![• We take X to be the angle at which the pointer
comes to rest, so we use the interval [0
360] 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.](https://image.slidesharecdn.com/statics-141224162209-conversion-gate02/85/Statics-distributions-questions-13-320.jpg)
Statics distributions questions
- 1. 6 QUESTIONS FOR 6 DISTRIBUTIONS
- 2. • 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
- 3. Binomial Distribution - Example Example A quality control engineer is in charge of testing whether or not 90% of the DVD players produced by his company conform to specifications. To do this, the engineer randomly selects a batch of 12 DVD players from each day’s production. The day’s production is acceptable provided no more than 1 DVD player fails to meet specifications. 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 DVD players actually conform to specification? (ii) What is the probability that the engineer unnecessarily requires the entire day’s production to be tested if in fact 90% of the DVD players conform to specifications?
- 4. • (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) = (0 12)(0,2)0(0.8)12+ (1 12)(0,2)1(0.8)11 = 0,069 + 0,206 = 0,275
- 5. • We now want P(X > 1) with parameters n = 12, p = 0.1. • P(X ≤ 1) = P(X = 0) + P(X = 1) = (0 12)(0,1)0(0.9)12+ (1 12)(0,1)1(0.9)11 =0,659 • So P(X > 1) = 0.341
- 6. 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?
- 7. • 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.
- 8. 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?
- 9. • 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`
- 10. • (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`
- 11. • (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
- 12. 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°?
- 13. • We take X to be the angle at which the pointer comes to rest, so we use the interval [0 360] 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.
- 14. f(x)=___1___= ___1___ b-a 360 P(5<=x<=300)=∫5 300 __1__dx = __295__ = 0.8194 360 360
- 15. 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 that the 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.
- 16. • (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
- 17. • Tuğçe Belek • Ayşenur Yüksel • Elif Demirok • Jülide Yatar