1) The concession manager of a local entertainment venue just had 2 cancellations on his crew. This means that if more than 72,000 people come to tonight’s event, the concession lines will be excessively long, and this will harm business. The manager knows from experience that the number of people who attend this type of event (X) is ~N(67,000, 4,000).

a) What is the probability that there will be more than 72,000 people at the event?

I Think i got Part A. I did (72000-67000)/4000 = 1.25 = Z

1-P(Z<1.25) = .1056 via ZTable

b) Suppose the manager can hire two temporary employees to make sure business won’t be harmed in the future at an additional cost of $200. If he believes the future harm to the business from having more that 72,000 people at the event is quantified at $5000, show that he should hire the employees?

for Part B i have no idea what the question means? isnt it obvious that if it costs him $200 for 2 additional employees, and he thinks having no extra employees is a $5000 potential cost.. Im just not sure what it means.

46. Mr. Bill invests $10,000 in a certain stock on January 1. Having examined past movements of this stock and consulting with his broker, Mr. Bill estimates that the annual return from this stock, X, is Normally distributed with mean 10% and standard deviation 4%. Here X (when expressed as a decimal) is the profit Howard receives per dollar invested. Because Mr. Bill is in the 33% tax bracket, he will then have to pay the Internal Revenue Service 33% of his profit.

a) Calculate the probability that Mr. Bill will have to pay the IRS at least $400.

I tried this one forever.. I kept getting z = 22.5 because i have no idea where to begin or what to do

b) Determine the value for Mr. Bill’s after-tax profit such that he is 90% certain after-tax profit will be less than this amount, i.e., determine the 90th percentile of his after-tax profit.

47. Approximate the following binomial probabilities by the use of normal approximation. Twenty percent of students who finish high school do not go to college. What is the probability that in a sample of 80 high school students

a) exactly 10 will not go to college? I know this is suppost to equal .0274 I have again no idea how to get it, i tried finding variance by 16*(1-.20) = 12.8- and taking the sqrt to get 3.58 then pluging in (10 -16)/3.58 but this is apparently wrong

b) 70 or more will go to college?

c) 14 or fewer will not go to college?

48. Suppose that X is the time (in minutes) required to complete one of two tasks in an assembly process, and Y is the time (in minutes) required to complete the second of two tasks in the assembly process. X and Y are independent. Further, suppose X~N(10, 4) and Y~N(15, 3). Let T = X + Y. What is the probability that it will take more than one-half hour to fully assemble one item?

Ive never seen this type of problem in my notes.. idk where to even begin or what to use.