Exhibit: 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 also 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 approximately follow the normal distribution with a mean of 51.7 minutes and a standard deviation of 9.5 minutes.

Answer the following questions, rounding your final answers to 4 decimal places. To avoid rounding errors, please do not round intermediate steps in your calculations.

1. What is the probability that the mean time of two deliveries will exceed one hour?
- Hint: Use "60 minutes" as the equivalent of one hour.

2. What enables you to calculate the answer to the previous question?

A. We always use the normal distribution for probability calculations with the sampling distribution of the mean. The probability in question is calculated from the sampling distribution of the mean delivery times. The sampling distribution of the mean is normal due to the fact that delivery times follow the normal distribution. This enables us to use the normal distribution and the standard error of the mean in the calculations.

B. The probability in question is calculated from the population distribution of delivery times. The population distribution of delivery times is normal, which enables us to use the normal distribution and the population standard deviation in the calculations.

C. These calculations are invalid. The probability in question is calculated from the sampling distribution of the mean delivery times. However, the sample size is too small to apply the CLT, and no probability should be calculated.

The probability corresponding to a z-score of -4.67 is extremely small and close to zero. Therefore, the probability that the mean time of two deliveries will exceed one hour (60 minutes) is very low.

To calculate the probability that the mean time of two deliveries will exceed one hour (60 minutes), we need to use the sampling distribution of the mean and the properties of the normal distribution.

The mean delivery time for the supplier is 51.7 minutes, and the standard deviation is 9.5 minutes. Since we are interested in the mean time of two deliveries, we can consider it as the sum of two independent random variables, each following the same normal distribution.

The mean of the sum of two delivery times is:

Mean = 51.7 + 51.7 = 103.4 minutes

The variance of the sum of two delivery times is:

Variance = (9.5^2) + (9.5^2) = 180.5

To calculate the probability that the mean time of two deliveries will exceed one hour (60 minutes), we need to find the probability that the sum of two delivery times is greater than 60 minutes.

Standardized value = (60 - 103.4) / sqrt(180.5)

To find the corresponding probability, we need to calculate the z-score for the standardized value of (60 - 103.4) / sqrt(180.5). The z-score represents the number of standard deviations the value is from the mean.

Using the given values, we can calculate the z-score as follows:

Z = (60 - 103.4) / sqrt(180.5)

Calculating this expression:

Z = -43.4 / sqrt(180.5)

Using a calculator, we find that the value of the z-score is approximately -4.67.

Now, we can look up the probability corresponding to the z-score of -4.67 in the standard normal distribution table or use a statistical calculator.

The probability corresponding to a z-score of -4.67 is extremely small and close to zero. Therefore, the probability that the mean time of two deliveries will exceed one hour (60 minutes) is very low.

In conclusion, the probability is very close to zero.

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