Question 1120173: The heights of 18-year-old females are approximately normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
a. What is the probability that an 18-year-old woman selected at random is between 63 and 65 inches tall?
b. Suppose samples of 25 18-year-old females are taken at a time. Describe the sampling distribution of the sample mean and compute the mean and standard deviation of this sampling distribution.
c. Find the z-score corresponding to a sample mean of 66 inches for a sample of 25 females.
d. Find the probability that a sample mean from a sample like this would be higher than 66 inches.
e. Based on the probability found in the previous part, would a sample like this be unusual?
f. If a random sample of 25 18-year-old females is selected, what is the probability that the mean height for this sample is between 63 and 65 inches?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
a. (63-64)/1=-1/3 and for 65, it would be +1/3
Probability between those two values is 0.2611
b. The sampling distribution of samples of size 25 is normally distributed with mean 64 inches and sd 3/sqrt(25) or 0.6 inches. The sampling distribution of a given sample of size n has a sd of sigma/sqrt(n)
c. For 66 inches, the SD is +2/3, and that is divided by sqrt of 25 or 5, so the z score is (x-mean)/sigma/sqrt(n) which is (x-mean)*sqrt(n)/sigma, which is 2*5/3 or 3.33
d. This probability would be 0.0004
e which would be unusual.
f. This would be between-5/3 and +5/3 sd, since the sd of the sampling distribution is 0.6. The probability is 0.9044.
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