# the following data list the average monthly snow fall for jan in 15 cities find the mean, variance and standard deviation 21 5 43 39 45 42 26 40 38 32 17 23 31 28 16 please show work I tried this and got confused (1) To find the mean we add all of the values and divide by 15 or `bar(x)=(sum_(i=1)^15 x_i)/n`

Adding all of the values we get 446. Then `bar(x)=446/15=29.7bar(3)`

Since the data values are whole numbers, typically you would report the mean as 29.7.

(2) We assume that this is...

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(1) To find the mean we add all of the values and divide by 15 or `bar(x)=(sum_(i=1)^15 x_i)/n`

Adding all of the values we get 446. Then `bar(x)=446/15=29.7bar(3)`

Since the data values are whole numbers, typically you would report the mean as 29.7.

(2) We assume that this is a sample. The formula for the variance is `s^2=(n(sum X^2)-(sum X)^2)/(n(n-1))`

Here n=15. We previously found the sum of the data points to be 446. Now we square each data point and add the results: `sum(X^2)=15188`

*** Note that `(sum X)^2 != sum X^2` For the left side you add all of the data together and then square the sum. For the right side you square each data point, and then add these together. ***

So the variance is `s^2=(15(15188)-446^2)/(15(14))=(227820-198916)/(210)~~137.6381`

So we report the variance as 137.64

(3) The standard deviation is the square root of the variance. Try to not round before taking the square root:

`s=sqrt(s^2)=sqrt(28904/210)~~11.7319`

So we report the standard deviation as 11.73

These values agree with the values obtained with a graphing calculator or a spreadsheet program. Note that algebra systems report two different standard deviations -- one is for the oppulation, the other for a sample. You must take care to use the correct one.

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