According to an IRS study it takes a mean of 330 minutes for

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? (Round your answer to 3 decimal places.)

b. What is the likelihood the sample mean is greater than 320 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)

c. What is the likelihood the sample mean is between 320 and 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)

d. What is the likelihood the sample mean is greater than 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Solution

a)
Mean ( u ) =330
Standard Deviation ( sd )=80/Sqrt(40)=12.6491
Number ( n ) = 40
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)  

standard error of the mean = 80/Sqrt(40) =    12.649
b)
P(X > 320) = (320-330)/80/ Sqrt ( 40 )
= -10/12.649= -0.7906
= P ( Z >-0.7906) From Standard Normal Table
= 0.7852                  
      
c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 320) = (320-330)/80/ Sqrt ( 40 )
= -10/12.6491
= -0.7906
= P ( Z <-0.7906) From Standard Normal Table
= 0.2146
P(X < 350) = (350-330)/80/ Sqrt ( 40 )
= 20/12.6491 = 1.5811
= P ( Z <1.5811) From Standard Normal Table
= 0.94308
P(320 < X < 350) = 0.94308-0.2146 = 0.7281              
d)
P(X > 350) = (350-330)/80/ Sqrt ( 40 )
= 20/12.649= 1.5811
= P ( Z >1.5811) From Standard Normal Table
= 0.0571