Supplier delivery times. Supplier on time delivery performance is crit

Osvaldo Apodaca 2021-12-11 Answered
Supplier delivery times. 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 random and well approximated by the Normal distribution with mean 48.2 minutes and standard deviation 13.4 minutes. What is the probability that the mean time of 5 deliveries will exceed one hour?

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Expert Answer

alkaholikd9
Answered 2021-12-12 Author has 4364 answers

Step 1
Solution:
Let X be the delivery times that follows normal distribution with mean \(\displaystyle\mu={48.2}\) minutes and standard deviation \(\displaystyle\sigma={13.4}\) minutes.
Step 2
Probability that the mean time of 5 deliveries will exceed one hour (60 minutes):
The probability that the mean time of 5 deliveries will exceed 60 minutes can be calculated as given below:
\(\displaystyle{P}{\left(\overline{{{x}}}{>}{60}\right)}={P}{\left({\frac{{\overline{{{x}}}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}{>}{\frac{{{60}-\mu}}{{{\frac{{\sigma}}{{\sqrt{{{n}}}}}}}}}\right)}\)
\(\displaystyle={P}{\left({z}{>}{\frac{{{60}-{48.2}}}{{{\frac{{{13.4}}}{{\sqrt{{{5}}}}}}}}}\right)}\)
\(\displaystyle={P}{\left({z}{>}{1.97}\right)}\)
\(\displaystyle={1}-{P}{\left({z}{<}{1.97}\right)}\)
\(\displaystyle={1}-{0.9756}\) (Using standard normal table, \(\displaystyle{P}{\left({z}{<}{1.97}\right)}={0.9756}{)}\)
\(\displaystyle={0.0244}\)
Thus, the probability that the mean time of 5 deliveries will exceed 60 minutes is 0.0244.

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