Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Normal distribution :
Mean (m) = 1049
Standard deviation (sd) = 189
Number of samples (n) = 16
A) What is the probability that one randomly selected student scores above 1100 on the GRE?
Obtain the z score
Zscore = (x - m) / sd
Zscore = (1100 - 1049) / 189
Zscore = 51 / 189 = 0.2698412 = 0.27
Using the z-table
P(X > 100) = 1 - P(X < 100) = 1 - 0.6064 = 0.3936
B) The sample mean = population mean
Hence sample mean = 1049
The sample standard deviation = standard error
Standard Error (SE) = sd / sqrt(n)
SE = 189/4 = 47.25
Hence shape = normal, standard deviation = 47.25, mean = 1049
C.) P(X < 1100)
Zscore = (x - m) / SE
Zscore = (1100 - 1049) / 47.25
Zscore = 51 / 47.25 = 1.0793650 = 1.08
Using the z-table :
P(X < 1100) = 0.8599
D.) P(X > 1100)
Zscore = (x - m) / SE
Zscore = (1100 - 1049) / 47.25
Zscore = 51 / 47.25 = 1.0793650 = 1.08
Using the z-table :
P(X > 1100) = 1 - P(X < 1100) = (1 - 0.8599) = 0.1401
E) if n = 64
SE = 189 / sqrt(64) = 189 / 8 = 23.625
P(X > 1100)
Zscore = (x - m) / SE
Zscore = (1100 - 1049) / 23.625
Zscore = 51 / 47.25 = 2.1587301 = 2.16
Using the z-table :
P(X > 1100) = 1 - P(X < 1100) = (1 - 0.9846) = 0.0154
The probability decreases as the same distribution hinges towards a Boal distribution with increasing sample size.
