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.