Question

A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chickens. 55 of the children preferred hamburger and 20 preferred chickens. Set up a 2x2 contingency table using this information and answer the following questions:

Age/ Food	Hamburger	Chicken	Total
Child
Adult
Total			200

What is the probability that a randomly selected individual is an adult?

What is the probability that a randomly selected individual is a child and prefers chicken?

Given the person is a child, what is the probability that this child prefers a

hamburger?

Assume we know that a person has ordered chicken, what is the probability that this individual is an adult?

Are food preference and age statistically independent?

2) Three messenger services deliver to a small town in Oregon. Service A has 60% of all the scheduled deliveries, service B has 30%, and service C has the remaining 10%. Their on-time rates are 80%, 60%, and 40% respectively. Define event O as a service delivers a package on time.

Calculate P(A and O)

Calculate P(B and O)

Calculate P(C and O)

Calculate the probability that a package was delivered on time.

If a package was delivered on time, what is the probability that it was service A?

If a package was delivered 40 minutes late, what is the probability that it was service A?

3) The number of power outages at a nuclear power plant has a Poisson distribution with a mean of 6 outages per year.

What is the probability that there will be exactly 3 power outages in a year?

What is the probability that there will be at least 1 power outage in a year?

What is the variance for this distribution?

What is the mean power outage for this nuclear power plant in a decade?

           

           

Answer 1

Solution 1:

We set up a 2 x 2 contingency table first.

Age/ Food	Hamburger	Chicken	Total
Child	55	20	75
Adult	65	60	125
Total	120	80	200

i. The probability that randomly selected individual is an adult = 125/200 or 0.625

ii. The probability that randomly selected individual is a child and prefers chicken = 20/200 or 0.10

iii. P (hamburger/child) = P (hamburger and child)/P (child)

P (hamburger/child) = 55/75

P (hamburger/child) = 11/15 or 0.7333

Given the person is a child, what is the probability that this child prefers a hamburger is 0.7333.

iv. P (adult/chicken) = P (adult and chicken)/P (chicken)

P (adult/chicken) = 60/80

P (adult/chicken) = 0.75

Assume we know that a person has ordered chicken, the probability that this individual is an adult is 0.75.

v. P (Hamburger and child) = P (hamburger)*p (child)

55/200 = (120/200)*(75/200)

55/200 = 0.6*0.375

0.275 = 0.225

Since they are not equal, food preference and age are not independent.

Solution 2:

i. P(Aand O) = P(A)P(O|A)

P (A and O) = (.60)(.80)

P (A and O) = 0.48

ii. P(Band O) = P(B) P(O|B)

P (B and O) = (.30)(.60)

P (B and O) = 0.1

iii. P(Cand O) = P(C)P(O|C)

P (C and O) = (.10)(.40)

P (C and O) = 0.04

iv. P(O) = P(Aand O) + P(Band O) + P(Cand O)

P (O) = .48 + .18 + .04

P (O) = 0.70

v. P(A|O) = P(Aand O) / P(O)

P(A|O) = 0.48 / 0.70

P (A|O) = 0.686

vi. P(A|Oc) = P(Aand Oc) / P(Oc)

P (A|Oc) = (0.60)(0.20) / 0.30

P (A|Oc) = 0.40

Solution 3:

Given that mean, = 6. Using poisson distribution formula,

P (X = x) = e^-^x/x!

i. P (X = 3) = e^-6 (6)^3/3!

P (X = 3) = 0.0892

ii. P (X 1) = 1 - P (X = 0)

= 1 - (e^-6 (6)^0/0!)

= 1 - 0.0025

= 0.9975

iii. The variance for the distribution is equal to the mean = 6.

iv. The mean power outage for this nuclear power plant in a decade is 6.