Question

Sparagowski & Associates conducted a study of service times at the drive-up window of fast-food restaurants. The average service time at McDonald's restaurants was 2.97 minutes. Service time such as these frequently follow an exponential distribution.

for x ³ 0 P( x <= x₀ ) = 1 - e^-x0/mu for x >= 0

e = 2.718282

µ = 2.97 minutes

a) What is the probability that a customer's service time is between 2 and 3 minutes? Recall that Lambda is the reciprocal of mu in Excel EXPON.DIST

P( x <= 2 ) = ______ f(x) ______ EXPON.DIST

P( x <= 3 ) = ______ f(x) _____ EXPON.DIST

P( 2 <= x <= 3 ) = ____ ______

b) What is the probability that a customer's service time is greater than 2 minutes?

P( x > 2 ) = ______ _______

c) What is the probability that a cusomter's service time is more than 5 minutes?

P( x <= 5 ) = ______ _______ EXPON.DIST

P( x > 5 ) = _______ ________

d) What is the probability that a customer's service time is more than 2.97 minutes?

P( x <= 2.97 ) = _____ _______ EXPON.DIST

P( x > 2.97 ) = _____ _______

Answer 1

a)

probability =

P(X<2)=

1-exp(-2/2.97)=

0.4900

probability =

P(X<3)=

1-exp(-3/2.97)=

0.6358

probability =

P(2<X<3)=

(1-exp(-3/2.97)-(1-exp(-2/2.97))=

0.1458

b)

probability =

P(X>2)=

1-P(X<2)=

1-(1-exp(-2/2.97))=

0.5100

c)

probability =

P(X<5)=

1-exp(-5/2.97)=

0.8143

probability =

P(X>5)=

1-P(X<5)=

1-(1-exp(-5/2.97))=

0.1857

d)

probability =

P(X<2.97)=

1-exp(-2.97/2.97)=

0.6321

probability =

P(X>2.97)=

1-P(X<2.97)=

1-(1-exp(-2.97/2.97))=

0.3679