Question

The scores on a certain test are normally distributed with a mean score of 60 and a standard deviation of 5. What is the probability that a sample of 90 students will have a mean score of at least 60.527?

Answer 1

Answer:  approximately 0.1587

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Work Shown:

mu = 60 is the population mean

sigma = 5 is the population standard deviation

n = 90 is the sample size

Compute the z score for the raw score xbar = 60.527

z = (xbar-mu)/(sigma/sqrt(n))

z = (60.527-60)/(5/sqrt(90))

z = 0.527/(5/9.48683298050514)

z = 0.527/0.52704627669473

z = 0.999912196145243

z = 1.00  

I'm rounding to two decimal places as this is common in many z tables.

So P(xbar > 60.527) is approximately the same as P(Z > 1.00)

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Use a z table to find that

P(Z > 1) = 1-P(Z < 1)

P(Z > 1) = 1-0.8413

P(Z > 1) = 0.1587

You can get a more accurate answer with a calculator (eg: the "normcdf" function on a TI83 or TI84); however, most stats books use tables to help make sure all students are on the same page (in terms of the final answer).