13. A sample of university students have an average GPA of 2.78 with a standard
ID: 3339579 • Letter: 1
Question
13. A sample of university students have an average GPA of 2.78 with a standard deviation of 0.45. What is the probability that a randomly selected student will have a GPAZ score Probability a. less than 3.40? _______ ____ b. less than 3.78? ______ ____ c. more than 3.50? ______ ____ d. more than 2.50? _______ ____ e. between 2.00 and 3.00? ______ ___ f. between 3.00 and 3.50? _______ ____
14. Is the variable Age When First Married roughly normally distributed? 15. Using the normal distribution calculations, 60% of respondents are married before what age?
16. Using the normal distribution calculations, what percentage of people are first married in their 20’s?
125 75 Mean 22.79 Std. Dev.-5.033 N-1,20 20 30 50 60 10 Age When First Married 14. Is the variable Age When First Married roughly normally distributed? 15. Using the normal distribution calculations, 60% of respondents are married before what age? 16. Using the normal distribution calcula 20's?
Explanation / Answer
As per the Chegg policy, we are advised to do one question at a time so i am attempting 13th.
13. a) z score for 3.40 = (3.40 - 2.78)/0.45 = 0.93
So,
Probability = P(X < 3.40) = P(z < 0.93) = 0.9159
b) z score for 3.78 = (3.78 - 2.78)/0.45 = 2.22
So,
Probability = P(X < 3.78) = P(z < 2.22) = 0.9869
c) z score for 3.50 = (3.50 - 2.78)/0.45 = 1.6
So,
Probability = P(X > 3.50) = P(z > 1.6) = 0.0548
d) z score for 2.50 = (2.50 - 2.78)/0.45 = -0.62
So,
Probability = P(X > 2.50) = P(z > -0.62) = 0.7331
e) z score for 2 = (2 - 2.78)/0.45 = -1.73
z score for 3 = (3 - 3.78)/0.45 = 0.49
So,
Probability = P(2 < X < 3) = P(-1.73 < z < 0.49) = 0.6460
f) z score for 3 = (3 - 3.78)/0.45 = 0.49
z score for 5 = (5 - 2.78)/0.45 = 4.93
So,
Probability = P(3 < X < 5) = P(0.49 < z < 4.93) = 0.3125
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