Notes

Example 1
Let X
be a random variable with mean μ=20
and standard deviation σ=4
. A sample of size 64 is randomly selected from this population. What is the approximate probability that the sample mean X¯
of the selected sample is less than 19
?
Solution to Example 1
No information about the population distribution is given. However, the mean and the standard deviation of the population are given. The sample size n=64
is greater than 30
and we are asked a question related to the
sample mean, we therefore may use the central limit theorem to answer the above question.
According to the central limit theorem, the distribution of the sample mean X¯
is close to a normal distribution with the mean μX¯
and standard deviation σX¯
given by
μX¯=μ=20

σX¯=σn−−√=464−−√

We are looking for the probability P(X¯<19)

The Z-score Z
corresponding to X¯=19
is given by
Z=X¯−μX¯σX¯=19−20464−−√=−2

Use a table or a normal probability calculator to obtain the probability that the mean of the sample is less than 19
.
P(X¯<19)=P(Z<−2)≈0.0228


Example 2
In the first semester of the year 2003, the average return for a group of 251 investing companies was 4.5%
and the standard deviation was 1.5%
. If a sample of 40 companies is randomly selected from this group, what is the approximate probability that the average return of the companies in this sample was between 4%
and 5%
in the first semester of the year 2003?
Solution to Example 2
The population is made up of 251 companies with average (mean) return equal to 4.5%
with standard deviation equal to 1.5%

The sample is large enough: n=40(≥30)
. We are looking for the probability concerning the average (mean) return, we therefore may use the central limit theorem.
Let X¯
be the random variable representing the mean. According to the central limit theorem, the distribution of X¯
is close to a normal distribution with the mean and standard deviation given by
μX¯=μ=4.5%

σX¯=σn−−√=1.5%40−−√

We are looking for the probability P(4%<X¯<5%)

The Z-scores Z1
and Z2
corresponding to X¯1=4%
and X¯2=5%
, respectively, are given by

Z1=X¯1−μX¯σX¯=4%−4.5%1.5%40−−√≈−2.10818


Z2=X¯2−μX¯σX¯=5%−4.5%1.5%40−−√≈2.10818

Use a table or a normal probability calculator to obtain the probability that average return of the companies in the sample was between 4%
and 5%
.
P(4%<X<5%)=P(−2.10818<Z<2.10818)≈0.965



Example 3
A pension fund company carries out a study of a large group of mutual funds and find that their average return over a period of 5 years was 80%
with a standard deviation equal to 30%
. If a sample of 50
mutual funds is randomly selected from the group, what is the approximate probability that the sample had an average return greater than 90%
over the 5 year period?
Solution to Example 3
The question is related to the average (mean) return and the sample size n=50
is large enough (≥30)
, we may therefore use the central limit theorem.
Let X¯
be the random variable representing the mean of the sample. According to the central limit theorem, the distribution of X¯
is close to a normal distribution with the mean and standard deviation given by
μX¯=μ=80%

σX¯=σn−−√=30%50−−√

We are looking for the probability P(X¯>90%)

The Z-scores Z
corresponding to 90%
is given by
Z=90%−80%30%50−−√≈2.35702

Use a table or a normal probability calculator to obtain the probability that average return of the companies in the sample was greater than 90%
.
P(X>90%)=P(Z>2.35702)≈0.0092



Example 4
The daily number of tools produced by a company is 2000. The average length of the tools is 10
centimeters with a standard deviation equal to 0.3
centimeters. If a sample of 200
tools is selected at random, what is the approximate probability that the average length of the tools in the sample is within 0.05
centimeter of the average length?
Solution to Example 4
The question is related to the average (mean) length of the tool and the sample size n=200
is large enough, we may therefore use the central limit theorem.
Let X¯
be the random variable representing the average (mean) of the sample. According to the central limit theorem, the distribution of X¯
is close to a normal distribution with the mean and standard deviation given by
μX¯=μ=10

σX¯=σn−−√=0.3200−−−√

We are looking for the probability that X¯
is within 0.05
centimeter of the average length means we are looking for the probability: P(10−0.05≤X¯≤10+0.05)

The Z-scores Z1
corresponding to to X¯=10−0.05=9.95
is given by
Z1=9.95−100.3200−−−√≈−2.35702

The Z-scores Z2
corresponding to to X¯=10+0.05=10.05
is given by
Z2=10.05−100.3200−−−√≈2.35702

Use a table or a normal probability calculator to obtain the probability that the average length of the tools in the sample is within 0.05
centimeter of the average length.
P(9.96≤X¯<10.05)=P(−2.35702≤Z≤2.35702)≈0.9816



Example 5
An airplane has a capacity of 200 seats and a total baggage limit of 6000 kilograms. Assume the total weight X
checked by each passenger is a random variable with a mean of 28 kilograms and standard deviation 15 kilograms. If 200 passengers board a flight, what is the approximate probability that the total weight of their baggage will not exceed the limit?
Solution to Example 5
For the luggage of the 200 passengers not to exceed 6000 kilograms, the average of the weight X
checked by each passenger must not exceed 6000200=30
kilograms. Therefore the problem is reduced to find the probability: P(X¯<30)
where X¯
is the sample mean of the weight X
.
Since the sample size is n=200
, the distribution of X¯
is close to a normal distribution with
mean: μX¯=μ=28

standard deviation: σX¯=σn−−√=15200−−−√

We are looking for the probability that X¯
is less 30
written as P(X¯≤30)

The Z-scores Z
corresponding to to X¯=30
is given by
Z=30−2815200−−−√≈1.88561

Use a table or a normal probability calculator to obtain the probability that the total weight of their baggage will not exceed the limit.
P(X¯≤30)=P(Z<1.88561)≈0.9703