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L. Discrete Distributions.

1. Binomial Distribution.

L1. Text 5.18-5.21, 5.23, 5.27[5.14a-d, 5.15, 5.16, 5.19, 5.20, 5.21] (5.15, 5.16, 5.18*, 5.19, 5.20).

2. Geometric Distribution.

L2-L5.

3. Poisson Distribution.

L6-L8. Text 5.30, 5.31, 5.33, 5.34, 5.35, 5.38[5.29 – 5.30, 5.34 - 5.37] (5.22 - 5.23, 5.26-5.29).

4. Hypergeometric Distribution.

Text 5.44-5.45, 5.49-5.50 [5.22 – 5.23, 5.27 – 5.28] (5.30, 5.31, 5.35, 5.36). L9, L10.

5. Review.

Note that problem L11 is a review of the continuous uniform distribution.

This document includes Binomial and Geometric distribution problems.

Binomial Distribution Problems

Exercise 5.18 [Exercise 5.14a-c in 8th and 9th]: Do the following .

a) when and . Solution: If we use the formula and recall that and , then .

b) when and . Solution: If we use the binomial table,

c) when and . Solution: If we use the binomial table,

d) when and . Solution: If we use the formula ,

e) (in 9th edition) when and . Solution: If we use the binomial table, . But we are in trouble because we do not have a table for . But 9 successes out of 10 tries means one failure out of 10 tries and the probability of failure is . . Since I don’t have an answer book for this problem, I will also do

Exercise 5.19 [Not in 8th and 9th]: If and find the following.

a) . Solution:

Problem L1: Assume that x has the binomial distribution with 10 tries and that the probability of

success on any one try is 25%. Find the following:

b. ,

c. The probability that is at least 1, .

d. If the probability of success on any one try is .75, and , what is ?.

e. Using the cumulative binomial table find the following:

(i) when ,

(ii) when ,

(iii) when ,

(iv) when ,

f. If ten people are selected at random from a large (continuous) population what is the probability that at least one makes more than the median income?

g. In f., what is the probability that between two and eight make more than the median

income?

h. What is the probability that all make more than the median income.?

i. For a binomial distribution with , , find the smallest symmetrical region

about the mean that contains (at least) 95% of the probability.

Solution: If and , we can write

a) If we use the formula and recall that and , then . You could also use the table: .

b) The formula for the mean is, (where ; in this case ), so =2.5, .

c) The probability that is at least 1, is the probability that is any number equal to or above 1 and is thus the sum of . But, remember, that 'at least one' is the complement of 'none', so .

d) and . .This is the same as the answer for a) and could be gotten from the table the same way. To see why, read on!

e) (i) when , .

(ii) when , . According to the table, this is .99363.

(iii) when , The outline says that for any discrete distribution (with integer

values for ) . So

(iv) when , . You cannot find this on the table, because you can use the

probability of failure, .3, instead of .7. In 10 tries, 3 successes correspond to 7 failures and 9 successes,

correspond to 1 failure. So find when , .

f) If people are selected from a relatively large population, the probability that each individual has an income above the median is .5. Since there are 10 people, .

g) again. .

h) again. .

i) For we are supposed to find the smallest symmetrical region about the mean that contains (at least) 95% of the probability. The particular intervals that we must include, should have equal probabilities above and below them. Since this means that 5% must be split between these two areas, we want so that and For 1, and . On the high side, if we try , , , . So the lowest number we can use is 1 and the highest number we can use is 7. The interval we have found is hardly symmetrical, but it is the best we can do.

Exercise 5.20 [Exercise 5.15 in 8th and 9th]: Find the mean and standard deviation if has the binomial distribution and a) and , b) and , c) and and d) and

Solution: According to the outline, for the binomial distribution , where is between zero and one and is the probability of success on an individual try, is the probability of failure and is the number of tries. The formula for the mean is.The variance is .

a) and So , , and .

b) and So , , and .

c) and So , , and

d) and So , , and

Exercise 5.21 [Exercise 5.16 in 8th and 9th]: Assume that there is a 50% chance of a stock yielding an increased closing price on any given day. What is the probability of an increase in 5 consecutive days?

Solution: This answer is copied from the Instructor’s Solutions Manual and is unchecked. Given p = 0.5 and n = 5, P(X = 5) = 0.0312. You should be able to get this by looking up and on the binomial table and subtracting them.

(Exercise 5.18 In 8th edition only): Suppose that a random sample of web surfers is selected and ask if they remember a specific internet banner. Use the binomial model and assume that

a. What is the expected value or mean of this binomial distribution?

b. What is the standard deviation of this binomial distribution?

c. What is the probability that none of the surfers will remember the banner ad?

d. What is the probability that exactly one of the surfers will remember the banner ad?

e. What is the probability that exactly two of the surfers will remember the banner ad?

f. What is the probability that three or more of the surfers will remember the banner ad?

Solution: and

b) .

c) . Better, if we use Table 15, the Cumulative Binomial Distribution, .

d ). Better, if we use the Cumulative Binomial Table ,

e) . Better, if we use the Cumulative Binomial Table ,

f) To find , compute . Better, using the table, find

[Exercise 5.19 in 8th and 9th edition only] A telephone company has a 70% chance of making a same day repair. For the first five repair requests find: a) the probability that all five repairs will be made on the same day; b) the probability that at least 3 repairs will be made on the same day; c) the probability that fewer than two repairs will be made on the same day. d) In a) – c) what assumptions are required? e) what are the mean and standard deviation of the distribution? f) Repeat a) – c) and e) with an 80% probability and e) compare the results with a) – c) and e).

Solution: This is a binomial problem with a sample size of 5 and a probability of success of .7.

a) All five repaired on the same day: To do this using the table, note that we cannot use , because the table only goes up to 5. But, if we want 5 successes out of 5 tries, we want 0 failures. The probability of failure is Look up on the Cumulative Binomial Table with and , and find .16807.

b) At least 3 repaired on the same day: We want , actually, the probability of between 3 and 5 successes. But, if we use the .3 column of the table again, we need to know that 3 successes correspond to 2 failures and that 5 successes correspond to 0 failures. We look up on the Cumulative Binomial Table with and , and find .83692.

c) Fewer than 2 repaired on same day: We want , actually between 0 and one successes. But, if we use the .3 column of the table again, we need to know that 1 success corresponds to 4 failures and that 0 successes corresponds to 5 failures. We look up .

Instead of answering the above, the Instructor’s Solutions Manual provides the following Excel table.

d) According to the Instructor’s Solutions Manual there are two assumptions:

(1) Independence of the repairs, (2) Only two outcomes - repair accomplished same day or repair not accomplished same day.

e) .

f) Now and Again, we must think in terms of failures instead of successes. The probability of failure is now .2.

(a) 5 successes corresponds to 0 failures and, on the Cumulative Binomial Table with and , we find

(b) We want , actually, the probability of between 3 and 5 successes. But, if we use the .2 column of the table again, we need to know that 3 successes correspond to 2 failures and that 5 successes correspond to 0 failures. We look up on the Cumulative Binomial Table with and , and find .94208.

(c) We want , actually between 0 and one successes. But, if we use the .2 column of the table again, we need to know that 1 success corresponds to 4 failures and that 0 successes corresponds to 5 failures. We look up .

(e) .

g) According to the Instructor’s Solutions Manual, “The larger p is, the more likely it is that

troubles reported on a given day will be repaired on the same day, and the less likely it is that troubles reported on a given day will not be repaired on the same day.”

Exercise 5.23 [Exercise 5.20 in 8th and 9th]: A student takes a multiple choice exam with 5 questions, each with four answers a, b, c and d. For each question she picks an answer randomly so that each answer has an equal chance of being picked. Find the following probabilities: a) five questions correct; b) at least four questions correct; c) no questions correct and d) no more than two questions correct.

Solution: This is a binomial problem with a sample size of 5 and a probability of success of .25. All answers are taken from the binomial table.

(a) P(X = 5) = = 0.0010

(b) P(X 4) =

(d) P(X 2) = 0.8965

Exercise 5.27 [Exercise 5.21 in 8th and 9th]: On one day 20 customers place orders. Each has a 5% probability of exceeding their credit limit. Find a) the mean and standard deviation of the number of customers exceeding the limit, b) the probability that no customers will exceed the credit limit c) the probability that one customer exceeds the credit limit and d) the probability that at least two will exceed the limit.

Solution: This is a binomial problem with a sample size of 5 and a probability of success of .05. All answers are copied from the Instructor’s Solutions Manual and are unchecked. If you cannot get them, please tell me.

(a) mean = 1 standard deviation = 0.9747

(b) P(X = 0) = 0.3585

(d) P(X 2) = 0.2642

Geometric Distribution Problems

Problem L2:.a. If , what is the chance of the first success on the 9th try? The 5th? The first?

b. If , what is the mean number of tries to a success? The variance?

c. What is the probability that the first success occurs between the 3rd and the 7th try?

Solution: The outline gives the formula: , which tells us the probability that the first success occurs on try . The cumulative distribution, which is used more frequently is . The mean is : . and the variance is .

a) If , and (Surprise?)

b) The mean is : . and the variance is

c) The cumulative distribution formula is , so .

Problem L3: a. If a bunch of fair coins are spilled and I pick them up randomly, what is the chance of the first head on try 1? Try 2? Try 3? If x is the first success, what is the probability that x is at least 3?

b. On the average, how many coins will I have to pick up to get a head?

Solution: a) and , , . In order to have our first success on the 3rd or later try, we must have failures on the 1st and 2nd try. , but remember that so .

Problem L4: a. If p = .01, Use the Chebyshef inequality to give a range in which three quarters of first successes must lie.

b. Try confirming this using the Cumulative probability for this distribution.