Before I get to the question itself, I have a concern about part (a) of this pro

Question

Before I get to the question itself, I have a concern about part (a) of this problem. There's a hint that says to use the normal approximation to the binomial distribution, however, I don't think this will work since the conditions for using the normal approximation to the binomial is not met. Is this a trick "hint" made by the professor? Note: this question comes from an old final exam.

Also, here are the conditions for using the normal approximation to the binomial (please, correct me if I am wrong):

Condition 1: np must be greater than 10 (in this problem, n = 1000 and p = 0.004?)

Condition 2: n(p-1) must be greater than 10

Anyways, that's the predicament I faced when attempting to solve this problem. I hope you can enlighten me on where I went wrong. Here's the question:

Problem 2 A large metropolitan transit authority has a fleet of 1000 buses, and it is observed that, on an average, there is a 0.4% chance that a bus will break down on any one day. (a) If the maintenance department of the transit organization has sufficient ca- pacity to cope with 6 breakdowns on any day, calculate the probability that on any day there will be insufficient staff to attend to all the breakdowns occurring on that day. Hint: Use normal approximations to Binomiall (b) If the probability of having insufficient staff on any day were actually 11.1%, let Y be the number of days that would have to go by for the organization to experience 3 days of insufficient maintenance capacity (note that Y 30. What are the expected value and the standard deviation of Y

Explanation / Answer

a) if you Professor say so, then I think we should use Normal approximation even we will give far away result.

here mean =np=1000*0.004=4

and std deviation =(np(1-p))1/2 =1.996

ehnce P(X>7) =1-P(X<=6) =1-P(Z<(6.5-4)/1.996)=1-P(Z<1.2525)=1-0.8948 =0.1052

b)expected value =r/p =3/0.111=27.027

std deviation =(r(1-p)/p2)1/2=14.713

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