Chapter 12.2, Problem 55E

(a)

To determine

To find: the probability.

Answer to Problem 55E

0.9050

Explanation of Solution

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

Let X is the length of researcher’s shower on a randomly selected day which follows a Normal distribution with mean 4.5 minutes and standard and standard deviation 0.9 minutes.

Finding the z-value for $P\left(3<X<6\right)$

Finding the associating the normal probabilities

  $\begin{array}{l}{z}_l=\frac{3-4.5}{0.9}=-1.67\Rightarrow p_l=0.0475\\ z_u=\frac{6-4.5}{0.9}=1.67\Rightarrow p_u=0.9525\end{array}$

Finding the difference in probabilities to find the probability:

  $p_u-p_l=0.9525-0.0475=0.9050$

(b)

To determine

To Explain: that the researcher took a 7 minute shower, would it be classified as an outlier by the 1.5IQR rule.

Answer to Problem 55E

Yes

Explanation of Solution

Given:

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

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

25^th percentile $Q_1$

Finding the z-score that associates with a probability of 0.25 in the normal probability and it is observed that the nearest probability is 0.2514 which lies in the row -0.6 and in the column .07 of the normal probability table and therefore the associating z-score is then -0.6+.07 = -0.67.

  $Q_1=-0.67$

75^th percentile $Q_3$

Finding the z-score that associates with a probability of 0.75 in the normal probability and is it is observed that the nearest probability is 0.7486 which lies in the row 0.6 and in the column .07 of the normal probability table and therefore the associating z-score is then 0.6+ .07 =0.67.

  $Q_3=0.67$

Outliers

The inter quartile range IQR is

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

Outliers are observations

  $\begin{array}{l}{Q}_3+1.5IQR=0.67+1.5\left(1.34\right)=2.68\\ Q_1-1.5IQR=-0.67-1.5\left(134\right)=-2.68\end{array}$

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

The z-score is

  $z=\frac{7-4.5}{0.9}=2.78$

Since 2.78 is above 2.68, the 7-minute shower would be classified as an outlier.

(c)

To determine

To find: the probability that her shower time is 7 minutes or larger on at least 2 of the days.

Answer to Problem 55E

0.000323

Explanation of Solution

Given:

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

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Multiplication and complement rule:

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

Calculation:

The z-score is

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

Finding the associating probability

  $\begin{array}{l}P\left(X>7\right)=P\left(Z>2.78\right)=P\left(Z<-2.78\right)=0.0027\\ P\left(X\le 7\right)=P\left(Z<2.78\right)=0.9973\end{array}$

Multiplication and complement rule:

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

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

(d)

To determine

To find: the probability that the mean length of her shower times on these 10 days exceeds 5 minutes.

Answer to Problem 55E

0.0392

Explanation of Solution

Given:

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

Formula used:

  $z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}$

Calculation:

The z-score is

  $z=\frac{\overline{x}-\mu }{\sigma /\sqrt{n}}=\frac{5-4.5}{0.9/\sqrt{10}}=1.76$

Finding the associating probability

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