Question

Suppose all customers visiting the local bank have to take a random number and wait in a line. There are 200 customers in the bank in any given day, however only 85 percent of the customers are served. The length of the time customers must wait to see a Bank Teller is uniformly distributed between 50 minutes and 4 hours. a) What is the probability that a customer would have to wait between 20 minutes and 2 hours? b) What is the probability that a customer would have to wait between 50 minutes and 3 hours? c) Compute the expected waiting time.d) Compute the standard deviation of the waiting time.

Answer 1

Let X is continuous random variable with uniform distribution U (a, b) .

The probability density function for X can be

defined as,

$fx (x) = 1\div b-a$ where $a< X < b$

The formula for mean is, $E (X) =\frac{(b+a)}{2}$

The formula for variance is,$V (X) = \frac{(b-a)^2}{12}$

The cumulative distribution function of x is given by:

$F(x)= P ( X <= x) = x - a \div b - a$

PDF of uniform distribution $f(x) = 1 \div ( b - a ) for a < x < b$

b = maximum Value

a = minimum Value

$f(x) = 1/(b-a) = 1 \div (240-50) = 1 \div190 = 0.0053$

a.

the probability that a customer would have to wait between 20 minutes and 2 hours

$P(a < X < b) =( b - a ) \times f(x)\\ P(20 < X < 120) = (120-20) \times f(x)\\ = 100\times0.0053\\ = 0.5263$

b.

the probability that a customer would have to wait between 50 minutes and 3 hours

$P(a < X < b) =( b - a ) \times f(x)\\ P(50 < X < 180) = (180-50) \times f(x)\\ = 130\times0.0053\\ = 0.6842$

c.

$mean = \frac{a + b }{ 2}\\ =\frac{(50+240)}{2}\\ =145$

the expected waiting time = 145 minutes

d.

standard deviation $= \sqrt {\frac{(( b - a ) ^ 2 }{ 12} }$

$= \sqrt {\frac{(( 240-50 ) ^ 2 }{ 12} }$

$=54.8483$

the standard deviation of the waiting time = 54.8483