile $Q_3$

The percentile reflects the data value below which has x percent of all data values, indicating that 75 percent of all data values are below the 75th percentile.

Calculating the z-score usual probability row, which corresponds to a probability of 0.75. it is noticed that the nearest probability is 0.7486 in row 0.6 and column 0.07 of the usual probability table and that the resulting z-score is 0.6+.07=0.67.

  $Q_3=0.67$

(b)

To determine

To find: the probability that her shower period on at least 2 of the day is 7 minutes or higher.

Answer to Problem 49E

0.000323

Explanation of Solution

Given:

  $\begin{array}{l}\mu =4.5\\ \sigma =0.9\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The z-score is

  $z=\frac{x-\mu }{\sigma }=\frac{7-4.5}{0.9}=2.78$

The corresponding probability

  $\begin{array}{c}P\left(X>7\right)=P\left(Z>2.78\right)\\ =P\left(Z<-2.78\right)\\ =0.0027\\ P\left(X\le 7\right)=P\left(Z<2.78\right)\\ =0.9973\\ z=\frac{x-\mu }{\sigma }\end{array}$

Rule of Multiplication and complement

  $\begin{array}{l}P\left(A\text{ and B}\right)=P\left(A\right)\times P\left(B\right)\\ P\left(\text{not A}\right)=1-P\left(A\right)\end{array}$

Then get

  $\begin{array}{l}P\left(\text{At least 2 of 10 days > 7}\right)=1-P\left(0\text{ of 10 days >7}\right)-P\left(\text{1 of 10 days >7}\right)\\ =1-{0.9973}^{10}-10\times {0.9973}^9\times 0.0027=0.000323\end{array}$

(c)

To determine

To find: the chance of reaching 5 minutes for the mean duration of her shower times on these 10 days.

Answer to Problem 49E

0.0392

Explanation of Solution

Given:

  $\begin{array}{l}\mu =4.5\\ \sigma =0.9\\ n=10\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

  $\begin{array}{c}P\left(X>5\right)=P\left(Z>1.76\right)\\ =P\left(Z<-1.76\right)\\ =0.0392\end{array}$

  $\begin{array}{c}P\left(X>5\right)=P\left(Z>1.76\right)\\ =P\left(Z<-1.76\right)\\ =0.0392\end{array}$