he298c

2021-08-17

A mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that 86 of these orders were shipped on time. If the company really ships 90% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is 0.86 or less?

Caren

Given:
p=90%=0.90
x=86
n=100
The sample proportion is calculated by dividing the number of successes by the sample size:
$\stackrel{^}{p}=\frac{x}{n}=\frac{86}{100}=0.86$
The man of the sampling distribution of hat p is equal to the population proportion p:
${\mu }_{\stackrel{^}{p}}=p=0.90$
The sampling distribution standard deviation of $\stackrel{^}{p}$ is:
${\sigma }_{\stackrel{^}{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}=\sqrt{\frac{0.90\left(1-0.90\right)}{100}}=0.03$
The z-score is the value decreased by the mean, divided by the standard deviation:
$z=\frac{x-\mu }{\sigma }=\frac{0.86-0.90}{0.03}\approx -1.33$
Using table A, calculate the corresponding probability:
$P\left(\stackrel{^}{p}\le 0.86\right)=P\left(z<-1.33\right)=0.0918$

Do you have a similar question?