6 QUESTIONS FOR 6 DISTRIBUTIONS

• Bernoulli Example

• Suppose our class passed (C or better) the last

exam with probability 0.75. Let the random

variable X be the probability that someone

passes the exam.

Solution : X ~ Bernoulli(.75) X 0 1

p(x) .25 .75

• E(X)=p=0.75

• Var(X)=p(1-p)=.75(.25)=0.1875

Binomial Distribution - ExampleExample

A quality control engineer is in charge of testing whether or not90% of the DVD players produced by his company conform to

specifications. To do this, the engineer randomly selects a batch of12 DVD players from each day’s production. The day’s productionis acceptable provided no more than 1 DVD player fails to meetspecifications. Otherwise, the entire day’s production has to be

tested.(i) What is the probability that the engineer incorrectly passes a

day’s production as acceptable if only 80% of the day’s DVDplayers actually conform to specification?

(ii) What is the probability that the engineer unnecessarilyrequires the entire day’s production to be tested if in fact 90%

of the DVD players conform to specifications?

• (i) Let X denote the number of DVD players in the sample that fail to meet specifications. In part (i) we want P(X ≤ 1) with binomial parameters n = 12, p = 0.2

• P(X ≤ 1) = P(X = 0) + P(X = 1)

= (012)(0,2)0(0.8)12+ (1

12)(0,2)1(0.8)11

= 0,069 + 0,206 = 0,275

• We now want P(X > 1) with parameters n = 12, p = 0.1.

• P(X ≤ 1) = P(X = 0) + P(X = 1)

= (012)(0,1)0(0.9)12+ (1

12)(0,1)1(0.9)11

=0,659

• So P(X > 1) = 0.341

Geometric Example

• A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And, let X denote the number of people he selects until he finds his first success. What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game?

• Solution. To find the desired probability, we need to find P(X = 4), which can be determined readily using the p.m.f. of a geometric random variable with p = 0.20, 1−p = 0.80, and x = 4:

• \(P(X=4)=0.80^3 \times 0.20=0.1024\)

• There is about a 10% chance that the marketing representative would have to select 4 people before he would find one who attended the last home football game.

Poisson Example

• A life insurance salesman sells on the average `3` life insurance policies per week. Use Poisson's law to calculate the probability that in a given week he will sell

a) Some policies

b)`2` or more policies but less than `5` policies.

c)Assuming that there are `5` working days per week, what is the probability that in a given day he will sell one policy?

• Here, μ = 3• (a) "Some policies" means "1 or more policies". We can

work this out by finding 1 minus the "zero policies" probability:P(X > 0) = 1 − P(x0)

Now `P(X)=(e^(-mu)mu^x)/(x!)` so `P(x_0)=(e^-3 3^0)/(0!)=4.9787xx10^-2`Therefore the probability of `1` or more policies is given by:`"Probability"=P(X>=0)``=1-P(x_0)``=1-4.9787xx10^-10``=0.95021`

• (b) The probability of selling 2 or more, but less than 5 policies is:

• `P(2<=X<5)`

• `=P(x_2)+P(x_3)+P(x_4)`

• `=(e^-3 3^2)/(2!)+(e^-3 3^3)/(3!)+(e^-3 3^4)/(4!)`

• `=0.61611`

• (c) Average number of policies sold per day: `3/5=0.6`

• So on a given day, `P(X)=(e^-0.6(0.6)^1)/(1!)=0.32929

Uniform Example

• Suppose that you spin the dial shown in the figure so that it comes to rest at a random position. Model this with a suitable probability density function, and use it to find the probability that the dial will land somewhere between 5° and 300°?

• We take X to be the angle at which the pointer comes to rest, so we use the interval [0360] as its range. Since all angles are equally likely, the probability density function should not depend on x and therefore should be constant. That is, we take f to be uniform.

f(x)=___1___= ___1___

b-a 360

P(5<=x<=300)=∫5

300__1__dx = __295__ = 0.8194

360 360

Exponential Example

• Jobs are sent to a printer at an average of 3 jobs per hour. (a)What is the expected time between jobs? (b) What is the probability thatthe next job is sent within 5 minutes?

• Solution: Job arrivals represent rare events, thus the time T between them is Exponential with rate 3 jobs/hour i.e. λ = 3.

• (a) Thus E(T) = 1/λ = 1/3 hours or 20 minutes.

• (b) Using the same units (hours) we have 5 min.=1/12 hours. Thus we compute

P(T < 1/12) = Exp3(1/12) = 1 − e 3· 1/12 = 1 − e −1/4 = 0.2212

• Tuğçe Belek

• Ayşenur Yüksel

• Elif Demirok

• Jülide Yatar