SOLUTION: This problem is giving me fits, because of the word "inclusive." I believe the set up for the problem is P(14<= u <= 18) with Sigma =4, N=25 and a mean of 16.
Restaurant custom
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-> SOLUTION: This problem is giving me fits, because of the word "inclusive." I believe the set up for the problem is P(14<= u <= 18) with Sigma =4, N=25 and a mean of 16.
Restaurant custom
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Question 912104: This problem is giving me fits, because of the word "inclusive." I believe the set up for the problem is P(14<= u <= 18) with Sigma =4, N=25 and a mean of 16.
Restaurant customers spend a mean of 16 minutes in the restaurant with a standard diviation of 4 minutes. What is the probability that a random sample of 25 customers will yield a mean of between 14 and 18 minutes "inclusive."
with a standard normal distribution.
Please provide formula used,
I tried z formula solving for 14-16/ sigma divided by the square root of N and then 18-16/ sigma divided by the square root of N, and then got the z value from the table and subtracted, but I am not sure that is correct.
Thanks to all who may be able to help! Answer by ewatrrr(24785) (Show Source):
