I am having difficulty with this problem, did I do my math correctly? And, in particular question d I do not understand how it would be influenced, any help would be great. Thank you.
Based on the results of a study, the mean =3.5 and st. deviation 0.5 for the GPA of all non traditional students. Suppose that random sample n=100 non traditional students is selected from the population of all the non traditional students, and GPA of each student is determined. Then the sample mean will be normally distributed.
a) Calculate Uxbar and O(sigma)xbar.
U=Uxbar = 3.5
O(sigma)xbar= square root of (sigma/n)
= square root of (0.5/100)
= 0.0707
b) What is the approximate probability that the nontraditional student sample has a mean GPA that exceeds 3.62?
x=3.62
z=x-U/O(sigma) = 3.62-3.5/.0707=1.4 We find z = 1.4 to be 0.4222 (from a z-score table)
P(z>3.62)= A2 = 0.5-A1 = 0.5-0.4554 = .04460
c)How would the sampling distribution of xbar change if the sample size n was doubled from 100 to 200? How do your answers in part b change with increased sample size.
The sampling distribution would expand in terms of z-scores by 0.6.
Standard deviation would become:
O(sigma)xbar= square root of (sigma/n)
= square root of (0.5/200)
= 0.05
Answers in part b would decrease.
Based on the results of a study, the mean =3.5 and st. deviation 0.5 for the GPA of all non traditional students. Suppose that random sample n=100 non traditional students is selected from the population of all the non traditional students, and GPA of each student is determined. Then the sample mean will be normally distributed.
a) Calculate Uxbar and O(sigma)xbar.
U=Uxbar = 3.5
O(sigma)xbar= square root of (sigma/n)
= square root of (0.5/100)
= 0.0707
b) What is the approximate probability that the nontraditional student sample has a mean GPA that exceeds 3.62?
x=3.62
z=x-U/O(sigma) = 3.62-3.5/.0707=1.4 We find z = 1.4 to be 0.4222 (from a z-score table)
P(z>3.62)= A2 = 0.5-A1 = 0.5-0.4554 = .04460
c)How would the sampling distribution of xbar change if the sample size n was doubled from 100 to 200? How do your answers in part b change with increased sample size.
The sampling distribution would expand in terms of z-scores by 0.6.
Standard deviation would become:
O(sigma)xbar= square root of (sigma/n)
= square root of (0.5/200)
= 0.05
Answers in part b would decrease.