the distribution of file size has a mean of $\bar{x}=3.2$ megabytes and a standard deviation of $s_x = 1.9$…

the distribution of file size has a mean of $\bar{x}=3.2$ megabytes and a standard deviation of $s_x = 1.9$ megabytes. choose the correct interpretation of $s_x = 1.9$ megabytes. the size of these 18 files typically varies by about 1.9 megabytes from the mean of 3.2 megabytes. the size of these 18 files typically varies by about 3.2 megabytes from the mean of 1.9 megabytes. standard deviation is calculated by finding the sum of the squared differences of each individual value and the mean. then, you divide by the sample size minus one. finally, take the square root. $s=sqrt{\frac{sum(x - \bar{x})^2}{n - 1}}$ you can expect these 18 files to vary in size from 1.3 to 5.1 megabytes.

the distribution of file size has a mean of $\bar{x}=3.2$ megabytes and a standard deviation of $s_x = 1.9$ megabytes. choose the correct interpretation of $s_x = 1.9$ megabytes. the size of these 18 files typically varies by about 1.9 megabytes from the mean of 3.2 megabytes. the size of these 18 files typically varies by about 3.2 megabytes from the mean of 1.9 megabytes. standard deviation is calculated by finding the sum of the squared differences of each individual value and the mean. then, you divide by the sample size minus one. finally, take the square root. $s=sqrt{\frac{sum(x - \bar{x})^2}{n - 1}}$ you can expect these 18 files to vary in size from 1.3 to 5.1 megabytes.

Answer

Brief Explanations:

The standard - deviation measures the typical amount by which data points differ from the mean. Here, the mean file size is 3.2 megabytes and the standard deviation $s_x = 1.9$ megabytes. This means the size of the files typically varies by about 1.9 megabytes from the mean of 3.2 megabytes.

Answer:

The size of these 18 files typically varies by about 1.9 megabytes from the mean of 3.2 megabytes.