Homework #6 Solution. Fall 2000, IE 230 Textbook: D.C ...

[Pages:7]Homework #6 Solution.

Fall 2000, IE 230

Textbook: D.C. Montgomery and G.C. Runger, Applied Statistics and Probability for Engineers, John Wiley & Sons, New York, 1999. Chapter 4, Sections 4.6-4.9. Pages 7-8 of the concise notes.

This assignment requires the use of MSExcel, but does not require an electronic submission.

1. Suppose that we are doing an presidential-election survey, selecting people at random from the Purdue University student body.

For each part below, name the relevant probability distribution, state any known parameter values, and list any parameter values that need to be determined.

(a) (Let the experiment be to choose one person.) The last digit of the student ID number.

-----------------------------------------------------------------------------------------Discrete (equally likely) uniform distribution with range 0, 1,..., 9. In the Concise Notes, the parameters are a = 0, b = 9, and c = 1.

------------------------------------------------------------------------------------------

(b) (Let the experiment be to choose 50 students.) The number of chosen students who are registered to vote in Tippecanoe County.

-----------------------------------------------------------------------------------------Binomial with n = 50 and p = P(success), where "success" = "a randomly chosen student is registered to vote."

------------------------------------------------------------------------------------------

(c) (Let the experiment be to choose students until a registered voter is obtained.) The number of students chosen.

-----------------------------------------------------------------------------------------Geometric with p = P(success), where "success" = "a randomly chosen student is registered to vote."

------------------------------------------------------------------------------------------

(d) (Let the experiment be to choose students until ten registered voters are obtained.) The number of students chosen.

-----------------------------------------------------------------------------------------Negative Binomial, with r = 10 successes and p = P(success), where "success" = "a randomly chosen student is registered to vote."

------------------------------------------------------------------------------------------

(e) (Suppose that of 37,000 students, 3400 are registered to vote in Tippecanoe County. Let the experiment be to choose 50 students.) The number of chosen students who are registered to vote in Tippecanoe County.

-----------------------------------------------------------------------------------------Hypergeometric, with population size N = 37,000 students, sample size n = 50 students, number of successes K = 3400 students, where "success" = " a randomly chosen student is registered to vote."

------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

(f) (Let the experiment be to choose a random October hour of business for the Tippecanoe County voter-registration office.) The number of Purdue students who register to vote during the hour.

-----------------------------------------------------------------------------------------Poisson, with mean = "the expected number of Purdue students who register to vote during such an hour."

------------------------------------------------------------------------------------------

2. Result: The mean of the binomial distribution is E(X) = np, where nis the number of Bernoulli trials and p is the probability of success on each trial. Show that this result is correct for the special case n = 2. (Hint: Write the definition and algebraically simplify.) (You should also think about the same exercise for the result that the variance is V(X) = n p (1-p).).

-----------------------------------------------------------------------------------------For each step, be sure you understand the reason.

E(X) = xi f (xi) all i

n

=x

n

p x (1 - p)n -x

x =0 x

= 0 2 p 0 (1 - p)2-0 + 1 2 p 1 (1 - p)2-1 + 2 2 p 2 (1 - p)2-2

0

1

2

= 0 (1) (1) (1 - p)2 + 1 (2) p 1 (1 - p)1 + 2 (1) p 2 (1)

= 2 p (1 - p) + 2 p 2 = 2p .

V(X) = E(X 2) - [E(X)]2

=

x

2 i

f

(xi )

- E2(X)

all i

=

n

x2

n

p x (1 - p)n -x

- (np)2

x =0

x

= 02 2 p 0 (1 - p)2-0 + 12 2 p 1 (1 - p)2-1 + 22 2 p 2 (1 - p)2-2 - (2p)2

0

1

2

= 0 (1) (1) (1 - p)2 + 1 (2) p 1 (1 - p)1 + 4 (1) p 2 (1) - (4p 2) = 2 p (1 - p) + 4 p 2 - 4p 2 = 2p 2 - 2p = 2p(1 - p) ------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

3. (From Problem 4-64.) Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting. Assume that your calls are independent.

(a) What is the probability that your first call to connect is your tenth call?.

-----------------------------------------------------------------------------------------Geometric, with p = 0.02 where "success" = "a call connects" and X = "number of calls until connection".

Then P(X = 10) = (1 - p)10-1 p = 0.989 (0.02) = 0.0167. ------------------------------------------------------------------------------------------

(b) What is the probability that more than five calls are required to connect?

-----------------------------------------------------------------------------------------Geometric, with p = 0.02 where "success" = "a call connects" and X = "number of calls until connection".

Then P(X > 5) = 1 - P(X 5) = 1 -

5 x =1

(1

-

p)x -1

p

= 1 - p [1 + (1 - p) + (1 - p)2 + (1 - p)3 + (1 - p)4]

= 1 - (0.02) [1 + 0.98 + (0.98)2 + (0.98)3 + (0.98)4] = 0.904 ------------------------------------------------------------------------------------------

(c) What is the mean number of calls required to connect?

-----------------------------------------------------------------------------------------Geometric, with p = 0.02 where "success" = "a call connects" and X = "number of calls until connection".

Then E(X) = 1 / p = 1 / 0.02 = 50 calls. ------------------------------------------------------------------------------------------

(d) Suppose that both you and a friend are calling the radio station. Use MSExcel to compute the probability that the number of calls required for you both to connect is greater than 50. (In addition to the numerical answer, handwrite the cell code.)

-----------------------------------------------------------------------------------------Negative Binomial, with r = 2 connections and p = 0.02 where "success" = "a call connects" and X = "number of calls until rth connection".

Then

P(X

>

50)

=

1

-

P(X

50)

=

1

-

F (50)

=

1

-

50 x =1

f

(x ).

The MSExcel function NegBinomDist asks for the distribution parameters as the Number of Failures, the Number of Successes, and the Probability of Success, which is different from how we have parameterized the distribution. For X = x, these parameter values are x -2, 2, and 0.02, respectively. The MSExcel code for f (x) is then "=NegBinomDist(c10-2,2,0.02)", where (for example) cell c10 contains the value of x.

This function is for the pmf only, so you need to build the cumulative probability yourself within the spread sheet. To do this, simply compute f (x) for x = 1, 2,...,50 and sum. ------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

4. A popular gambling-casino game is Keno. On a sheet of paper, eighty rectangles, arranged in an 8 ? 10 array, are numbered 1 through 80, with numbers 1 through 10 in the first row. The gambler marks n rectangles on the paper. The paper and the gambler's bet are then given to a casino runner. A few minutes after the game is closed, TV screens show the 80 rectangles with exactly 20 rectangles lit up. The gambler wins if all of his or her n marked rectangles are lit. (The size of the payoff depends up on n, but that does not concern us now.)

(a) What is the probability of winning for n = 1?

-----------------------------------------------------------------------------------------Hypergeometric, with population size N = 80, sample size n = 1, and number of successes K = 20, where "success" is a lit number. The random variable is X = the number of successes in the sample."

Then P(X = 1) = 20 60 / 80 = (20)(0) / (80) = 1 / 4.

10

1

Alternatively, for n = 1, the distribution is the discrete (equally spaced) uniform distribution over the values {1, 2,..., 80},

where X = "the number marked". Let W = "the gambler wins." Then P(W) = P(X = a~lit~number) = 20 / 80. ------------------------------------------------------------------------------------------

(b) For n = 3, write the expression for the probability of winning. (You do not need to compute the numerical value.)

-----------------------------------------------------------------------------------------Hypergeometric, with population size N = 80, sample size n = 3, and number of successes K = 20, where "success" is a lit number. The random variable is X = the number of successes in the sample."

20 Then P(X = 3) =

60

/

80 = [(20 ? 19 ? 18) / (3 ? 2)] ?(1) = 0.0139.

30

3

[(80 ? 79 ? 78) / (3 ? 2)]

------------------------------------------------------------------------------------------

(c) For general n, write the expression for the probability of winning. (You do not need to compute the numerical value.)

-----------------------------------------------------------------------------------------Hypergeometric, with population size N = 80, sample size n, and number of successes K = 20, where "success" is a lit number. The random variable is X = the number of successes in the sample."

Then P(X = 3) = 20 60 / 80 .

n0

n

------------------------------------------------------------------------------------------

(d) Handwrite here the MSExcel cell code to compute the answer to Part (b).

-----------------------------------------------------------------------------------------"=HypgeomDist(3,3,20,80)"

------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

(e) For n = 1, 2,..., 20, compute the probability of winning. (Use MSExcel, but you can just handwrite a table of probabilities.)

------------------------------------------------------------------------------------------

x P(X = x) MSExcel code

1 0.250 "=HypgeomDist(1,1,20,80)" 2 0.601 "=Hypgeomdist(2,2,20,80)" 3 0.0139 ... 4 0.0031 5 0.00065 6 0.00013 7 2.44e-5 8 4.35e-6 9 7.24e-7 10 1.12e-7 11 1.6e-8 12 2.1e-9 13 2.5e-10 14 2.6e-11 15 2.3e-12 16 1.8e-13 17 1.1e-14 18 5.4e-16 ... 19 1.7e-17 "=HypgeomDist(19,19,20,80)" 20 2.8e-19 "=Hypgeomdist(20,20,20,80)" ------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

5. (From Problem 4-93.) The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your call waiting times are independent.

(a) If you call 10 times, what is the probability that exactly 9 calls are answered within 30 seconds.

-----------------------------------------------------------------------------------------Binomial, with n = 10 calls and p = P(success) = 0.75, where "success" is having a call answered in less than 30 seconds. Then X = "the number of calls answered in less than 30 seconds."

P(X = 9) = n p 9 (1 - p)n -9 = (10)(0.75)9 (1 - 0.75)1 = 0.188. 9

------------------------------------------------------------------------------------------

(b) Suppose that you call 20 times.

-----------------------------------------------------------------------------------------Binomial, with n = 20 calls and p = P(success) = 0.75, where "success" is having a call answered in less than 30 seconds. Then X = "the number of calls answered in less than 30 seconds."

P(X = 9) = 20 (0.75)9 (1 - 0.75))20-9. 9

From MSExcel, =BinomDist(9,20,0.75,False) = 0.0030. ------------------------------------------------------------------------------------------

(i) Write the expression for the probability that at least 16 calls are answered in less than 30 seconds?

------------------------------------------------------------------------------------------

20

P(X 16) =

20 (0.75)x (1 - 0.75))20-x.

x =16 x

------------------------------------------------------------------------------------------

(ii) Use MSExcel to compute the numerical value for Part (i). (Handwrite the cell code that you used, as well as the numerical answer.)

-----------------------------------------------------------------------------------------P(X 16) = 1 - P(X 15) = 1 - BinomDist(16,20,0.75,True)

------------------------------------------------------------------------------------------

(c) If you call 20 times, what is the mean number of calls that are answered in less than 30 seconds?

-----------------------------------------------------------------------------------------E(X) = n p = 20 ? 0.75 = 15 calls

------------------------------------------------------------------------------------------

(d) If you call 20 times, what is the mean number of calls that are unanswered in less than 30 seconds?

-----------------------------------------------------------------------------------------Let Y denote the number of calls unanswered in 30 seconds. Then Y is binomial with n = 20 and p = 0.25, where now a "success" is a call that is unanswered in 30 seconds.

E(Y) = n p = 20 ? 0.25 = 5 calls. ------------------------------------------------------------------------------------------

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Schmeiser

Homework #6 Solution.

Fall 2000, IE 230

6. (From Problem 4-107.) An installation technician for a specialized communication system is dispatched to a city only when three or more orders have been placed. Suppose that orders follow a Poisson distribution with a mean of 0.25 orders per week for a city of 100,000. Suppose that your city contains a population of 800,000.

(a) What is the probability that a technician is required sometime during a one-week period? (Provide the expression, the numerical answer, and the MSExcel cell code.)

-----------------------------------------------------------------------------------------Let R = "technician required sometime during the week". Let X = "number of orders during the week", where X is Poisson with mean = (0.25 / 100000) ? (100,000 / 800,000) = 2 orders.

Then P(R) = P(X 3) = 1 - P(X 2) = 1 - [f (0) + f (1) + f (2)], where f (x) = e- x / x !.

The MSExcel code is "1 - Poisson(2,2,True)", which yields 0.323. ------------------------------------------------------------------------------------------

(b) If you are the first person in the city to place an order, what is the probability that you have to wait more than two weeks from the time that you place your order until a technician is dispatched?

-----------------------------------------------------------------------------------------Let W = "you have to wait more than two weeks." Let Y = "number of orders during the two weeks following the order", where Y is Poisson with mean = (2) ? (0.25 / 100000) ? (100,000 / 800,000) = 4 orders.

Then P(W) = P(Y < 2) = f 0) + f (1), where f (x) = e- x / x !.

Therefore, P(W) = e-4 0 / 0! + e-4 1 / 1! = e-4 (1 + 4) = 0.0916. ------------------------------------------------------------------------------------------

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