Exercise 14 Page 311

Let's start by reviewing the definition for standard deviation σ. Standard deviation is a measure of spread that represents the average distance between each data value and the mean. The formula for the standard deviation of a data set containing n elements { x_1, x_2, ..., x_n } with a mean value x is shown below. Standard deviation [0.8em] σ = sqrt((x_1-x)^2+(x_2-x)^2+... +(x_n-x)^2/n) The first step to calculate the standard deviation is to find the mean. x=x_1+x_2+x_3+ ... +x_n/n

Then, we find the square of the distance between each data value and the mean. (x_n-x)^2 We continue by finding the mean of the squared values. (x_1-x)^2+(x_2-x)^2+... +(x_n-x)^2/n Finally, the standard deviation is the square root of mean of squared deviations. σ = sqrt((x_1-x)^2+(x_2-x)^2+... +(x_n-x)^2/n)

Example

Find the standard deviation for the data set 3, 4, 5, 5, 5, 6, 7. We will solve this by following the steps previously mentioned. First we find the mean.

x=x_1+x_2+x_3+ ... +x_n/n

SubstituteValues

Substitute values

x=3+4+5+5+5+6+7/7

AddTerms

Add terms

x=35/7

CalcQuot

Calculate quotient

x=5

Now, we can calculate the square of the distance between each data value and the mean.

Data value x	Deviation from the mean x-x	Squared deviation (x-x)^()2
3	3-5=-2	(-2)^2 = 4
4	4-5=-1	(-1)^2 = 1
5	5-5=0	(0)^2 = 0
5	5-5=0	(0)^2 = 0
5	5-5=0	(0)^2 = 0
6	6-5=1	(1)^2 = 1
7	7-5=2	(2)^2 = 4
		Total : 10

Next, we proceed to find the mean of the squared values. Notice that we already know that the sum of all squared deviations is 10.

(x_1-x)^2+(x_2-x)^2+... +(x_n-x)^2/n

SubstituteValues

Substitute values

10/7

CalcQuot

Calculate quotient

1.428571...

Finally we calculate the square root of the mean of the squared deviations.

σ = sqrt((x_1-x)^2+(x_2-x)^2+... +(x_n-x)^2/n)

SubstituteValues

Substitute values

sqrt(1.428571...)

CalcRoot

Calculate root

1.195228 ...

RoundDec

Round to 1 decimal place(s)

1.2

We found that the standard deviation for this data value is approximately 1.2.