Question

The IQ scores of MBA students follow a normal distribution with a population mean of 120 points and a population standard deviation of 12. A random sample of 36 MBA students is chosen.

1. What is the probability that a randomly chosen sample of 36 MBA students has an average IQ less than 115?

2. What is the 91st percentile of sample average IQ’s of size 36 taken from the population of MBA students?

3. Calculate the bounds that determine the middle 72% of sample average IQ’s of size 36 taken from the population of MBA students?

1. 120 + 0.5284
2. 120 + 0.72
3. 120 + 1.28
4. 120 + 2.16
5. 120 + 12.96

Answer 1

Let X follows normal distribution with mean ( ) = 120

and standard deviation = = 12

Let's find the mean and standard deviation of sample mean ( )

n = sample size = 36

so sample mean follows normal distribution with mean = = 120

and standard deviation of sample mean is  

a) Here we want to find P( < 115)

Let's use excel:

P( < 115) = "=NORMDIST(115,120,2,1)" = 0.0062

b)  Here we want to find 91st percentile ( P₉₁) of sample average IQ’s of size 36 taken from the population of MBA students

such that P ( <  P₉₁ ) = 0.91

Let's use excel:

P₉₁ = "=NORMINV(0.91,120,2)" = 122.68

c) let c = 0.72

therefore = 1 - c = 1 - 0.72 = 0.28

Let's find critical z ( z_c ) corresponding to probability 0.86 using excel:

z_c = "=NORMSINV(0.86)" = 1.08

So correct answer is 120 +/- 2.16 that is ( d) .