Question

The number of customer arrivals at a bank's drive-up window in a 15-minute period is Poisson distributed with a mean of seven customer arrivals per 15-minute period. Define the random variable x to be the time (in minutes) between successive customer arrivals at the bank's drive-up window. (a) Write the formula for the exponential probability curve of x. (c) Find the probability that the time between arrivals is (Round your answers to 4 decimal places.) 1.Between one and two minutes. 2. Less than one minute 3. More than three minutes. 4. Between 1/2 and 3½ minutes. (d) Calculate μx, σ2x , and σx. (e) Find the probability that the time between arrivals falls within one standard deviation of the mean; within two standard deviations of the mean. (Round your answers to 4 decimal places.)

Answer 1

a) here expected interarrival time =15/7 minutes

exponential probability curve of x f(x)=(7/15)e^-7x/15 for x >0

1

P(.Between one and two minutes)

P(1<X<2)=(1-exp(-2/2.14285714285714)-(1-exp(-1/2.14285714285714))=

0.2338

2)

P( Less than one minute):

P(X<1)=1-exp(-1/2.14285714285714)=

0.3729

3)

P(More than three minutes):

P(X>3)=1-P(X<3)=1-(1-exp(-3/2.14285714285714))=

0.2466

4)

P(0.5<X<3.5)=(1-exp(-3.5/2.14285714285714)-(1-exp(-0.5/2.14285714285714))=

0.5966

d)

mean μ =	1/λ=β =	2.1429
varaince =σ2=	1/λ2=β2 =	4.5918
standard deviation σ=	1/λ=β =	2.1429

e)

  probability that the time between arrivals falls within one standard deviation of the mean:

P(0<X<4.28571428571429)=(1-exp(-4.28571428571429/2.14285714285714)-(1-exp(-0/2.14285714285714))=

0.8647

f)

P(0<X<6.42857142857143)=(1-exp(-6.42857142857143/2.14285714285714)-(1-exp(-0/2.14285714285714))=

0.9502