According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers. a. What is the standard error of the mean in this example? b. What is the likelihood the sample mean is greater than 360 minutes?

Question

contestada

Respuesta :

JeanaShupp JeanaShupp · Answer 1 · 2019-07-17T05:41:57+02:00

Answer: 0.0089

Step-by-step explanation:

Given : Mean : [tex]\mu=\ 330[/tex]

Standard deviation :[tex]\sigma= 80[/tex]

Sample size : [tex]n=40[/tex]

We know that the standard error of the mean is given by :-

[tex]\text{S.E.}=\dfrac{\sigma}{\sqrt{n}}\\\\\Rightarrow\text{S.E.}=\dfrac{80}{\sqrt{40}}\\\\\Rightarrow\ \text{S.E.}=12.6491106407\approx12.65[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x = 360

[tex]z=\dfrac{360 -330}{\dfrac{80}{\sqrt{40}}}=2.37170824513\approx2.37[/tex]

The p-value = [tex]P(z>2.37)=1-P(z<2.37)= 1- 0.9911059\approx0.0089[/tex]

Hence, the likelihood the sample mean is greater than 360 minutes= 0.0089