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

Answer:

$P(\bar X$

And we can use the following z score:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

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

$P(\bar X$

C.0.8413

Step-by-step 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

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

$P(\bar X$

And we can use the following z score:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

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

$P(\bar X$

C.0.8413

Answer 2

Answer:

(C) 0.8413

Step-by-step 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