The daily sales at a convenience store have a mean of $1350 and a standard deviation of $150. Assuming n/N is less than or equal to .05, that standard deviation of the sampling distribution of the mean sales of a sample of 25 days for this convenience store is ?
The daily sales at a convenience store have a mean of $1350 and a standard deviation of $150. The mean of the sampling distribution of the mean sales of a sample of 25 days for this convenience store is
To find the standard deviation of the sampling distribution of the mean sales, you can use the formula:
Standard Deviation (sampling distribution) = Standard deviation (population) / √(sample size)
First, let's find the standard deviation of the sampling distribution:
Standard deviation (sampling distribution) = 150 / √25
Standard deviation (sampling distribution) = 150 / 5
Standard deviation (sampling distribution) = 30
Therefore, the standard deviation of the sampling distribution of the mean sales of a sample of 25 days for this convenience store is 30.
To calculate the standard deviation (σ) of the sampling distribution of the mean, we can use the formula:
σ = (Standard Deviation of Population) / √(Sample Size)
Given that the standard deviation of the daily sales at the convenience store is $150 and the sample size (n) is 25, we can substitute these values into the formula:
σ = 150 / √25
Now, let's calculate:
σ = 150 / 5 = 30
Therefore, the standard deviation of the sampling distribution of the mean sales for a sample of 25 days at this convenience store is $30.