Answer:
Kindly check explanation
Step-by-step 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
