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Calculating the probabilities under a standard normal curve

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The time spent using e-mail per session is normally distributed, µ = 8 minutes and σ = 2 minutes. If you select a random sample of 25 sessions, then:

a. What is the probability that the sample mean is between 7.8 and 8.2 minutes?
b. What is the probability that the sample mean is between 7.5 and 8.2 minutes?
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 7.8 and 8.2 minutes?
d. Explain the difference in the results of (a) and (c)

2.
The amount of time a bank teller spends with each customer has a population mean,µ, of 3.10 minutes and standard deviation, σ , of 0.40 minute. If you select a random sample of 16 customers,
a) What is the probability that the mean time spent per customer is at least 3 minutes?
b) There is an 85 % chance that the sample mean is less than how many minutes?
c) What assumption must you make in order to solve a) and b)?
d) If you select a random sample of 64 customers, there is an 85% chance that the sample mean is less than how many minutes?

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Solution Summary

The solution gives detailed steps on calculating the probabilities under the assumption of normal distribution. All formula and calculations are shown and explained.

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1. First we need to know that Z= (Xbar-μ)/(σ/√n) where xbar is the sample mean, µ = 8, σ = 2 and n=25.

a) P(7.8 ≤ sample mean ≤ 8.2) = P ( (7.8-8)/(2/√25) ≤ Z ≤ (8.2-8)/(2/√25) )

P(-.2/.4 ≤ Z ≤ .2/.4) = P(-.5 ≤ Z ≤ .5) = .3830 From the Z table

b) P(7.5 ≤ sample mean ≤ 8) = P((7.5-8)/(2/√25) ≤ Z ≤ (8-8)/(2/√25) )

= ...

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