Suppose the time it takes for a purchasing agent to complete an online ordering process is normally distributed with a mean of 8 minutes and a standard deviation of 2 minutes. Suppose a random sample of 25 ordering processes is selected.
The standard deviation of the sampling distribution of mean times is
I think you are asking about the Standard Error of the mean (SEm).
SEm = SD/√n
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To find the standard deviation of the sampling distribution of mean times, we can use the formula:
Standard deviation of the sampling distribution of the mean = Standard deviation of the population / Square root of the sample size
Given:
Standard deviation of the population (σ) = 2 minutes
Sample size (n) = 25
Using the formula mentioned above, we can calculate the standard deviation of the sampling distribution of the mean:
Standard deviation of the sampling distribution of mean times = 2 / √25
= 2 / 5
= 0.4 minutes
Therefore, the standard deviation of the sampling distribution of mean times is 0.4 minutes.
To find the standard deviation of the sampling distribution of mean times, also known as the standard error of the mean, we can use the formula:
Standard Error of the Mean = Standard Deviation / √(Sample Size)
Given that the standard deviation of the purchasing agent's completion time is 2 minutes, and the sample size is 25, we can calculate the standard error of the mean as follows:
Standard Error of the Mean = 2 / √(25)
Simplifying the calculation:
Standard Error of the Mean = 2 / 5
Therefore, the standard deviation of the sampling distribution of mean times is 0.4 minutes.