Question

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12. What is the probability Professor Elderman’s class of 36 has a class average below 78?

A.0.5675
B.0.1587
C.0.8413
D.Cannot be determined.

Answer 1

(C) 0.8413

Explanation:

Test statistic (z) = (sample mean - population mean) ÷ (sd/√n)

sample mean = 78

population mean = 76

sd = 12

z = (78 - 76) ÷ (12/√36) = 2 ÷ 2 = 1

The cumulative area of the test statistic is the probability that Professor Elderman's class of 36 has a class average below 78. The cumulative area is 0.8413

Therefore, the probability is 0.8413

Answer 2

`P(\bar X <78)`

And we can use the following z score:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got using the normal standard distribution table or excel:

`P(\bar X <78)= P(Z<(78-76)/((12)/(√(36)))) = P(Z<1)=0.8413`

C.0.8413

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the records of a population, and for this case we know the distribution for X is given by:

`X \sim N(76,12)`

Where
`\mu=76` and
`\sigma=12`

Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We select a sample of size n=36, and we want to calculate this probability:

`P(\bar X <78)`

And we can use the following z score:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got using the normal standard distribution table or excel:

`P(\bar X <78)= P(Z<(78-76)/((12)/(√(36)))) = P(Z<1)=0.8413`

C.0.8413