Question

1. Suppose the exam scores are normally distributed with a population mean of 77.2% and a standard deviation of 16.8%.

The highest mean: 84.5 The lowest mean: 70.4

What is the probability of a student getting a score of 90% or better?  (Round to four decimal places.)

What is the probability of a class of 22 students having a mean of 90% or better?  (Round to six decimal places.)

Explain, in your.own words, why the answers to these two questions are drastically different.

_____________

2. People generally believe that the mean body temperature is 98.60°F.

Researchers at a university took a simple random sample of 139 healthy individuals and found these results:

• The distribution is approximately normal
• The sample mean is 98.23°F
• The standard deviation is 0.64°F

We want to determine whether these sample results differ from 98.60°F by a significant amount. One way to make that determination is to study the behavior of samples drawn from a population with a mean of 98.60. In other words, we’re going to assume the population mean really is 98.60. We’re going to see what normal sample means would look like if the population mean really was 98.60.

What is the lowest sample mean you obtained? 98.48

Based solely on the simulations, would it be reasonable to obtain a sample mean of 98.23?

• yes
• no

If we assume the population mean is 98.60 and the standard deviation is 0.64, what is the probability of a single person having a temperature of 98.23 or less?  (Round to four decimal places.)

If we assume the population mean is 98.60 and the standard deviation is 0.64, what is the probability of obtaining a sample mean of 98.23 or less from a sample of 139 individuals?  (Round to six decimal places.)

What can we say about the assumption that the population mean is 98.60? Explain your answer completely. Keep in mind:

• The simulation you completed was under the assumption that the population mean is 98.60.
• The calculations you completed in the last two portions were assuming the population mean is 98.60.
• The sample that produced a mean of 98.23 degrees was the actual, real data.

Answer 1

#1.
mean = 77.2 and sd = 16.8

P(X > 90)
= P(z > (90 - 77.2)/16.8)
= P(z > 0.7619)
= 0.2231

P(X > 90)
= P(z > (90 - 77.2)/(16.8/sqrt(22)))
= P(z > 3.5737)
= 0.00017602

Answer in two cases are different because of the sample size used. The SE becomes s/sqrt(n)