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Normal Distribution - Confidence Limits

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The mean amount of a purchase at the Smith Hardware store is $23.50 with a standard deviation of $5.00. The amounts purchased follow a normal distribution. 50 customers were sampled:
a.What is the likelihood the sample mean is at least $25.00?
b.What is the likelihood the sample mean will fall into the $22.50 to $25.00 range?
c.Within what limits will 90% of the sample means occur?

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The expert examines the normal distribution and confidence limits. Step-by-step and clear explanation of the solution.

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M = 23.50, SD = 5, N = 50

SE = SD/sqrt(N) = 5/sqrt(50) = 0.707

(a) x = 25

z = (x - M)/SE = (25 - 23.5)/0.707 = 2.12

P(x >= 25) = ...

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