City Parking Systems, Inc., operates a parking garage in the downtown area. Duri

Question

City Parking Systems, Inc., operates a parking garage in the downtown area. During the rush hours, cars arrive at the garage at the rate of 24 every 60 minutes. Three garage attendants can handle this level of arrivals, but not too much more.

P(x)=(xe-)/x!

P=probability given the random variables.

x=value of the random variable X

=the average number of occurrences per unit time or space.

e=the base of the natural log system (2.71828)

e-4=.01832

You need to develop on excel formulas to answer these questions:

1. P(x=a specific number)=POISSON(,x,FALSE);

2. P(xa specific number)=POISSON(,x,TRUE);

3. P(x>a specific number)=1-POISSON(,x,TRUE)

P(x=0) is 0.01832

P(x=1) is 0.14653

P(x=2) is 0.14653

P(x=3) is 0.14653

P(x=4) is 0.19537

P(x1) is much more than 0.09158

P(x2) is 0.23810

P(x3) is 0.62884

P(x4) is 0.62884

P(x>1) is 0.76190

P(x>2) is 0.76190

P(x>3) is not 0.56653

P(x>4) is 0.37116

a.

P(x=0) is 0.01832

b.

P(x=1) is 0.14653

c.

P(x=2) is 0.14653

d.

P(x=3) is 0.14653

e.

P(x=4) is 0.19537

f.

P(x1) is much more than 0.09158

g.

P(x2) is 0.23810

h.

P(x3) is 0.62884

i.

P(x4) is 0.62884

j.

P(x>1) is 0.76190

k.

P(x>2) is 0.76190

l.

P(x>3) is not 0.56653

m.

P(x>4) is 0.37116

Explanation / Answer

lambda = 24/60 = 0.4 if we take every minute.

0 P(X=0) 0.67032 1 P(X=1) 0.268128 2 P(X=2) 0.053626 3 P(X=3) 0.00715 4 P(X=4) 0.000715 0 P(X<=0) 0.67032 1 P(X<=1) 0.268128 2 P(X<=2) 0.053626 3 P(X<=3) 0.00715 4 P(X<=4) 0.000715 0 P(X>0) 0.32968 1 P(X>1) 0.731872 2 P(X>2) 0.946374 3 P(X>3) 0.99285 4 P(X>4) 0.999285

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