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 45.8 minutes and standard deviation 13.4 minutes. A random sample of 7 deliveries is selected. a) What is the probability that a particular delivery will arrive in less than one hour? Round your answer to four decimal places. b) What is the probability that the mean time of 7 deliveries will exceed one hour? Round your answer to four decimal places. c) Between what two times do the middle 60% of the average delivery times fall? and Round your answers to two decimal places. d) What is the probability that, in a random sample of 7 deliveries, more than three will arrive in less than an hour? Round your answer to four decimal places.

Answer 1

a) P( < 60)

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

= P(Z < (60 - 45.8)/(13.4/))

= P(Z < 2.8)

= 0.9974

b) P( > 60)

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

= P(Z > (60 - 45.8)/(13.4/))

= P(Z > 2.8)

= 1 - P(Z < 2.8)

= 1 - 0.9974 = 0.0026

c) P( < x) = 0.2

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

or, P(Z < (x - 45.8)/(13.4/)) = 0.2

or, (x - 45.8)/(13.4/) = -0.84

or, x = -0.84 * (13.4/) + 45.8

or, x = 41.55

P( > x) = 0.2

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

or, P(Z > (x - 45.8)/(13.4/)) = 0.2

or, P(Z < (x - 45.8)/(13.4/)) = 0.8

or, (x - 45.8)/(13.4/) = 0.84

or, x = 0.84 * (13.4/) + 45.8

or, x = 50.05

d) P(X < 60)

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

= P(Z < (60 - 45.8)/13.4)

= P(Z < 1.06)

= 0.8554

n = 7

p = 0.8554

It is binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 3) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

             = 7C4 * (0.8554)^4 * (0.1446)^3 + 7C5 * (0.8554)^5 * (0.1446)^2 + 7C6 * (0.8554)^6 * (0.1446)^1 + 7C7 * (0.8554)^7 * (0.1446)^0 = 0.9894