Question

The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 15. 9 minutes. Based on a random sample of 40 orders, a reporter determines a mean order time of 63 minutes. Let µ represent the average number of minutes to fill an order. Z = StartFraction x overbar minus mu Over (StartFraction sigma Over StartRoot n EndRoot EndFraction) EndFraction To the nearest hundredth, the z-statistic is z =.

Answer 1

By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

What is standard deviation?

Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean), or expected value

Here given that

The owners of an eyeglass store claim that they fill customers' orders, on average, in 60 minutes with a standard deviation of 15. 9 minutes. Based on a random sample of 40 orders, a reporter determines a mean order time of 63 minutes

mean=μ=60

standard deviation=σ=15.9

number of sample=n=40

$\bar{x}=63$

Now we can calculate z by z-score formula as

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}} \\\\&=\frac{63-60}{\frac{15 \cdot 9}{40}} \\ \\=\frac{3}{2.5140} \\ \\=1.19\end{aligned}$

By applying z-score formula we got that if a reporter determines a mean order time of 63 minutes then value o Z is 1.19

To learn more about standard deviation visit :brainly.com/question/12402189