(Open This Document in 'Page Layout' View!) Class Days and Time

251solngr3-081 4/15/08

Graded Assignment 3 Name:

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

1. Find for the following distributions (Use tables in Binomial and Poisson problems.). You will have to substitute one distribution for another in some cases and will get at least one zero probability.

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

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

c. Continuous Uniform with

d. Binomial Distribution with .

e. Binomial Distribution with .

f. Binomial distribution with (Approximate Solution)

g. Geometric Distribution with

h. Poisson Distribution with parameter of 12.

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

j. Hypergeometric Distribution with (Approximate Solution)

k. For a Hypergeometric Distribution with , find . (Extra credit if you actually evaluate this.)

l. (Extra credit) Findfor an exponential distribution with .

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

a. Continuous Uniform Distribution with .

b. Binomial Distribution with .

c. Geometric Distribution with

d. Poisson Distribution with parameter of 12.

e. Hypergeometric Distribution with .

f. Compare means and standard deviations for a Binomial distribution with and a Hypergeometric Distribution with .

g. (Extra credit) Exponential distribution with

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

a. At a convention you observe that the waiting time for an elevator has a continuous uniform distribution between 0 and 20 minutes.What is the probability that you will wait more than 15 minutes?

b. Assume that the distribution in a) applies to your waiting time when you go down to breakfast on the four mornings of the convention. What is the chance that you will wait for more than 15 minutes all four mornings? 3 out of four mornings? (Use the probability from the last problem with another distribution.)

c. Assume that the distribution in a) applies to your waiting time when you go down to breakfast on the four mornings of the convention. What is the chance that the first morning on which you will wait more than 15 minutes is the fourth morning? The third? The second? The first?

d. Assume that you know that there are 4 errors in each 1000 lines of computer code. What is the mean number of errors you would expect in a job with 2500 lines? What is the probability of over 15 errors in 2500 lines?

e. A bank can serve up to 10 customers in a 15 minute period without a significant delay. The average number of customers in a 15-minute period is 7.5. What is the probability that there will be fewer than 5 customers in a 15-minute period? What is the probability that there will be a significant delay in a given 15 minute period? In view of the rather high probability of having a significant delay , the bank wants to revise its procedures so the chance of a significant delay is below 1%, in other words to raise the number of customers it can process in 15 minutes above 10 so that the chance of a significant delay is below 1%. Using your tables what does that number have to be?

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 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. Suppose I claim that 30% of a shipment of batteries is defective.

(i) Assume that the shipment only consists of 10 batteries and I look at four. What is the probability that at least one is defective? What is the probability that at least two are defective?

(ii) Assume that the shipment consists of many batteries and I look at four. What is the probability that at least one is defective? What is the probability that at least two are defective?

(iii) Assume that the shipment consists of a humongous number of batteries (more than 1200). I take a sample of 60 and want to figure out the probability that more than half will be defective. I do not have a table that goes up to , but I do have a Poisson table with a parameter of 18. What would I look for on this table? Why can’t I use it?

You will find out soon that the Normaldistribution will work.

h. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the jorcillator only works as long as both components work. The life of each component has a continuous uniform distribution between zero and 12 months., so the jorcillator will not last more than one year. What is the probability that it dies before the end of June?

i. (Extra credit) According to Bowerman and O’Connell, 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 4 minutes.

j. A coin is tossed 5 times. Define the following events. ,

. (i) Find - Use a table in all three parts of this problem. (ii) Find . (iii) Find .

Solution to Question 1

1. Find for the following distributions.(Use tables in Binomial and Poisson problems.) You will have to substitute one distribution for another in some cases and will get at least one zero probability.

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

. In the diagram below, shade the area between 11 and 15. (There is no area between 15 and 17.) The horizontal line has a height of .

0 11 15 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 12 and 17. There is no probability between 11 and 12. The horizontal line has a height of .

0 11 12 17 20

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

c. Continuous Uniform with

If you make the box as above, and then try to shade the area between 11 and 17, you will find

d. 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 .

e. 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 - .65 = .35.) 11 successes correspond to 20 – 11 = 9 failures out of 20 tries. 17 successes correspond to 3 failures. So try 3 to 11 successes when

and .

f. Binomial distribution with (Approximate Solution)

can't be done directly with our tables, since is missing. However, note that is small, so that . The test for using a Poisson distribution is that if this ratio is above 500 we can use the Poisson distribution with a parameter of

g. Geometric Distribution with

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

h. Poisson Distribution with parameter of 12.

i. 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 11 to 17.

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 the outline. The formula is and with and this becomes . So , etc. So if we compute , the rest is fairly easy. 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.

j. Hypergeometric Distribution with (Approximate Solution)

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

k. For a Hypergeometric Distribution with , find . (Extra credit if you actually evaluate this.)

We already have and so and

. Anybody who stuck this one out deserves a virtue badge with fronds.

l. (Extra credit) Findfor an 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 .

Solution to Question 2

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 20 tries, when the probability of a success in an individual try is 60% is 12.

c. Geometric Distribution with

If , so. If we play a game repeatedly and the probability of a success in an individual try is 10% , on the average our first success will occur on the 10th try.

d. Poisson Distribution with parameter of 12.

,, so .

e. Hypergeometric Distribution with .

If and then and .So. Note that this variance is about 20% of the variance of a corresponding Binomial distribution. If we take a sample of 20 from a population of 100 of which 45% are successes, on the average we will get 9 successes.

f. Compare means and standard deviations for a Binomial distribution with and a Hypergeometric Distribution with .

If and , then for both distributions. For the binomial distribution so that . For the Hypergeometric distribution,

.So . There is a difference of less than 1% between these two. 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 in place of the hypergeometric distribution here.

g. (Extra credit) Exponential distribution with ,

We have and .

Solution to Question 3

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

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. At a convention you observe that the waiting time for an elevator has a continuous uniform distribution between 0 and 20 minutes. What is the probability that you will wait more than 15 minutes?

We have a continuous uniform distribution with with (Make a diagram!).

. In the diagram below, shade the area between 15 and 20. The horizontal line has a height of .

0 15 20

So .

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

This is a Binomial distribution problem with and . The binomial table with follows.

n x .01 .05 .10 .15 .20 .25 .30 .35 .40 .45 .50

4 0 0.96060 0.81451 0.6561 0.52201 0.4096 0.31641 0.2401 0.17851 0.1296 0.09151 0.0625

1 0.99941 0.98598 0.9477 0.89048 0.8192 0.73828 0.6517 0.56298 0.4752 0.39098 0.3125

2 1.00000 0.99952 0.9963 0.98802 0.9728 0.94922 0.9163 0.87352 0.8208 0.75852 0.6875

3 1.00000 0.99999 0.9999 0.99949 0.9984 0.99609 0.9919 0.98499 0.9744 0.95899 0.9375

4 1.00000 1.00000 1.0000 1.00000 1.0000 1.00000 1.0000 1.00000 1.0000 1.00000 1.0000

and .

This is a geometric distribution problem with . So we can write , , , . I should have asked what the probability is that you will never wait. Remember that , and . But why does this look so much like on the table above?

Since you were given an average number of hits per unit space and you want the probability of a certain number or number of hits you have a Poisson distribution. If there are an average of 4 errors in 1000 lines, there must be an average of in 2500 lines. According to the piece of the table below, if the parameter is 10,

Poisson 9.0Poisson 9.5Poisson 10.0

k P(x=k) P(xk) k P(x=k) P(xk)k P(x=k) P(xk)

0 0.000123 0.00012 0 0.000075 0.00007 0 0.000045 0.00005

1 0.001111 0.00123 1 0.000711 0.00079 1 0.000454 0.00050

2 0.004998 0.00623 2 0.003378 0.00416 2 0.002270 0.00277

3 0.014994 0.02123 3 0.010696 0.01486 3 0.007567 0.01034

4 0.033737 0.05496 4 0.025403 0.04026 4 0.018917 0.02925

5 0.060727 0.11569 5 0.048266 0.08853 5 0.037833 0.06709

6 0.091090 0.20678 6 0.076421 0.16495 6 0.063055 0.13014

7 0.117116 0.32390 7 0.103714 0.26866 7 0.090079 0.22022

8 0.131756 0.45565 8 0.123160 0.39182 8 0.112599 0.33282

9 0.131756 0.58741 9 0.130003 0.52183 9 0.125110 0.45793

10 0.118580 0.70599 10 0.123502 0.64533 10 0.125110 0.58304

11 0.097020 0.80301 11 0.106661 0.75199 11 0.113736 0.69678

12 0.072765 0.87577 12 0.084440 0.83643 12 0.094780 0.79156

13 0.050376 0.92615 13 0.061706 0.89814 13 0.072908 0.86446

14 0.032384 0.95853 14 0.041872 0.94001 14 0.052077 0.91654

15 0.019431 0.97796 15 0.026519 0.96653 15 0.034718 0.95126

16 0.010930 0.98889 16 0.015746 0.98227 16 0.021699 0.97296

17 0.005786 0.99468 17 0.008799 0.99107 17 0.012764 0.98572

18 0.002893 0.99757 18 0.004644 0.99572 18 0.007091 0.99281

19 0.001370 0.99894 19 0.002322 0.99804 19 0.003732 0.99655

20 0.000617 0.99956 20 0.001103 0.99914 20 0.001866 0.99841

21 0.000264 0.99983 21 0.000499 0.99964 21 0.000889 0.99930

22 0.000108 0.99993 22 0.000215 0.99985 22 0.000404 0.99970

23 0.000042 0.99998 23 0.000089 0.99994 23 0.000176 0.99988

24 0.000016 0.99999 24 0.000035 0.99998 24 0.000073 0.99995

25 0.000006 1.00000 25 0.000013 0.99999 25 0.000029 0.99998

26 0.000002 1.00000 26 0.000005 1.00000 26 0.000011 0.99999

27 0.000001 1.00000 27 0.000002 1.00000 27 0.000004 1.00000

28 0.000000 1.00000 28 0.000001 1.00000 28 0.000001 1.00000

This is a Poisson distribution with a parameter of 7.5. You can find it in your book of tables. . The probably of a significant delay is . As you crawl down the Poisson table with a mean of .75, note that and , so just tool up so that we can serve up to 15 people.

This is a binomial problem. We want when and We should test to see if we can use the Poisson distribution before we start, because in cases where probabilities are very small and the sample size is large (or ) we can substitute the Poisson distribution for the binomial. Note that is above 500. The mean is . According to the Poisson(1.8) table. This is an extremely small probability and should make us suspect that something is wrong. What you are doing here is a hypothesis test. Usually we reject a hypothesis if the probability of getting the results we observe or something more extreme is below 5% (or 1% if you think 5% is too high). In this case we have tested the hypothesis that and found it extremely unlikely.

g. Suppose I claim that 30% of a shipment of batteries is defective.

(i) Assume that the shipment only consists of 10 batteries and I look at four. What is the probability that at least one is defective? What is the probability that at least two are defective?

This is a Hypergeometric problem. , and

We know so , and .This means and .

(ii) Assume that the shipment consists of many batteries and I look at four. What is the probability that at least one is defective? What is the probability that at least two are defective?

If we use the binomial distribution with and . and .

You will find out soon that the Normal distribution will work.