Download Statistics for Engineers Tutorial-03

Statistics for Engineers
Tutorial-03-Solution
Problem 3.1- Assume that the population of human body temperatures has reported to have a
mean of 37.0 oC and standard deviation of 0.62 oC. If a sample of size 108 is randomly selected,
find the probability of getting a mean of 36.4 oC or lower. (The value of 36.4 oC was actually
obtained).
Solution:
The distribution of the population was not given, but because the sample size 108 exceeds 30, we
use the central limit theorem and conclude that the distribution of sample means is a normal
distribution with the following parameters:
𝜇𝑋 = 𝜇 = 37.0
𝜎𝑋 =
𝜎
𝑛
=
0.62
108
= 0.059659
We can now find the area corresponding to the probability we seek.
𝑧=
𝑥 − 𝜇𝑋 36.4 − 37.0
=
= −10.06
𝜎𝑋
0.059659
𝑃(𝑍 < −10.06) ≈ 𝑃(𝑍 < −3.9) = 0.000033
The results shows that if the mean of our body temperature is really 37.0 oC, then there is an
extremely small probability of getting a sample mean of 36.4 oC or lower when 108 subjects are
randomly selected.
Either the population mean really is 37.0 and the sample represents a chance event that is
extremely rare, or the population mean is actually lower than 37.0 oC so the sample is typical.
Because the probability is so low, it seems more reasonable to conclude that the population mean
is lower than 37.0 oC.
Problem 3.2- The amount of nicotine in Dytusoon cigarettes has a mean of 0.941 g and a
standard deviation of 0.313 g. The Huntington Tobacco Company, which produces Dytusoon
cigarettes, claims that it has now reduced the amount of nicotine. The supporting evidence
consists of a sample of 40 cigarettes with a mean nicotine amount of 0.882 g.
3.3.a- Assuming that the given mean and standard deviation have not changed, find the
probability of randomly selecting 40 cigarettes with a mean of 0.882 g or less.
3.3.b- Based on the result from part (a), is it valid to claim that the amount of nicotine is
lower (Is it a rare event)? Why or why not?
Solution:
The distribution of the population was not given, but because the sample size 40 exceeds 30, we
use the central limit theorem and conclude that the distribution of sample means is a normal
distribution with the following parameters:
𝜇𝑋 = 𝜇 = 0.941
𝜎𝑋 =
𝜎
𝑛
=
0.313
40
= 0.04949
We can now find the area corresponding to the probability we seek.
𝑧=
𝑥 − 𝜇𝑋 0.882 − 0.941
=
= −1.192160
𝜎𝑋
0.04949
𝑃(𝑍 < −1.192160) = 0.1166
3.3.b- No, because the probability of 0.1170 shows that it is easy to get a mean such as
0.882 g, assuming that the nicotine amounts have not been changed.