Question

The combined (verbal + quantitative reasoning) score on the GRE (Graduate Record Exam) is normally distributed with mean 1049 and standard deviation 189. Suppose n = 16 randomly selected students take the GRE on the same day.

a. What is the probability that one randomly selected student scores above 1100 on the GRE?
b. Describe the sampling distribution of the sample mean for the 16 students (description should include mean, standard deviation and shape).
c. What is the probability that a random sample of 16 students has a mean GRE score that is less than 1100?
d. What is the probability that a random sample of 16 students has a mean score that is more than 1100?
e. If the sample size in (d) above was increased to 64, what affect would it have on the probability (more, less, the same, can't tell) and why?

Answer 1

Kindly check explanation

Explanation:

Given The following :

Assume score is normally distributed with;

Mean (m) = 1049

Standard deviation (sd) = 189

Numbw rof observations (n) = 16

A) What is the probability that one randomly selected student scores above 1100 on the GRE?

Zscore = (x - m) / sd

X = 1100

Zscore = (1100 - 1049) / 189

Zscore = 51 / 189 = 0.27

From z-table ; 0.27 correspond to 0.6064

P(X > 1100) = 1 - P(X < 1100) = 1 - 0.6064 = 0.3936

2) sample mean will be equal to population mean = 1049

Sample Standard deviation : (population standard deviation / sqrt (n))

= 189 / sqrt(16) = 189/4 = 47.25

Shape of distribution is approximately normal

3)

P(Z < 1100)

X = 1100

Zscore = (x - m) / standard error

Standard Error = population standard deviation / sqrt (n)

= 189 / 4 = 47.25

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

From z-table :

P(Z < 1100) = 0.8599

4)

D.)

P(Z > 1100)

X = 1100

Zscore = (x - m) / SE

Zscore = (1100 - 1049) / 47.25

Zscore = 51 / 47.25 = 1.0793650 = 1.08

From z-table :

P(X > 1100) = 1 - P(X < 1100) = (1 - 0.8599) = 0.140

E)

With a sample size of 64

Standard Error becomes:

189 / sqrt(64) = 189/ 8 = 23.63

Zscore = (1100 - 1049) / 23.63

Zscore = 51 / 23.63 = 2.16

P(Z > 2.16) = 1 - P(Z < 2.16) = 1 - 0.9846 = 0.0154

Probability is less due to increase in sample size