Question

Part A: The average adult female is 60.35 inches tall with a standard deviation of 2.1999. Based on this information, 12.29% of adult females are less than what height? Assume the distribution is approximately normal.

Part B: Suppose that the probability of a baseball player getting a hit in an at bat is 0.3189. If the player has 26 at bats during a week, what's the probability that he gets at least 10 hits?

Part C: In a recent survey of 130 WMU graduates, 11 students said that parking was too limited on campus. What is the estimate of the population proportion? What is the standard error of this estimate?

Part D: Suppose that in the United States the typical adult male is 67.98 inches tall with a standard deviation of 6.718. You take a random sample of 50 adult males. What is the probability that the mean height of the sample is less than 67.18?

Answer 1

Part A: The average adult female is 60.35 inches tall with a standard deviation of 2.1999. Based on this information, 12.29% of adult females are less than what height? Assume the distribution is approximately normal.

Let X = Height of a adult female in inches

We want to find x score such that P( X < x ) = 0.1229

Formula of x score is as follows

From z table the z score corresponding to 0.1230 ( because it is close to 0.1229) is -1.16

therefore ,

If we use excel then we get

x = "=NORMINV(0.1229,60.35,2.1999)" = 57.79677

Part B: Suppose that the probability of a baseball player getting a hit in an at bat is 0.3189. If the player has 26 at bats during a week, what's the probability that he gets at least 10 hits?

Let X = number of hits

So X follows binomial distribution with n = 26 and p = 0.3189

Here we want to find P( X >= 10) = 1 - P( X < 10) = 1 - P( X <= 9 ) .........( 1 )

Using excel:

P( X <= 9 ) = "=BINOMDIST(9,26,0.3189,1)" = 0.700724

Plug this in equation ( 1 ), we get

P( X >= 10) = 1 -  0.700724 = 0.299276

Part C: In a recent survey of 130 WMU graduates, 11 students said that parking was too limited on campus. What is the estimate of the population proportion? What is the standard error of this estimate?

the sample proportion( ) is an estimate of population estimate

The formula of standard error of sample proportion ( ) is as follows :

Part D: Suppose that in the United States the typical adult male is 67.98 inches tall with a standard deviation of 6.718. You take a random sample of 50 adult males. What is the probability that the mean height of the sample is less than 67.18?

The mean of sample mean ( ) is same as mean of individual x = 67.98

The formula of standard deviation of sample mean ( ) is as follows:

We want to find  P( X < 67.18 )

P( X < 67.18 ) = "=NORMDIST(67.18,67.98,0.95,1)" = 0.1999