Question

The GPAs of all 5540 students enrolled at a university has a distribution with a mean of 3.02 and a standard deviaiton of 0.29. Let x-bar be the mean GPA of a random sample of 48 students selected from this university.

a. Find the mean and standard deviation of x-bar and comment on the shape of its sampling distribution (1 point for each).

b. What is the probability that the sample mean of such a sample will be greater than 3.10?

Answer 1

Solution :

Given that,

mean = = 3.02

standard deviation = = 0.29

n=48

mean = = =3.02

standard deviation = = / n = 0.29/ 48 = 0.04186

P( > 310) = 1 - P( < 3.10)

= 1 - P[( - ) / < (3.10-3.02) / 0.04186 ]

= 1 - P(z < 1.91)

Using z table

= 1 - 0.9719

= 0.0281

probability=0.0281