Chapter 7.2, Problem 45E

(a)

To determine

To Calculate: the probability that the proportion in an SRS of 100 orders is 0.86 or less.

Answer to Problem 45E

9.18%

Explanation of Solution

Given:

  $\begin{array}{c}p=90\%=0.40\\ \widehat{p}=0.86\\ n=100\end{array}$

Formula used:

  ${\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}$

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.90$

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.90\left(1-0.90\right)}{100}}\\ =0.03\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.86-0.90}{0.03}\\ =-1.33\end{array}$

The associating probability using the normal probability table in $P\left(Z<-1.33\right)$ is given in the row starting with $-1.3$ and in the column starting with $.03$ of the standard normal probability table.

  $\begin{array}{c}P\left(\widehat{p}>0.86\right)=P\left(z>-1.33\right)\\ =0.0918\\ =9.18\%\end{array}$

(b)

To determine

To Explain: that there is convincing evidence that less than 90% of all orders from this company are shipped within three working days.

Answer to Problem 45E

No

Explanation of Solution

Given:

  $\begin{array}{c}p=90\%=0.40\\ \widehat{p}=0.86\\ n=100\end{array}$

Formula used:

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ z=\frac{x-\mu }{\sigma }\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is equal to the population proportion $p$ .

  ${\mu }_{\widehat{p}}=p=0.90$

The 10% condition requires that the sample is less than 10% of the population size. Since the ample of 100 orders is less than 10% of the population of all orders in the past week (as there are 5000 orders in the past week), the 10% condition is satisfied.

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is the square root of the product of $p$ and $\left(1-p\right)$ divided by the sample sine $n$ .

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.90\left(1-0.90\right)}{100}}\\ =0.03\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.86-0.90}{0.03}\\ =-1.33\end{array}$

The associating probability using the normal probability table $P\left(Z<-1.33\right)$ is given in the row starting with $-1.3$ and in the column starting with $.03$ of the standard normal probability table

  $\begin{array}{c}P\left(\widehat{p}>0.86\right)=P\left(z>-1.33\right)\\ =0.0918\\ =9.18\%\end{array}$

Since the probability is not small, the event is likely to occur by chance and therefore there is no convincing evidence that less than 90 of all orders from this company are shipped within 3 working days.