Question

A study shows that at a local bank, the time bank customers spent from the moment they walk into the bank until they leave the bank can be modeled as a NORMAL random variable, with the mean of 20 minutes and the standard deviation of 4 minutes.

(a) If you just walk into the bank, what is the probability that you will stay in the bank longer than 25 minutes?

(b) For the next 10 customers walking into the bank, what is the probability that exactly 4 of the customers stay in the bank longer than 25 minutes? (Note that you don't have to calculate out the fi nal number; just leave the formula there with the corresponding numbers plugged in.)

(c) Let *t*denote the time that the next customer stays
in the bank. What is the value of *t*such that *t*
will be exceeded (i.e., the customer will stay longer than
*t*) with a probability of 0.01?

Answer #1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 20

S.d = 4

A)

P(x>25)

Z = (25-20)/4 = 1.25

From z table, P(z>1.25) = 0.1056

B)

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.1056

N = number of trials = 10

R = desired success = 4

P(4) = 10c4*(0.1056^4)*(1-0.1056)^10-4 = 0.01336797765

C)

From z table, P(z>2.33) = 0.01

2.33 = (x - 20)/4

X = 29.32

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