20 15 17 24 18 17 15 17 14 17 19 17 24 18 20 18 19 19 15 15 SS Subscript xSSxequals=Upper Sigma x squared minus StartFraction left parenthesis Upper Sigma x right parenthesis squared Over n EndFractionΣx2−(Σx)2n LOADIN ​(a) Use the shortcut formula to calculate the sample standard deviation. sequals=nothing ​(Simplify your answer. Round to one decimal place as​ needed.)

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We are asked to find the sample standard deviation of the following twenty numbers:

20, 15, 17, 24, 18, 17, 15, 17, 14, 17, 19, 17, 24, 18, 20, 18, 19, 19, 15, 15

The standard deviation is the square root of the variance. The formula for the variance is `s^2=sum(x-bar(x))^2/(n-1)` where x is a data value, `bar(x)` is the sample mean, and n is the size of the sample. Thus the standard deviation is given by `s=sqrt(sum(x-bar(x))^2/(n-1)`

For each data value we would subtract the mean and square the result. Then we would add each of these values together and divide by 19. Thus

`s=sqrt((2.1^2+(-2.9)^2+(-.9)^2+6.1^2+.1^2+(-.9)^2+(-2.9)^2+(-.9)^2+(-3.9)^2+(-.9)^2+1.1^2+(-.9)^2+6.1^2+.1^2+2.1^2+.1^2+1.1^2+1.1^2+(-2.9)^2+(-2.9)^2)/19)` `=sqrt(699/95)~~2.71254396`

There is a shortcut method.

First add up all of the data values: `sum x=358`

Then square each data value and get the sum of these: `sum X^2=6548`

**Note that `sum x^2 ne (sum x )^2` **

Now the shortcut formula is: `s=sqrt((n(sum x^2)-(sumx)^2)/(n(n-1)))`

These two formulas can be shown to be algebraically the same.

Now to compute s we get `s=sqrt((20(6548)-(358)^2)/(20*19))`


So the sample standard deviation is `s~~2.7`

Note that the formula for the population standard deviation divides by n instead of (n-1). This is because a sample taken from a population is assumed to be "biased" in the sense that the spread will be less than that of the total population. (Imagine selecting fifteen people from a school. It is highly unlikely that a truly random sample includes the shortest and tallest person.) Dividing by (n-1) increases the size (dividing by a smaller number makes a fraction larger) of the deviation to correct for the sample.

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