Statistics Help!!! standard error problem?

So I really can't figure this out...someone please help me!18. According to an IRS study, it takes an average of 330 minutes for taxpayers to prepare,


copy, and electronically file a 1040 tax form. A consumer watchdog agency selects a random


sample of 40 taxpayers and finds the standard deviation of the time to prepare, copy,


and electronically file form 1040 is 80 minutes.


a. What assumption or assumptions do you need to make about the shape of the


population?


b. What is the standard error of the mean in this example?


c. What is the likelihood the sample mean is greater than 320 minutes?


d. What is the likelihood the sample mean is between 320 and 350 minutes?


e. What is the likelihood the sample mean is greater than 350 minutes?





its due tonight

Statistics Help!!! standard error problem?
a.





What does it mean the shape of the population?





The shape of the distribution of the time should be gaussian bell (normal distribution).





b.





std of the mean = standard deviation / √n = 80 / √(40) =


= √(160) = 12.65





c.





T_m ~ N(330, 160)





=%26gt; X = (T_m - 330) / √(160) ~ N(0,1)





P(T_m %26gt; 320) = P(X %26gt; (320-330)/√(160)) =


= P(X %26gt; -0.79057) = 1 - Φ(-0.79057) = Φ(0.79057) =


= 0.78541





d)





P(320 %26lt; T_m %26lt; 350) = P( -0.79057 %26lt; X %26lt; 1.66020) =


= Φ(1.66020) - Φ( -0.79057) = Φ(1.66020) -1+ Φ( -0.79057) =


= 0.94738 -1 + 0.78541 = 0.73279





e)





P(T_m %26gt; 350) = P(X %26gt; 1.66020) = 1- Φ(1.66020) =


= 0.05262