Leo Almazora August 9, 2012Alria Ventanilla ABE 112Minch Cerrero

Exercises:Discrete Probability Distributions

The

table above shows the different formulas associated with each type of distribution.

I. Problems involving the binomial distribution:27. Given a binomial experiment with n=20 and p=0.70

a. Compute f(12).

f (12 )=(2012)(0.70)12(1−0.70)20−12

¿( 20 !12 ! (20−12 )! )(0.70)12¿

¿0.1143… ≈ 0.114

b. Compute f(16).

f (16 )=(2016)(0.70)16(1−0.70)20−16

¿( 20!16 ! (20−16 ) ! )(0.70)16(0.30)4

¿0.1304… ≈ 0.130

c. Compute P(x≥16).

P ( x≥16 )=f (16 )+ f (17 )+f (18 )+ f (19 )+f (20 )

→f (16 )=0.130

Distribution Probability Function Expected Value VarianceBinomial f ( x )=(nx ) px(1−p)n−x E ( x )=μ=np Var ( x )=σ2=np(1−p)

Poissonf ( x )= μxe−μ

x !E ( x )=μ Var ( x )=μ

Hypergeometric

f ( x )=( rx)(N−r

n−x )(Nn )

E ( x )=μ=n( rN ) Var ( x )=σ2=n( rN )(1− rN )(N−n

N−1 )

→f (17 )=(2017)(0.70)17(1−0.70)20−17

¿( 20!17 ! (20−17 ) ! )(0.70)17¿

¿0.0716… ≈ 0.072

→f (18 )=(2018)(0.70)18(1−0.70)20−18

¿( 20 !18 ! (20−18 )! )(0.70)18(0.30)2

¿0.0278… ≈ 0.028

→f (19 )=(2019)(0.70)19(1−0.70)20−19

¿( 20 !19 ! (20−19 )! )(0.70)19(0.30)1

¿0.0068… ≈ 0.007

→f (20 )=(2020)(0.70)20(1−0.70)20−20

¿( 20!20 ! (20−20 ) ! )(0.70)20 ¿

¿0.00079… ≈ 0.001

∴P ( x≥16 )=0.130+0.072+0.028+0.007+0.001=0.238

d. Compute P(x≤15).

P ( x≤15 )=1−P ( x ≥16 )=1−0.238=0.762

e. Compute E(x).

E ( x )=μ=20 (0.70 )=14

f. Compute Var(x) and σ.

Var ( x )=σ2=np (1−p )=20 (0.70 ) (1−0.70 )=4.2

σ=√Var ( x )=√4.2=2.0493…≈2.1

29. Given that in San Francisco, 30% of workers take public transportation every day: a. Take the probability that 3 workers from a sample of 10 take public transportation daily.

We shall define success as the event where a worker takes public transportation every day.

f (3 )=(103 )(0.30)3(1−0.30)10−3={ 10 !3 ! (10−3 )! }(0.30)3 ¿

b. Take the probability that at least 3 workers take public transportation daily.

P ( x≥3 )=1−P (x<3 )=1− {f (0 )+ f (1 )+ f (2)}

→f (0 )=(100 )(0.30)0(1−0.30)10−0={ 10 !0 ! (10−0 ) !}(0.30)0¿

→f (1 )=(101 )(0.30)1(1−0.30)10−1={ 10 !1 ! (10−1 )! }(0.30)1¿

→f (2 )=(102 )(0.30)2(1−0.30)10−2={ 10 !2 ! (10−2 )! }(0.30)2(0.70)8=0.23347…≈0.233

∴P ( x≥3 )=1−{0.028+0.121+0.233 }=1−0.382=0.618

31. A survey revealed that nine percent of undergraduate students carry credit card balances greater than $7000. Suppose 10 undergraduate students are selected randomly to be interviewed about credit card usage.

a. Is the selection of 10 undergraduate students a binomial experiment? Explain.

Yes. There are four properties of a binomial experiment:

1) The experiment consists of n identical trials: In this case, each student can be considered as a trial. Since they are selected from the same sample set (undergraduate students), each unique “trial” can be considered identical to the others.

2) Two outcomes are possible on each trial: One could argue that there are many possible ways undergraduate students can be classified with respect to credit card use. They may have credit card balances greater than $7000, equal to $7000, or less than $7000. Some may not even have credit cards at all. However, as long as we limit the condition of success to be “has a credit card balance greater than $7000,” all other possible scenarios for undergraduate students can be lumped under “does not have a credit card balance greater than $7000.” Of course, we assume that the original survey also included all undergraduate students in sampling, not just those with credit cards.

3) The probability of success does not change from trial to trial: The probability of success was determined from a past survey, which we assume to represent the entire undergraduate student population. We further assume the size of that population to be several orders of magnitude greater than the sample size of ten. Therefore, the likelihood of a student selected fulfilling the conditions of success should not change significantly with each unique student interviewed.

4) The trials are independent: Since the students are surveyed randomly, the possibility that one participant influences another’s credit card usage is very minute. Therefore, the probability of a student having a credit card balance exceeding $7000 should not be affected by whether or not another student has that level of debt.

b. What is the probability that 2 of the students will have a credit card balance greater than $7000?

p=thechances that anundergraduate student has acredit card balance greater than $7000=9%=0.09

f (2 )=(102 )(0.09)2(1−0.09)10−2={ 10!2! (10−2 ) !}¿

c. What is the probability that none will have a credit card balance greater than $7000?

f (0 )=(100 )(0.09)0(1−0.09)10−0={ 10!0 ! (10−0 ) ! }¿

d. What is the probability that at least 3 will have a credit card balance greater than $7000?

P ( x≥3 )=1−P (x<3 )=1− {f (0 )+ f (1 )+ f (2)}

→f (0 )=0.389

→f (1 )=(101 )(0.09)1(1−0.09)10−1={ 10 !1 ! (10−1 ) ! }(0.09 )1 ¿

→f (2 )=0.171

∴P ( x≥3 )=1−{0.389+0.385+0.171 }=1−0.945≈0.055

33. Fifty percent of Americans believed the country was in a recession, although technically the economy had not shown two straight quarters of negative growth. Given a sample of 20 Americans:

a. Compute the probability that exactly 12 people believed the country was in a recession.

p=thechances that an Americanbelieves theUS¿be∈arecession=50%=0.5

f (12 )=(2012)(0.5)12(1−0.5)20−12={ 20 !12 ! (20−12 )! }¿

b. Compute the probability that no more than 5 believed the country was in a recession.

P ( x≤5 )=f (0 )+ f (1 )+ f (2 )+ f (3 )+ f (4 )+ f (5)

→f (0 )=(200 )(0.50)0(0.50)20−0={ 20 !0 ! (20−0 )! }¿

→f (1 )=(201 )(0.50)1(0.50)20−1={ 20 !1 ! (20−1 ) ! }¿

→f (2 )=(202 )(0.50)2(0.50)20−2={ 20 !2 ! (20−2 )! }¿

→f (3 )=(203 )(0.50)3(0.50)20−3={ 20!3 ! (20−3 ) ! }¿

→f (4 )=(204 )(0.50)4(0.50)20−4={ 20 !4 ! (20−4 )! }(0.50)20=4.620552×10−3

→f (5 )=(205 )(0.50)5(0.50)20−5={ 20!5 ! (20−5 ) ! }(0.50)20=1.4785766×10−2

∴P ( x≤5 )=(9.53×10−7)+(1.9073×10−6)+(1.81198×10−4 )+(1.087188×10−3 )+(4.620552×10−3 )+(1.4785766×10−2 )≈0.0207

c. How many people would be expected to believe the country was in a recession?

E ( x )=μ=np=20 (0.50 )=10

d. What is the variance and standard deviation of people who believed the country was in a recession?

Var ( x )=σ2=np (1−p )=20 (0.5 ) (1−0.5 )=5

σ=√Var ( x )=√5=2.236…≈2.24

35. In a university, 20% of the students withdraw without completing the introductory statistics course. Assuming 20 students registered for the course:

a. Compute the probability that 2 or fewer will withdraw.

p=thechances that a student willwithdrawwithout completing theintroductory statistics course=20%=0.2

P ( x≤2 )=f (0 )+f (1 )+ f (2 )

→f (0 )=(200 )(0.20)0(1−0.20)20−0={ 20!0 ! (20−0 ) ! }(0.20 )0¿

→f (1 )=(201 )(0.20)1(1−0.20)20−1={ 20 !1 ! (20−1 )! }(0.20 )1¿

→f (2 )=(202 )(0.20)2(1−0.20)20−2={ 20!2! (20−2 )! }(0.20 )2 ¿

∴P ( x≤2 )=0.0115+0.0576+0.1369=0.206

b. Compute the probability that exactly 4 will withdraw.

f (4 )=(204 )(0.2)4 (1−0.2)20−4={ 20!4 ! (20−4 ) !}¿

c. Compute the probability that more than 3 will withdraw.

P ( x>3 )=1− {P(x≤2)+ f (3 ) }

→f (3 )=(203 )(0.20)3(1−0.20)20−3={ 20 !3 ! (20−3 )! }(0.20 )3¿

→P ( x ≤2 )=0.206

∴P ( x>3 )=1−(0.206+0.205)=0.589

d. Compute the expected number of withdrawals.

E ( x )=μ=np=20 (0.20 )=4

37. Twenty-three percent of automobiles are not covered by insurance. On a particular weekend, 35 automobiles are involved in traffic accidents.

a. What is the expected number of these automobiles that are not covered by insurance?

Assuming there is no correlation between the likelihood that an automobile is not covered by insurance and the likelihood that it will be involved in a traffic accident:

E ( x )=μ=np=35 (0.23 )=8.05

b. What are the variance and standard deviation?

Var ( x )=σ2=np (1−p )=35 (0.23 ) (1−0.23 )=6.1985

σ=√Var ( x )=√6.1985=2.48967…≈2.490

II. Problems involving the Poisson distribution:41. During the period of time when a local university takes phone-in registrations, calls come in at the rate of 1 every 2 minutes.

a. What is the expected number of calls in 1 hour?

E ( x )=( 1call2minutes )(60minutes1hour )(1hour )=30calls

b. What is the probability of 3 calls in 5 minutes?

μ=( 1call2minutes ) (5minutes )=2.5calls

f (3 )=μ3e−μ

3 !=

(2.5 )3 (e )−2.5

3!=0.21376…≈0.214

c. What is the probability of no calls in a 5-minute period?

f (0 )=μ0 e−μ

0 !=

(2.5 )0 (e )−2.5

0 !=0.082084…≈0.082

43. Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute.

a. Compute the probability of no arrivals in a one-minute period.

f (0 )=μ0 e−μ

0 !=

(10 )0 ( e )−10

0 !=0.000045399…≈0.0000454

b. Compute the probability that 3 or fewer passengers arrive in a 1-minute period.

P ( x≤3 )=f (0 )+ f (1 )+ f (2 )+ f (3)

→f (0 )=0.0000454

→f (1 )= μ1 e−μ

1 !=

(10 )1 (e )−10

1!=0.00045399…≈0.000454

→f (2 )= μ2 e−μ

2 !=

(10 )2 ( e )−10

2!=0.0022699…≈0.00227

→f (3 )= μ3 e−μ

3!=

(10 )3 (e )−10

3 !=0.0075666…≈0.00757

∴P ( x≤3 )=0.0000454+0.000454+0.00227+0.00757≈0.0103

c. Compute the probability of no arrivals in a 15-second period.

μ=( 10arrivals1minute )( 1minute60 seconds )(15 seconds )=2.5arrivals∈a15−second period

f (0 )=μ0 e−μ

0 !=

(2.5 )0 (e )−2.5

0 !=0.082084…≈0.0821

d. Compute the probability of at least 1 arrival in a 15-second period.

P ( x≥1 )=1−f (0 )=1−0.0821=0.9179

45. Companies with 50 employees are expected to average 3 employee off-the-job accidents per year. Answer the following for companies with 50 employees.

a. What is the probability of no off-the-job accidents in a 1-year period?

f (0 )=μ0 e−μ

0 !=

(3 )0 ( e )−3

0 !=0.049787…≈0.0498

b. What is the probability of at least 2 off-the-job accidents during a one-year period?

P ( x≥2 )=1−{f (0 )+ f (1)}

→f (0 )=0.0498

→f (1 )= μ1 e−μ

1 !=

(3 )1 (e )−3

1 !=0.14936…≈0.1494

P ( x≥2 )=1−(0.0498+0.1494 )≈0.8008

c. What is the expected number of off-the-job accidents during six months?

μ=( 3accidents1 year )( 1 year12months ) (6months )=1.5accidents∈asix−month period

d. What is the probability of no off-the-job accidents during the next six months?

f (0 )=μ0 e−μ

0 !=

(1.5 )0 ( e )−1.5

0 !=0.22313…≈0.223

III. Problems involving the hypergeometric distribution:48. In a game of blackjack, two cards are dealt to a player, and within a normal deck, face cards (jacks, queens, and kings) and tens are assigned a value of ten points each. Given that:

a. What is the probability that both cards dealt are aces or 10-point cards?

There are 4 aces, 4 tens, and 12 face cards in a deck of 52 cards. Given that information, we know that if we define success to be the event of getting a ten-point card or an ace, there are a total of 20 possible unique occurrences of success in a 52-card deck. The general formula for a hypergeometric probability function is:

f ( x )=( rx)(N−r

n−x )(Nn )

where n is the number of trials, x is the number of successes, N is the number of elements in the population, and r is the number of elements in the population labeled success. For blackjack, there are two draws, each of which can be considered a trial. Therefore, assuming we are dealing with the odds for only the first player being dealt a hand from a fresh deck:

f (2 )=(202 )(52−202−2 )

(522 )=

( 20 !2 !18 ! )( 32!0 !32 ! )( 52!2 !50 ! )

=0.14328≈0.1433

b. What is the probability that both cards are aces?

For this next scenario, we redefine the terms of success. In this case, a successful draw is one where an ace is obtained. The number of successes within the population is therefore reduced to 4 (only 4 aces in a deck). Applying the general formula for hypergeometric probability function for the case where one player is dealt a hand from a full deck:

f (2 )=(42)(52−42−2 )

(522 )=

( 4 !2!2 ! )( 48 !0! 48 ! )( 52 !2 !50! )

=0.0045248…≈ 0.0045

c. What is the probability that both cards have a value of 10?

Again, the terms of success have to be redefined. A successful draw would be one where a ten-point card is obtained. The number of possible successes within a 52-card deck would be 16. Therefore:

f (2 )=(162 )(52−162−2 )

(522 )=

( 16 !2 !14 ! )( 36 !0! 36 ! )( 52 !2!50 ! )

=0.090497…≈0.0905

d. What is the probability that a player is dealt blackjack (an ace and a 10-point card)?

The answer to item 48.a includes any situation where both the cards dealt are either an ace or a ten-point card. This covers three situations: both cards are aces; both cards are ten-point cards; and one is a ten-point card and the other is an ace. Given that the answer to 48.b is the likelihood of the first situation, and the answer to 48.c is the likelihood of the second situation, the probability of a blackjack can be obtained by subtracting the answers to 48.b and 48.c from the answer in 48.a:

P (blackjack )=0.1433−(0.0045+0.0905 )=0.0483

50. A shipment of 10 items contains 2 defective and 8 non-defective items. A sample of items will be taken from the shipment, and the whole shipment will be rejected if 1 defective item is found.

a. If a sample of 3 items is taken, what is the probability that the shipment will be rejected?

For this problem, we define success to be the event where a defective item is found. The population size is 10 (the number of items in a shipment). If the sample size n=3:

f (1)=(21)(10−23−1 )

(103 )=

( 2 !1!1 ! )( 8 !2!6 ! )( 10 !3! 7 ! )

=0.466666…≈0.46667

In a sample of 3 items, it is also possible that 2 of the items sampled are found to be defective:

f (2)=(22)(10−23−2 )

(103 )=

( 2!2 !0 ! )( 8!1 !7 ! )( 10 !3 !7 ! )

=0.066666…≈0.06667

Therefore, the likelihood that a shipment will be rejected based on a sample size of 3 items is:

P (rejection)=f (1 )+f (2 )=0.46667+0.06667≈0.5333

b. If a sample of 4 items is taken, what is the probability that the shipment will be rejected?

Similar to the previous problem, we must take the probability that either one or both of the defective items are included in the sample:

P (rejection)=f (1 )+f (2 )

→f (1)=(21)(10−24−1 )

(104 )=

( 2!1 !1 ! )( 8!3 !5 ! )( 10 !4 ! 6! )

=0.53333…≈0.53333

→f (2)=(22)(10−24−2 )

(104 )=

( 2 !2!0 ! )( 8 !2!6 ! )( 10 !4 !6 ! )

=0.13333…≈0.13333

∴P (rejection )=0.53333+0.13333≈0.6667

c. If a sample of 5 items is taken, what is the probability that the shipment will be rejected?

P (rejection)=f (1 )+f (2 )

→f (1)=(21)(10−25−1 )

(105 )=

( 2!1 !1 ! )( 8!4 !4 ! )

( 10!5 !5 ! )=0.555555…≈0.55556

→f (2)=(22)(10−25−2 )

(105 )=

( 2 !2!0 ! )( 8 !3! 5! )( 10 !5!5 ! )

=0.222222…≈0.22222

∴P (rejection )=0.55556+0.22222≈0.7778

d. If management would like to have a 0.90 probability of rejecting a shipment that has 2 defective and 8 non-defective items, what sample size should be taken?

If the probability of rejecting a shipment is 0.90, that means there should only be at most a 0.10 probability that all the items sampled are non-defective. Taking the general formula for hypergeometric probability function, we can say:

f (0)=(20)(10−2n−0 )

(10n )≤0.1

Since we already know that a sample size of 5 yields only around a 0.7778 probability of rejecting a shipment of 10 items with 2 defective ones and 8 non-defective ones, we know the sample size should be higher. At n=6:

f (0 )=(20)(10−26−0 )

(106 )=

( 2 !0! 2! )( 8 !6 !2! )( 10 !6 ! 4 ! )

=0.13333…≈0.1333

We find that a sample size of 6 is still slightly too small. At n=7:

f (0 )=(20)(10−27−0 )

(107 )=

( 2 !0! 2! )( 8 !7 !1! )( 10 !7 !3 ! )

=0.06666…≈0.0667

Therefore, for management’s requirement to be fulfilled, a sample size of at least 7 should be used.

51. Axline Computers has two computer manufacturing plants: one in Texas, which has 40 employees; and one in Hawaii, which has 20. A random sample of 10 employees is to be asked to fill out a benefits questionnaire.

a. What is the probability that none of the employees in the sample work at the plant in Hawaii?

We define success in this case to be the event where an employee comes from Hawaii. The total population size is 60 (the number of employees Axline has), and the sample size is 10. Given that:

f (0 )=(200 )(60−2010−0 )

(6010)=

( 20 !0 !20 ! )( 40 !10 !30 ! )

( 60 !10 !50! )

=0.011243…≈0.01

b. What is the probability that 1 of the employees in the sample works at the plant in Hawaii?

f (1 )=(201 )(60−2010−1 )

(6010)=

( 20!1 !19 ! )( 40 !9 !31 ! )( 60 !10 !50 ! )

=0.072535…≈0.07

c. What is the probability that 2 or more of the employees in the sample work at the plant in Hawaii?

P ( x≥2 )=1−{f (0 )+ f (1 ) }=1−(0.01+0.07 )≈0.92

d. What is the probability that 9 of the employees surveyed work at the plant in Texas?

When 9 of the employees surveyed come from the plant at Texas, it means the remaining employee surveyed comes from Hawaii. Therefore, the probability that 9 employees surveyed work at the Texas plant is the same is as the probability of just 1 employee being surveyed being from the Hawaii plant.

P (9employees surveyed come¿the Texas plant )≈0.07