the owners of an eyeglass store claim that they fill customers orders, on average, in 60 minutes with a…

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 $mu$ represent the average number of minutes to fill an order.\n$z = \frac{\bar{x}-mu}{(\frac{sigma}{sqrt{n}})}$\nto the nearest hundredth, the z - statistic is $z=$

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 $mu$ represent the average number of minutes to fill an order.\n$z = \frac{\bar{x}-mu}{(\frac{sigma}{sqrt{n}})}$\nto the nearest hundredth, the z - statistic is $z=$

Answer

Explanation:

Step1: Identify given values

$\bar{x}=63$, $\mu = 60$, $\sigma=15.9$, $n = 40$

Step2: Substitute values into z - statistic formula

$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{63 - 60}{\frac{15.9}{\sqrt{40}}}$ First, calculate $\sqrt{40}\approx6.3246$. Then $\frac{15.9}{\sqrt{40}}\approx\frac{15.9}{6.3246}\approx2.514$. Next, $63 - 60=3$. So $z=\frac{3}{2.514}\approx1.19$

Answer:

$1.19$