Please show your work on how you got your answer Suppose that at a hotel\'s regi
ID: 2949347 • Letter: P
Question
Please show your work on how you got your answer
Suppose that at a hotel's registration desk there are 5 guests who check-in and need their luggage delivered to their rooms. Suppose that the probability that the luggage will be delivered within five minutes is 0.66, and that you would like to know how likely certain outcomes are, based on those assumptions. Please think for a moment about what framework you would use to model this You should come to the conclusion that you could let T be a random variable that counts the number of times that that the luggage is delivered on time. Then Tis a binomial random variable with n-5,p 0.66, and q 0.34 Using the notation B(n, p) for a binomial random variable with n trials with a probability of "success" of p, we have T B(5,p 0.66) (b) For each value of t, calculate p(t). (Round final answers to 4 decimal places.) p(0)0.1252, p(1) p(2) ,p (3) p(5) p(4)- (c) What is the likelihood that exactly three sets of luggage are delivered on time? Find P(T- 3). (Round final answer to 4 decimal places.) P(T-3) (d) What is the likelihood that up to three sets of luggage are delivered on time? Find P(TS 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.) P(TS 3) (e) How likely are up to two sets of luggage delivered on time? Find P(T3). (Do not round intermediate calculations. Round final answer to 4 decimal places.) P(T 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.) PIT> 2)Explanation / Answer
Answer to the question is as follows. Please don't hesitate to give a "thumbs up" to the solution in case you're satisfied with the answer.
b.
P(0) = 5C0*.66^0*.34^5 = 0.004543542 = .0045
P(1) = 5C1*.66^1*.34^5 =0.044099088 = .0441
P(2) = 5C2*.66^2*.34^5 =0.171208224 = .1712
P(3) = 5C3*.66^3*.34^5 =0.332345376 = .3323
P(4) = 5C4*.66^4*.34^5 =0.322570512 = .3223
P(5) = 5C5*.66^5*.34^5 =0.125233258 = .1252
c. P(T=3) = P(3) = 0.3323
P(T<=3) = P(1,2,3) = =0.044099088+0.171208224+0.332345376 = .5477
e. P(T<3) = P(T=1,2) = P(1,2) = .2153
f. P(T>=4) = P(T=4,5) = .4478
g. P(T>2) = 1-P(T=0,1) = 1-0.004543542-0.044099088 = .9514