91Ó°ÊÓ

$$\begin{gathered} p=90\%=0.40 \\ \hat{p}=0.86 \\ n=100 \end{gathered}$$

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}} \\ z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \end{gathered}$$

The sampling distribution of the sample proportions p has a mean of

${\mathrm{μ}}_{\hat{p}}=p=0.90$

The sampling distribution of the sample proportion p has a standard deviation of

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}} \\ =\sqrt{\frac{0.90(1−0.90)}{100}} \\ =0.03 \end{gathered}$$

The $z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.86−0.90}{0.03} \\ =−1.33 \end{gathered}$$

The associating probability using the normal probability table in $P(Z<−1.33)$ is given in the standard normal probability table in the row starting with $−1.3$ and the column starting with $0.03$

$$\begin{gathered} P(\hat{p}>0.86)=P(z>−1.33) \\ =0.0918 \\ =9.18\% \end{gathered}$$

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

${\mathrm{μ}}_{\hat{p}}=p=0.90$

The $10\%$ requirement states that the sample size must be less than $10\%$ of the total population size. The ten percent requirement is met because the abundance of $100$ orders is less than $10\%$ of the population of all orders in the previous week (there are $5000$ orders in the previous week).

${\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}}$

$$\begin{gathered} =\sqrt{\frac{0.90(1−0.90)}{100}} \\ =0.03 \end{gathered}$$

The square root of the product of $p$ and $(1-p)$ divided by the sample sine $n$ is the standard deviation of the sampling distribution of the sample proportion $p$

The $z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.86−0.90}{0.03} \\ =−1.33 \end{gathered}$$

In the conventional normal probability table, in the row beginning with $-1.3$and in the column beginning with$.03$ the associating probability is given using the normal probability table $P(Z<-1.33)$

$$\begin{gathered} P(\hat{P}>0.86)=P(z>−1.33) \\ =0.0918 \\ =9.18\% \end{gathered}$$

Because the probability is not insignificant, the event is likely to happen by chance, and there is no solid proof that less than $90$ percent of all purchases from this company are dispatched within three working days.