Question

Supplier on-time delivery performance is critical to enabling the buyer's organization to meet its customer service commitments. Therefore, monitoring supplier delivery times is critical. Based on a great deal of historical data, a manufacturer of personal computers finds for one of its just-in-time suppliers that the delivery times are well approximated by the normal distribution with mean 47.7 minutes and standard deviation 14.5 minutes. A random sample of 8 deliveries is selected.

Round all probabilities to four decimal places and times to two decimal places.

a) What is the probability that a particular delivery will arrive in less than one hour?  

b) What is the probability that the mean time of 8 deliveries will exceed one hour?  

c) Between what two times do the middle 60% of the average delivery times fall?  and

d) What is the probability that, in a random sample of 8 deliveries, more than three will arrive in less than an hour?  

Answer 1

a) P(X < 60)

= P((X - )/ < (60 - )/)

= P(Z < (60 - 47.7)/14.5)

= P(Z < 0.85)

= 0.8023

b) P( > 60)

= P(( - )/() > (60 - )/())

= P(Z > (60 - 47.7)/(14.5/))

= P(Z > 2.40)

= 1 - P(Z < 2.40)

= 1 - 0.9918

= 0.0082

c) P(X < x) = 0.2

or, P((X - )/ < (x - )/) = 0.2

or, P(Z < (x - 47.7)/14.5) = 0.2

or, (x - 47.7)/14.5 = -0.84

or, x = -0.84 * 14.5 + 47.7

or, x = 35.52

P(X > x) = 0.2

or, P((X - )/ > (x - )/) = 0.2

or, P(Z > (x - 47.7)/14.5) = 0.2

or, P(Z < (x - 47.7)/14.5) = 0.8

or, (x - 47.7)/14.5 = 0.84

or, x = 0.84 * 14.5 + 47.7

or, x = 59.88

d) n = 8

p = 0.8023

P(X > 3) = 1 - P(X < 3)

             = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))

            = 1 - (8C0 * (0.8023)^0 * (1 - 0.8023)^8 + 8C1 * (0.8023)^1 * (1 - 0.8023)^7 + 8C2 * (0.8023)^2 * (1 - 0.8023)^6 + 8C3 * (0.8023)^3 * (1 - 0.8023)^5)

          = 1 - 0.0099

          = 0.9901