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.
(a) What is the probability that a particular delivery will exceed one hour?
(b) Based on part (a), what is the probability that a particular delivery arrives in less than one hour?
(c) What is the probability that the mean time of 5 deliveries will exceed one hour?
in
http://davidmlane.com/normal.html
I entered:
mean -- 48.2
SD ---- 13.4
click on "above" and enter 60 (for 1 hour) to get
.1893
etc
78
To solve these questions, we will use the Normal distribution with the given mean and standard deviation.
(a) To find the probability that a delivery will exceed one hour, we need to find the area under the Normal distribution curve to the right of one hour (60 minutes).
Step 1: Calculate the z-score for one hour:
z = (x - mean) / standard deviation
z = (60 - 48.2) / 13.4
z = 0.8806
Step 2: Use a Z-table or a calculator to find the probability associated with the z-score of 0.8806.
The probability is P(Z > 0.8806) = 1 - P(Z < 0.8806)
By looking up the probability value associated with the z-score of 0.8806, we find it to be approximately 0.1908.
Therefore, the probability that a particular delivery will exceed one hour is 0.1908 or 19.08%.
(b) To find the probability that a delivery arrives in less than one hour, we can subtract the probability from part (a) from 1.
P(delivery arrives in less than one hour) = 1 - P(delivery exceeds one hour)
P(delivery arrives in less than one hour) = 1 - 0.1908
P(delivery arrives in less than one hour) = 0.8092 or 80.92%
Therefore, the probability that a particular delivery arrives in less than one hour is 0.8092 or 80.92%.
(c) To find the probability that the mean time of 5 deliveries will exceed one hour, we use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means from a large sample size will be approximately normally distributed, regardless of the shape of the original population distribution.
For the mean time of 5 deliveries, the mean would still be the same (48.2 minutes), but the standard deviation would change. According to the Central Limit Theorem, the standard deviation of the sample mean can be calculated by dividing the original standard deviation by the square root of the sample size.
Standard deviation of sample mean = standard deviation / √(sample size)
Standard deviation of sample mean = 13.4 / √5
Standard deviation of sample mean ≈ 5.9929
Now, we need to find the probability that the sample mean exceeds one hour, which is 60 minutes.
Step 1: Calculate the z-score for one hour:
z = (x - mean) / standard deviation
z = (60 - 48.2) / 5.9929
z = 1.9608
Step 2: Use a Z-table or a calculator to find the probability associated with the z-score of 1.9608.
The probability is P(Z > 1.9608) = 1 - P(Z < 1.9608)
By looking up the probability value associated with the z-score of 1.9608, we find it to be approximately 0.0266.
Therefore, the probability that the mean time of 5 deliveries will exceed one hour is 0.0266 or 2.66%.
To answer these questions, we will use the properties of the Normal distribution. The Normal distribution is a continuous probability distribution that is symmetric and bell-shaped.
(a) To find the probability that a delivery will exceed one hour, we need to calculate the area under the Normal curve to the right of 60 minutes (1 hour).
We know that the mean delivery time is 48.2 minutes and the standard deviation is 13.4 minutes. To calculate the probability, we can use a standard Normal distribution table or a statistical calculator.
Using a standard Normal table, we find the corresponding z-score for 60 minutes by calculating:
z = (60 - 48.2) / 13.4
This gives us a z-score of approximately 0.8811.
Looking up the z-score in the standard Normal table, we find that the area to the left of 0.8811 is 0.8078. Thus, the area to the right of 0.8811 (or 60 minutes) is 1 - 0.8078 = 0.1922.
Therefore, the probability that a particular delivery will exceed one hour is approximately 0.1922 or 19.22%.
(b) To find the probability that a particular delivery arrives in less than one hour, we can subtract the probability from part (a) from 1.
P(< 1 hour) = 1 - P(> 1 hour)
P(< 1 hour) = 1 - 0.1922
P(< 1 hour) = 0.8078
Therefore, the probability that a particular delivery arrives in less than one hour is approximately 0.8078 or 80.78%.
(c) To find the probability that the mean time of 5 deliveries will exceed one hour, we need to calculate the distribution of the sample mean.
The distribution of the sample mean is also Normal, with a mean equal to the population mean (48.2 minutes) and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
For 5 deliveries, the standard deviation of the sample mean is 13.4 / √5 = 5.9973 minutes.
We can use a similar approach as in part (a) to find the probability that the mean time of 5 deliveries will exceed one hour.
Using the z-score formula:
z = (60 - 48.2) / 5.9973
This gives us a z-score of approximately 1.9612.
Looking up the z-score in the standard Normal table, we find that the area to the left of 1.9612 is 0.9744.
Therefore, the probability that the mean time of 5 deliveries will exceed one hour is approximately 1 - 0.9744 = 0.0256 or 2.56%.