Graded Assignment 3 Name:Solution

251solngr3-071 4/10/07

Graded Assignment 3 Name:Solution

(Open this document in 'Page Layout' view!) Class days and time:

For full credit assignment must be neat and stapled and class day and time must be clearly indicated. Show your work! Look at the solved problems for Section L and ‘Great Distributions’ before you start.

1. Find for the following distributions (Use tables in c,d,f and h. All probabilities should show 4 places to the right of the decimal point):

a. Continuous Uniform with (Make a diagram!).

b. Continuous Uniform with (Make a diagram!).

c. Binomial Distribution with .

d. Binomial Distribution with .

e. Geometric Distribution with

f. Poisson Distribution with parameter of 15.

g. Show how you would do this for a Hypergeometric Distribution with Remember .

h. (Extra credit) Hypergeometric Distribution with

i. (Extra credit) Exponential distribution with .

j. For the hypergeometric distribution in g) find

2. Find the Mean and Standard deviation for the following distributions.

Some of these averages could be expressed in words – try it! (For example, The average number of successes when …… is ……)

a. Continuous Uniform Distribution with .

b. Binomial Distribution with .

c. Geometric Distribution with

d. Poisson Distribution with parameter of 15.

e. Hypergeometric Distribution with .

f. Hypergeometric Distribution with .

g. (Extra credit) Exponential distribution with .

3. Identify the distribution and do the following problems. Use tables when possible.

a. This problem uses all four discrete distributions.

(i) An IRS auditor is about to audit 6 returns from a group of returns for improper deductions.

Assume that these returns come from a group of 80 returns in which 25% have improper deductions, what is the probability that exactly one of the 6 has improper deductions?

(ii) What is the probability that at least one of the six has improper deductions?

(iii) If two or more of the 6 returns have improper deductions, the entire group of 80 will be audited. What is the chance that that happens?

(iv) Assume that the 6 returns come from a large group of returns (more than 120) that is 25% improper, and if two or more are improper, the whole group will be audited, what is the probability that the whole group will be audited? (Answers that use a specific population size will not be counted.)

(v) Assume that a large number of returns are audited from this group that is 25% improper, what is the chance that the second return is the first one found with improper deductions?

(vi) What is the chance that the first improper return will be among the first ten audited?

(vii) Assume that only 2% of the returns are improper and a sample of 100 taxpayers is taken, what is the chance that at least 3 have improper returns?

b. A fast food hamburger must be cooked a minimum of 90 seconds to be safe. At my lunch spot the time that burgers are on can be represented by a continuous uniform distribution with limits of 80 and 120 seconds. (i) What is the chance that a hamburger will be safe? (ii) What is the chance that, if I order 2 hamburgers both will be safe? (iii) What is the chance that if I order 3 hamburgers, at least one will be unsafe?

c. The number of students who come to my office has a Poisson distribution with a parameter (mean) of 2.5 a day. (i) What is the chance that no students come in a day? (ii) In a 5-day week? (iii) What is the chance that more than 5 come in a day (iv) In a 5-day week?

d. Complaints come in to my website at an average of seven an hour and take an hour to adjust. How many complaint handlers do I need to keep the probability of a complainer having to wait below 1%?

e. If 80% of shoppers at a website abandon their shopping carts without buying and there are currently 20 shoppers on my site, what is the chance that at least one will buy something? What is the average number that will buy something? What is the most likely number that will buy something?

f. According to Bowerman and O’Connell, in the movie Coma a young nurse finds that 10 of the last 30000 patients of a BostonHospital went into a coma during anesthesia. Upon investigation she finds that nationally 6 out of 100000 patients go into comas after anesthesia. Assume that the national probability is correct and that the Binomial distribution is appropriate, what is the mean number of patients out of 30000 that will go into a coma? In a hypothesis test, a researcher finds the probability of getting a number as large as or larger than what actually happened. So, find the probability of 10 or more people out of 30000 going into a coma. Obviously, we don’t have a binomial table that will do this. Show that this is an appropriate place to use a Poisson distribution and use the Poisson tables to find the probability that On the basis of your result, do you think that something strange is going on? (The answer is shorter than the problem.)

g. (Extra credit) A telemarketer finds that the length of a call has an exponential distribution with a mean of minutes. Find the probability that the length of a call will be (i) No more than 3 minutes, (ii) Between 1 and 2 minutes, (iii) More than 5 minutes.

1. Find for the following distributions (Use tables in c, d, f and h. All probabilities should show 4 places to the right of the decimal point):

a. Continuous Uniform with (Make a diagram!).

. In the diagram below, shade the area between 12 and 14. (There is no area between 14 and 16.) The horizontal line has a height of .

0 1 12 14 17

So .

Another way to do a problem of this type is to remember that for any continuous distribution, we can use differences between cumulative distributions, where the cumulative distribution is and and . If we use the cumulative distribution method, remember . So .

b. Continuous Uniform with (Make a diagram!).

. In the diagram below, shade the area between 15 and 17. There is no probability between 12 and 15. The horizontal line has a height of .

0 12 15 17 25

So . If we use the cumulative distribution method, remember and . So

Let’s try again!!!

Findwhen we have a Continuous Uniform distribution with (Make a diagram!). If you make the box as above, and then shade the area between 12 and 17, you will find

c. Binomial Distribution with .

. Remember that for a discrete distribution, the usual way to do a problem of this type is to remember that for any discrete distribution, we can use differences between cumulative distributions, where the cumulative distribution is .

d. Binomial Distribution with .

can't be done directly with tables that stop at , so try to do it with failures. (The probability of failure is 1 - .85 = .15.) 12 successes correspond to 25 – 12 = 13 failures out of 25 tries. 17 successes correspond to 8 failures. So try 8 to 13 successes when

and .

e. Geometric Distribution with

Remember that , because success at try or earlier implies that there cannot have been failures on the first tries.

f. Poisson Distribution with parameter of 15.

g. Show how you would do this for a Hypergeometric Distribution with

Remember .

We have no cumulative tables or cumulative distribution formula for this distribution, so the only available method is to add together probabilities over the range 10 to 15.

and so

and . This is certainly the answer that I expected, but we could take this further by using the recursive formula at the end of theoutline. The formula is and with and this becomes . So , etc. So if we compute , the rest is fairlyeasy.On the other hand, as you will see below, even computing , which is easier than computing is a pain and a good demonstration of why we have computers. The computer says that = 0.998852 - 0.549357 = .44950

h. (Extra credit) Hypergeometric Distribution with

Since is more than 20 times we can use the binomial distribution with and .

Note that if we actually computed this using the hypergeometric distribution, we would get

i. (Extra credit) Exponential distribution with .

In ‘Great Distributions I Have Known, we have the following information.

Exponential / is usually the amount of time you have to wait until a success. / and
when and the mean time to a success is . Both are zero if . / /

Since this is a continuous distribution,

Note: and .

j. For the hypergeometric distribution in g) find

. and . So

=.000001134

So the answer is .999998866. Anybody who stuck this one out deserves a virtue badge with fronds. The computer just says the answer is zero.

2. Find the Mean and Standard deviation for the following distributions.

Some of these averages could be expressed in words – try it! (For example, The average number of successes when …… is ……)

a. Continuous Uniform Distribution with .

b. Binomial Distribution with .

(Remember that and that both and must be between 0 and 1.) , so . The average number of successes in 25 tries, when the probability of a success in an individual try is 45% is 11.25.

c. Geometric Distribution with .

If , so . When the probability of a success in an individual try is 15% and we play a game repeatedly, on the average our first success will occur between the 6th and 7th try.

d. Poisson Distribution with parameter of 15.

,, so

e. Hypergeometric Distribution with .

If and then and so . If we take a sample of 25 from a population of 80 of which 45% are successes, on the average we will get 11.25 successes.

f. Hypergeometric Distribution with .

If and then and so . Note that the finite population correction factor in the variance formula has relatively little effect this time, which is why we can justify using the binomial distribution here.

g. (Extra credit) Exponential distribution with .

We have and .

3. Identify the distribution and do the following problems.

Remember:

(i) If you are looking for numbers of successes when the number of tries is given

and the probability of success is constant, you want the Binomial

distribution. This distribution can also be used to replace the Hypergeometric when the population size is large .

(ii) If you are looking for the try on which the first success occurs out of many

possible tries when the probability of success is constant, you want

the Geometric distribution.

(iii) If you are looking for numbers of successes when the average number of

successes per unit time or space is given, you want the Poisson

distribution. This distribution can be used to replace the binomial when the population is large and the probability of success is small.

(iv) If you are looking for numbers of successes when the number of tries is given

and the probability of success is not constant because the total number

of successes in the population is limited, you want the Hypergeometric

distribution.

a. This problem uses all four discrete distributions.

(i) An IRS auditor is about to audit 6 returns from a group of returns for improper deductions.

Assume that these returns come from a group of 80 returns in which 25% have improper deductions, what is the probability that exactly one of the 6 has improper deductions?

20 out of 80 returns are improper. This is Hypergeometric because we want to know the probability of a number of ‘successes’, but the total number in the population is limited.

. and . If we wrongly used the Binomial distribution here, we would have found when and The binomial formula says , so if , but we would be better off using a binomial table with and .

(ii) What is the probability that at least one of the six has improper deductions?

. This can be argued more directly if we realize that on the first try, the probability of not getting an improper return is .If we do not get an improper return on our first try, there are now 59 ‘proper returns left out of a total remaining of 79, so on the next try the probability becomes and so on, so that on the last try the probability is . Since these are conditional probabilities, we can multiply them together to get . If we had wrongly used the Binomial distribution, we would have found when and If , but we would have been better off using a binomial table with and .

(iii) If two or more of the 6 returns have improper deductions, the entire group of 80 will be audited. What is the chance that that happens?

Hypergeometric: . Fortunately, we have already computed these.

Binomial: when and If we use the binomial table with and .

(v) Assume that a large number of returns are audited from this group that is 25% improper, what is the chance that the second return is the first one found with improper deductions?

This is Geometric because it asks about the probability of a first success.. ,

(vi) What is the chance that the first improper return will be among the first ten audited?

Because this asks about a first success, it is Geometric., so .

(vii) Assume that only 2% of the returns are improper and a sample of 100 taxpayers is taken, what is the chance that at least 3 have improper returns?

I’m assuming that we still are talking about a large population. If this is true the number of possible successes is essentially unlimited and this is Binomial. when and But we do not have the appropriate table. Unless we want to do the problem by hand, call on the Poisson distribution, since which is safely over 500. Our parameter is

(i) What is the chance that a hamburger will be safe?

Continuous Uniform: The cumulative distribution is and .You could make a continuous uniform distribution diagram, with a rectangle with height . The base of the rectangle goes from to Shade the area in the rectangle above 90. Its area will be

(ii) What is thechance that, if I order 2 hamburgers both will be safe?

Binomial: when and The binomial formula says , so if It’s probably not worthwhile to use the binomial table, since we don’t have one for But, if the probability of success is .75, the probability of failure is .25 and 2 successes is no failures. Sure enough, for and

(iii) What is the chance that if I order 3 hamburgers, at least one will be unsafe?

Binomial: when and Note that if the probability of a safe hamburger is .75, the probability of an unsafe one is .25. The binomial formula says , so if It’s probably not worthwhile to use the Binomial table, since we don’t have one for But, if the probability of success (a safe burger) is .75, the probability of failure is .25 and we want 1 or more failures. So we have the same thing from the table

c. The number of students who come to my office has a Poisson distribution with a parameter (mean) of 2.5 a day.

(i) What is the chance that no students come in a day?

From the Poisson table with ,

(ii) In a 5-dayweek?

From the Poisson table with ,

(iii) What is the chance that more than 5 come in a day ?

From the Poisson table with ,

(iv) In a 5-day week?

From the Poisson table with ,

d. Complaints come in to my website at an average of seven an hour and take an hour to adjust. How many complaint handlers do I need to keep the probability of a complainer having to wait below 1%?

Poisson. If we look at the table with parameter of 7, the first cumulative probability above 99% is We need to hire 14 people. In other words, we want a number, call it , so that . 14 is the smallest number on the table that has If we only had 13 complaint handlers

e. If 80% of shoppers at a website abandon their shopping carts without buying and there arecurrently 20 shoppers on my site,

What is the chance that at least one will buy something?

Binomial: , . . Note that it seems more natural to define a success as buying something than failing to buy something. The probability of buying is

Whatis the average number that will buy something?

Binomial: , .

What is the most likely number that will buy something?

If we check the rest of the table we find that no difference between subsequent values of the cumulative distribution is more than .15, so 4 seems to be the mode. Since the Binomial distribution is not usually symmetrical for low values of we do have to check individual probabilities. We should be able to do this by calculus since

should be differentiable, and the derivative could be set equal to zero, but I can’t remember how to differentiate . One of my manuals says that the mode is any integer value or values between and . This seems to work when the mean is not an integer.