SOLUTION: I am stuck on this problem. I believe it is a Z formula, but cannot find the formula to do a Z with a variable "between". Here is the problem.
A normally distributed variabl
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A normally distributed variabl
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Question 911727: I am stuck on this problem. I believe it is a Z formula, but cannot find the formula to do a Z with a variable "between". Here is the problem.
A normally distributed variable with a mean of 16 minutes and a standard deviation of 4 minutes. what is the probability that a random sample of 25 will yield a mean between 14 and 18 minutes inclusive? What formula did you use?
Thanks to anyone that can help! Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! If you are Using a TI
P(14 ≤ x ≤ 18) = normalcdf(14,18,16,4)
Re TY
z1 = (14-16)/(4/sqrt(25)) = -10/4 = -2.5
z2 = (18-16)/(4/sqrt(25) )= 10/4 = 2.5
P(14 ≤ x ≤ 18) = P(z2) - P(z1)= .9938 - .0062
