A customer wants delivery to be ensured between 10am and 2pm. Your truck leaves

Question

A customer wants delivery to be ensured between 10am and 2pm. Your truck leaves the factory at 4am and the time taken to reach the customer is normally distributed with an average of 8 hours and a standard deviation of 1 hour.

a. What is the process capability ratio of the process?

b. What is the probability that the current delivery process will be on delivered between 10am and 2pm (in other words, what is the service level)?

c. What specific action(s) (in terms of mean and standard deviation targets to be achieved) would you take to improve the ability of the delivery process to be a 6-sigma process?

d. If the mean delivery time was delayed until 12:30pm, but the standard deviation improved from 1 hour to 0.5 hours, what would the service level be for this new mean and standard deviation?

Explanation / Answer

This is very tricky Question

The truck starts at 4 am , they are asking what is the probability that truck will be delivered between 10 am and 2 pm

This means truck is to be delivered with 6 Hours and 10 Hours

P(6<X<10)

=P((6-Mean)/SD<(X-Mean)/sd<(12-mean)/sd) = P((6-8)/1<Z<(10-8)/1) = P(-2<Z<2) = 0.9546

The area between -2 and +2 are considered to be 2 Sigman Qualiity.

c) The targets to be achieved to say 6 Sigma Quality

X-6*sd<Z<X+6*SD = 8-6*1<Z<8+6*1 = 2<Z<14

This means, The jobs will be done with in 2 to 14 hours. In terms our time, the truck is starting 4 am

If the wants to have a six sigma 6 am to 6 pm .

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