Question 921812: I am desperate! please help!
The average salary for graduates entering the actuarial field is $40,000 ( ). If the salaries are normally distributed with a standard deviation of $5000 (σ = 5000), find the probability that:
a. An individual graduate (n = 1) will have a salary over $45,000: P(X > 45,000).
b. A group of nine graduates (n = 9) will have an average over $45,000: P( )
c. A group of nine graduates (n = 50) will have an average over $45,000: P( )
d. A group of nine graduates (n = 125) will have an average over $45,000: P( )
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! For normal distributions we use z-value to determine probabilities (Pr),
we are given that the mean is 40000 and standard deviation is 5000
a) Pr(X>45000) = 1 - Pr(X<45000)
calculate the z-value = (45000 - 40000) / 5000 = 1, consult z-tables
Pr(X<45000) = 0.8413
Pr(X>45000) = 1 - 0.8413 = 0.1587
b) with a sample space of 9 students, we calculate a new standard deviation
sample standard deviation = standard deviation / square root (n)
sample standard deviation = 5000 / square root (9) = 1666.666666667
Pr(X>45000) = 1 - Pr(X<45000)
calculate the z-value = (45000 - 40000) / 1666.666666667 = 3 , consult z-tables
Pr(X<45000) = 0.9987
Pr(X>45000) = 1 - 0.9987 = 0.0013
c)n = 50, for n > 40 use the population standard deviation
sample standard deviation = 5000
calculate the z-value = (45000 - 40000) / 5000 = 1 , consult z-tables
Pr(X>45000) = 1 - 0.8413 = 0.1587
d) n = 125, for n > 40 use the population standard deviation
sample standard deviation = 5000
calculate the z-value = (45000 - 40000) / 5000 = 1 , consult z-tables
Pr(X>45000) = 1 - 0.8413 = 0.1587
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