Electricalc has determined that the assembly time for a particular electrical component is normally distributed with a mean of 20 minutes and a variance of 9 minutes.
a. What is the probability that an employee in the assembly division takes longer than 22 minutes to assemble one of these component?
b. What is the probability that the average assembly time for 15 such employees exceeds 22 minutes?
a. Use same process as previous post.
b. Because you are dealing with means (average) rather than individual scores:
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
To solve these probability questions, we need to use the properties of the normal distribution and the standard normal distribution table (also known as the Z-table).
To begin, we need to standardize the values using the formula:
Z = (X - μ) / σ
Where:
Z is the standard score,
X is the value we want to find the probability for,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.
a. To find the probability that an employee takes longer than 22 minutes to assemble one component, we need to find the area under the normal curve to the right of 22 minutes.
First, we calculate the standard score (Z):
Z = (22 - 20) / √9
Z = 2 / 3
Z ≈ 0.67
Then, we look up this Z-score in the standard normal distribution table (Z-table) to find the corresponding area. Since we want the probability to the right of 22 minutes, we subtract the Z-score from 1:
P(X > 22) = 1 - P(Z ≤ 0.67)
Using the Z-table, we find that P(Z ≤ 0.67) ≈ 0.7486.
Therefore, P(X > 22) = 1 - 0.7486 ≈ 0.2514.
So, there is approximately a 0.2514 probability that an employee takes longer than 22 minutes to assemble one component.
b. To find the probability that the average assembly time for 15 employees exceeds 22 minutes, we need to calculate the sample mean and the sample standard deviation.
Since the distribution of the assembly time is normally distributed, the distribution of the sample mean is also normally distributed. The mean of the sample mean is equal to the population mean, μ, which is 20 minutes in this case. The standard deviation of the sample mean, also known as the standard error, is calculated as σ / √n, where n is the number of observations (15 in this case).
Standard deviation of the sample mean = σ / √n
Standard deviation of the sample mean = 3 / √15 ≈ 0.7746
Next, we calculate the standard score (Z) for the value 22 minutes:
Z = (22 - 20) / 0.7746
Z ≈ 2.5819
Finally, we look up this Z-score in the Z-table to find the probability:
P(X̄ > 22) = 1 - P(Z ≤ 2.5819)
Using the Z-table, we find that P(Z ≤ 2.5819) ≈ 0.9950.
Therefore, P(X̄ > 22) = 1 - 0.9950 ≈ 0.0050.
So, there is approximately a 0.0050 probability that the average assembly time for 15 employees exceeds 22 minutes.