Population Distribution Problems; Sampling Distribution Problem; and Both.
Population (Normal)
If the average person watches 2 hours of TV per day with a standard deviation of 30 minutes, and the
distribution of TV watching is normal, what are the chances that a particular individual will watch more than
3 hours of TV per day?
 1-.9772 = .0228
 There is a 2.28 percent chance a person will watch TV for more than 3 hours a day
σ = .5
.9772
2
3
x
Sampling
If the average person watches 2 hours of TV per day with a standard deviation of 30 minutes, and the
distribution of TV watching is normal, what are the chances that the average of a random sample of 16
individuals will be more than 3 hours of TV per day?
day?
-- There is certainty that the average of the random sample will be less than 3 hours of TV per day
σ = .5
1
2
3
Population (Normal)
In the U.S., the average price of a gallon of regular gas last week was $1.55 with a standard deviation of $.10.
Assuming the distribution was normal what percentage of gas stations were selling gas for less than $1.30?
-- 0.6 percent of the gas stations are selling gas for less than $1.03.
Sampling
In the U.S., the average price of a gallon of regular gas last week was $1.55 with a standard deviation of $.10.
Would it be unusual to randomly select 36 gas stations and find an average less than $1.30? How about less
than $1.50?
There is no chance of finding an average of 130 in 36 stations.

-- It is still unusual (less than 1 %) to find average less than $1.50 in the 36 randomly selected gas
stations.
Population (Normal)
On average an officer in the Kennett Police Department issues 13 traffic tickets per week with a standard
deviation of 4 tickets. Officer Bob issued 20 tickets last week. Assuming the ticket distribution is normal is
this unusual?
-- This is unusual because 95.99 percent of the time under 20 tickets are written in a week. This puts
Bob in the top 5 Percent of ticket writers.
Sampling
On average an officer in the Kennett Police Department issues 13 traffic tickets per week with a standard
deviation of 4 tickets. Six officers averaged 17 tickets last week. Assuming the ticket distribution is normal is
this unusual?
-- 99 percent of the averages are under 17 tickets per week so this is very unusual.
Population
QVC averages 27 bad checks per day; yesterday they got 30 bad checks. Kevin thinks this is evidence of a
problem. Is he right?
There is a 31 percent chance of getting 30 or more bad checks. This is not particularly unusual.
Sampling
QVC averages 27 bad checks per day, Last month (over 30 days), they averaged 30 bad checks per day. Kevin
thinks this is evidence of a problem. Is he right? (Note the standard deviation of a Poisson distribution is the
square root of the average)
There is a 99% chance that the 30 day sample average would be less than 30, so this is definitely evidence of a
problem.
Explain why it would be impossible to answer this question accurately if the average was taken for February.
(If the sample is less than 30 and the population is not normal (poisson), then the sampling distribution
is not normal.
Harder Problems
George Jibberjaw is testing out a machine designed to put a product into a bag. He set the machine up and let it
run for a while then he sampled 49 bags to see how much was getting in to each bag. In the sample, the
average was 4.2 ounces with a standard deviation of .5 ounces. The next day he took another sample (49 bags)
and the average was only 4 ounces, but the standard deviation was still .5 ounces. In his report to the boss, he
states that based on his results the new machine is unreliable and definitely not a good investment. Is he right?
Explain why or why not.
if the average is 4, it is very unusual to see a sample of 4.2 ( only 3 percent of samples are that high). these
two samples are 2.85 standard errors apart. However if the true average were 4.1 (midway), then these two
samples would still be fairly unusual. Each would have less than 10 percent chance of happening.
. A local company searching for ways to save more money is considering leasing instead of buying the company cars for its sales
representatives. The rules and charges for leases vary and are often complicated, so depending on the number of miles driven; the
lease may or may not save money. Nationally, the cost per mile driven in any (owned or leased) vehicle is 28 cents with a standard
deviation of 7 cents. The company’s current costs are in line with the national average. It is still possible, however, that the lease
could be better for them. As a test, the company leases 40 cars for one year and finds the average cost per mile was 25 cents. The
company’s executives disagree as to what this indicates. Bob Spendalot, says that that the test average is pretty close to the their
previous average, and so the lease is not a better deal.
There is less than 1 percent chance that 40 cars would average 25 cents if the true average were 28 cents. This indicates that the lease
is the better deal.
1. A company in the U.S. is considering investing in a new small company in another country. One executive
is concerned about bankruptcy rates in that country. Unfortunately, the country does not keep accurate
bankruptcy rates. However, companies that exist are required to register with the government each year. A
person with some time can follow the companies until they disappear from the lists. Usually this means they
went bankrupt. The company does not have time to do this for every firm that ever existed, so they try a
sample. Of 36 randomly selected firms they followed through the lists, 4 were bankrupt within 10 years. In the
U.S. approximately 8 percent of all companies go bankrupt within 10 years. Does this indicate that the risk of
bankruptcy is higher than it is in the U.S.? Explain your answer.
oops you can’t do this one yet.
Answer only one of the following two questions (either 2a or 2b not both)
2a. (5 points) Without reference to a formula, explain why the standard error is less than the standard
deviation.
2b. (5 points) Compare and contrast sampling distributions and population distributions. What are they, how
are they the same and how are they different? I want you to think about this one
6. A pizza restaurant makes deliveries in an average of 45 minutes with a standard deviation of 6 minutes and
the delivery times are normally distributed. When a customer orders a pizza, how long should they say the
delivery will take if they only want to be late 25 percent of the time?
σ =6
.75
.25
45
45+.67(6) x
49.02
Tell them you will arrive in 49.02 minutes. (What the heck round it to 50 minutes)
To monitor performance, the manager occasionally samples 6 delivery times. What averages should he
expect, if the restaurant is performing as usual? What average might indicate a problem? Explain why a
small sample is o.k. in this case.
σ = 6/√6 =2.45
.90
45-1.64*2.45
41
45
1.64(2.45) x̄
49
If the average is 45, his sample should usually fall between 41 and 49. Anything slower than 49 is
indicative of a problem. (Faster than 41 may be a problem too. You may not want your drivers speeding)
On average, employees at QVC rack up 4.6 sick days per year. Would it be unusual to find an employee
out more than 7 days in one year?
Poisson Population
There is a less than a ten- percent chance that a person would be out more than 7 days
In a sample of 50 employees would it be unusual to see an average of 5 sick days per year?
σ = √4.6/√50=.30
1
4.6
5
x̄
There is a 90 percent chance that the average would be less than 5, so 5 and up is unusual.
In a sample of 5 employees would it be unusual to see an average of 5 sick days per year?
We can’t answer this question accurately because the sample is too small and the population is not normal.
7. The Federal Trade Commission mandates that the actual weight of a food product in any container should
not be less than 95 percent of the advertised weight. For example a sixteen-ounce box of Cheerios must
have at least 15.2 ounces inside. If a firm consistently violates this standard, the firm can be charged with
false advertising. One company, seizing on an opportunity to increase profits, adjusts its filling machine to
average 15.5 ounces with a standard deviation of .12 ounces. Assuming the weights are normally
distributed, what are the chances that a randomly selected box will violate FTC standards?
σ = .12
.006
15.2
15.5
x
There is less than one- percent chance that they will be caught.
Recognizing the potential for fraud, the FTC changes its policy so that the average of 30 boxes should not be
less than 98% of the advertised weight. Will this strategy catch the fraud perpetrated above?
σ = .12/√30=.02
1
15.5
15.68
(98% of 16)
x̄
Using this strategy, we can virtually guarantee that offenders will be caught.
8. Suppose a machine fills a package with 4 cookies. The weight of 4 cookies is normally distributed with a
mean of 12 ounces and a standard deviation of 1 ounce. What weight should the company advertise on the
package, so than 95 percent of the packages are above the advertised weight?
σ =1
.05
12-1.64(1)
10.36
12
x
If the advertised weight is 10.36 ounces, 95% of the packages will meet or exceed this weight.
Tim Tightfist is always wary of fraud. Lately he has been concerned with his toilet paper. To save money,
he has been buying a brand that advertises 1000 sheets. Just to make sure, he counts them. At the end of
his first roll he is alarmed to find there were only 978 sheets! First thing in the morning he calls the toll
free number 1-800-SWI-PEME. “There were only 978!” he yells “You didn’t think I’d count them, but I
did” “I’m sorry for the confusion,” replies the customer service rep. We get these calls all the time, she
lies. The 1000 sheets is just an average. I’m sure that if you keep counting we’ll measure up to our claim.
“Back to the drawing board so to speak,” he grumbles. Tim keeps counting, and several weeks later he has
counted the sheets on 10 different rolls and finds an average of 988.5 sheets. After a few quick calculations
he mutters “Now I’ve got the weasels!”
A short toll-free conversation later he’s back in the bathroom counting. What did the customer service rep
say this time? Tim was probably doing a calculation like this:
σ = √1000/√10 =10
.12
988.5
1000
x̄
Based on this analysis there is less than 13 % chance that 10 rolls would average less than 988.5 sheets
when the true average is 1000.Tim probably thought he had them dead to rights. No doubt, though, that
the customer service rep had a brilliant statistics instructor in college and she probably explained to Tim
that his analysis was based on the assumption that the population was normal. In this case, the population
probably follows a Poisson distribution.
What should Tim do next if he wants to uncover the fraud? What would constitute conclusive evidence of
malice aforethought on the part of the toilet paper magnates?
Tim needs to get back on the pot. Count the sheets on 30 rolls or more then we have a normal distribution
for the sample averages:
σ = √1000/√30 =5.77
.05
1000-1.64(5.77)
1000
x̄
990.5
If he finds a 30 roll average less than 990.5 sheets, then he’ll have significant evidence of Toilet paper
fraud!
10. A local Wal-Mart typically loses $578 per day to theft. The losses are normally distributed with a
standard deviation of $58. One month after hiring a new employee, the 30-day average was $585, and on
one of those days it was actually $650! The manager wants to call the new employee in and fire him. Do
you think there anything unusual about either of these figures?
The distribution of the sample averages should look like this if nothing has changed (μ= 578)
σ
578
585
= 58/√30 =10.5
x̄
So sample averages this large or larger will occur 25% of the time even if there is no problem. It’s too soon to
fire the new employee
As part of a master plan to reduce losses, by increasing employee vigilance, the manager is planning to
offer the staff a bonus if they show significant improvement (better than $578) over the next 30 days. What
would constitute significant evidence of improvement?
The distribution of the sample averages should look like this if nothing has changed (μ= 578)
σ = 58/√30 =10.5
.10
578-1.28(10.5)
570
578
x̄
If nothing has changed the sample average will most likely be more than 570. If the sample average is less
than 570, then this would indicate that the true average has probably dropped.