Question

Two questions

For each of the following questions: clearly indicate the probability distribution being used to solve the problem solve by hand, and verify your answer using MATLAB. 1. Two teams, A and B, play a series of games. If team B has a probability 0.4 of winning each game, is it to their advantage to play the best three out of five games or the best four out of seven, and why? Assume the outcomes of successive games are independent. 2. A quality control engineer works at a factory that bottles engine oil. The amount of oil is normally distributed with mean 40 litres and standard deviation of 0.05 litres. If the amount is less than 39.9 litres or more than 40.1 litres it is deemed to fail the quality control process. (a) Find the probability that a bottle will fail quality control (b) If the factory want to improve so that a bottle will fail 2% of the time, what must the quality control amounts be extended to?

Answer 1

(1) here we use Binomial distribution ,P(X=r)=ⁿC_rp^r(1-p)^n-r  

for the case I, here n=5, p=0.4

P(winning of 3 game out of 5)=P(X=3)=0.2304

for case 2, here n=7, p=0.4

P(winning of 4 game out of 7)=P(X=4)=0.1935

so case I is advantegeous i.e best 3 out of 5.

(2a) required probability=P(X<39.9)+P(X>40.1)=P(Z<-2)+P(Z>2)=P(Z<-2)+(1-P<-2)=1-2(P(Z<-2)=1-2*0.023=0.954

for x=39.9, z=(39.9-40)/0.05=-2

for x=40.1, z=(40.1-40)/2=2

(2b) here we want to find x such that P(|X|>x)=2%=0.02

or, P(X>x)+P(X<-x)=0.02

or, 1-P(X<x)+P(X<-x)=0.02

or, 1-(1-P(X<-x)+P(X<-x)=0.02 ( due to symmetrical properties of normal distribution)

or, P(X<-x)=0.01

and corresponding z=-2.3264

and higher limit of x=mean+sd*z=40+0.05*2.3264=40.1163

and lower limit of x=mean-sd*z=40-0.05*2.3264=39.8837

the new range should be (39.8837,40.1163)