Find $\mathrm { P } ( 395.4

Question

Find $\mathrm { P } ( 395.4 < \bar { X } < 404.6 )$ If the population mean = 400,?_x = 20,and n = 100. A).9786 B).9999 C).0214 D).9893

Quizplus · Accepted Answer

The standard error of the mean is $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{100}} = 2$. The z-scores for 395.4 and 404.6 are $\frac{395.4 - 400}{2} = -2.3$ and $\frac{404.6 - 400}{2} = 2.3$, respectively. The probability $\mathrm{P}(395.4 < \bar{X} < 404.6)$ is the area between these z-scores, which is approximately 0.9786.

Find P(395.4<Xˉ<404.6)\mathrm { P } ( 395.4 < \bar { X } < 404.6 )P(395.4<Xˉ<404.6)

Find $\mathrm { P } ( 395.4 < \bar { X } < 404.6 )$