Question 1046120: The heights of 18 year old American females are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
a. If 25 18 years old females are taken. Compute the mean and standard deviation of the sampling distribution.
b Find the z-score corresponding to a sample mean of 66 in. for a sample of 25 females.
c. Find the probability that a sample mean from a sample like this would be more that 66 in.
d. Based the the probability found in letter "c", would this sample be normal.
e. If a random sample of 25 18 year old female is selected, what is the probability that the mean height for this sample is between 63 and 65 in?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(xbar-mean)/sigma/sqrt (n)
a. The mean is 64 inches, and the sd is (3/5).
b. z=(66-64)/(3/5), the 5 being the sqrt (25).
z=3.33 (10/3)
c. Sample mean >66 is probability something is greater than z of (10/3). That is 0.0004.
d. It would be highly unlikely that such a sample would be normal or expected, since the probability of having this occur by chance is 4 in 10,000 or 1 in 2500.
e. That would be a z score of -1/3/5 to +1/3/5, or a z between -5/3 and +5/3, and that is 0.9044. As the sample size increases, the normal curve becomes compressed, with the same mean but a standard deviation that shrinks in proportion to the square root of the sample size.
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