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# Standard Deviation of a Data Set

## Percentages and the square root of the variance

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Calculating Standard Deviation

#### Objective

Here you will learn to calculate the standard deviation of sample sets and populations.

#### Concept

What is standard deviation? How is the standard deviation of a set related to variance? Is the standard deviation of a sample different from that of a population, the way it is with variation?

This lesson details the process of calculating standard deviation, and introduces a few examples of its use. After the lesson we’ll review the questions above, using the knowledge we have gained.

#### Watch This

http://youtu.be/HvDqbzu0i0E Khan Academy – Statistics-Standard Deviation

#### Guidance

Standard deviation   $(\sigma)$ is a very common term in statistics and it is not particularly difficult to calculate, particularly if you have already identified the variance of a set. The standard deviation is sort of a “reference difference from the mean” that you can use to evaluate the spread of the data in a set.

For instance, assume the mean of a particular set is 6 and the standard deviation is 4. If you are considering a value of 21, it is probably a very rare occurrence in that set since 21 is nearly 4 standard deviations away from the mean ( $4 \times 4 = 16$ and 16 more than the mean would be 22). However, a value of 8 is much more likely, given that it is only  $\frac{1}{2}$ of a standard deviation (SD) away from the mean.

Recall from the lesson Calculating Variance that calculating the variance of a set involves finding the arithmetic mean, subtracting each data point value from the mean and squaring the result, then finding the sum of the squared results and dividing by either the number of members of the set (population) or the number of members -1 (sample). See the first part of Example B for a review of finding the variance.

• Once you have the variance of a population, you are practically done finding the SD.
• To find the SD, simply take the square root of the variance . That’s it!
• One important difference between the variance and the standard deviation is that the units associated with variance are the square of the units of the original values, but the units associated with the standard deviation are the same as the units in the original set.

We will return to SD in our chapter on “Normal Distribution”, when we will further discuss the uses of the SD of both samples and populations.

Example A

What is the standard deviation of a set with  $\sigma^2$ of 14.6?

Solution: The standard deviation  $(\sigma)$ is simply the square root of the variance $(\sigma^2)$ . As a formula: $\sqrt{\sigma^2}$ .

In this case we have:  $\sqrt{14.6}=3.821 \ \therefore \sigma=3.821$

Example B

What is the  $\sigma$ of set $x$ ?

$x=\left \{3,4,5,6,7,8,9\right \}$

Solution: First find the variation of the set:

• $\mu (mean)=\frac{3+4+5+6+7+8+9}{42}=6$
• Deviations and squared deviations:
• $3-6=-3 \rightarrow (-3)^2=9$
• $4-6=-2 \rightarrow (-2)^2=4$
• $5-6=-1 \rightarrow (-1)^2=1$
• $6-6=0 \rightarrow (0)^2=0$
• $7-6=+1 \rightarrow (+1)^2=1$
• $8-6=+2 \rightarrow (+2)^2=4$
• $9-6=-3 \rightarrow (-3)^2=9$
• $\text{Sum of squared deviations}= 9+4+1+0+1+4+9=28$
• $\text{Variation}=\frac{28}{7}=4$

$\therefore \ Standard \ deviation \ of \ set \ x=\sqrt{4}=2$

Example C

Katrina wants to use the average scores of the top long jumpers at the 5 schools in her district to predict the average long jumps for top competitors at all schools in her state. Data for her district is below. Find the appropriate variance and standard deviation of the jumps.

$& \text{School} \ \#1 \quad 24^\prime 10.5^{\prime \prime} \\& \text{School} \ \#2 \quad 24^\prime 8.5^{\prime \prime} \\& \text{School} \ \#3 \quad 24^\prime 4.25^{\prime \prime} \\& \text{School} \ \#4 \quad 24^\prime 1.75^{\prime \prime} \\& \text{School} \ \#5 \quad 23^\prime 10.5^{\prime \prime}$

Solution: Since Katrina intends to generalize from her sample data back to the population of jumpers in her state; we need to find the sample variance and corresponding sample standard deviation.

• Start by finding the mean distance:  $\mu=\frac{24^\prime 10.5^{\prime \prime}+24^\prime 8.5^{\prime \prime}+24^\prime 4.25^{\prime \prime}+24^\prime 1.75^{\prime \prime}+23^\prime 10.5^{\prime \prime}}{5}=24^\prime 4.7^{\prime \prime}$
• As a decimal:  $24.39^\prime$
• Deviations and squared deviations of each value:
• $24^\prime 10.5^{\prime \prime}= 24.875^\prime: 24.875-24.39=.485 \rightarrow (.485)^2=.235$
• $24^\prime 8.5^{\prime \prime} = 24.71^\prime: 24.71^\prime-24.39^\prime=.32 \rightarrow (.32)^2=.102$
• $24^\prime 4.25^{\prime \prime} = 24.35^\prime: 24.35^\prime-24.39^\prime=-.04 \rightarrow (-.04)^2=.001$
• $24^\prime 1.75^{\prime \prime} = 24.15^\prime: 24.15^\prime-24.39^\prime=-.24 \rightarrow (-.24)^2=.058$
• $23^\prime 10.5^{\prime \prime}=23.875^\prime: 23.875-24.39^\prime=-.52 \rightarrow (-.52)^2=.270$
• $\text{Sum of squared deviations}= .235+.102+.001+.058+.270= .666$
• $\text{Sample variance}= \frac{.666}{4}=.167^\prime$ (Remember to divide by $n-1$ , since this is a sample)
• $\text{Standard deviation}= \sqrt{.167}=.409^\prime = 4.9^{\prime \prime}$

Concept Problem Revisited

What is standard deviation? How is the standard deviation of a set related to variance? Is the standard deviation of a sample different from that of a population, the way it is with variation?

By now you should know that standard deviation is a measure of the spread of data, and is calculated as the square root of the variance. Since variance is calculated slightly differently for a sample than for a population, the deviation will differ similarly.

#### Vocabulary

Standard deviation $(\sigma)$ is calculated by finding the square root of the variance. The standard deviation acts as a reference unit of difference from the mean in a set of data.

The variance $(\sigma^2)$ is calculated as the sum of the squared differences from the mean, divided by either the number of values (for populations) or the number of values minus one (for samples).

#### Guided Practice

1. Find the mean $(\mu)$ , variance $(\sigma^2)$ , and standard deviance  $(\sigma)$ of set $z$ .

$z=\left \{12.3,12.5,12.2,11.9,12.6,12.35 \right \}$

2. Find the mean $(\mu)$ , variance $(\sigma^2)$ , and standard deviance  $(\sigma)$ of set $y$ .

$y=\left \{ 9.1,10.1,8.27,7.9,8.6,10.0\right \}$

3. Which set has the greater standard deviation, $x$ or $y$ ?

$x=\left \{2,4,6,8,10 \right \} \ y=\left \{3,5,7,9,11,13 \right \}$

4. Kevin takes a random sample of ages of students in his class, and gets the following values, what is the sample variance and standard deviation of the set?

$a=\left \{ 15,16,16,15,17,17,18,16,17,16,18,18,15\right \}$

Solutions:

1. Let’s start by finding the mean, since we will need it to calculate the others:

$\frac{12.3+12.5+12.2+11.9+12.6+12.35}{6}&=12.30833 \\The \ mean \ (\mu)&=12.30833$

Now we calculate the deviation of each value from the mean and square it:

$12.3-12.30833 = -0.00833 \rightarrow -0.00833^2=0.00007$

$12.5-12.30833= 0.19167 \rightarrow 0.19167^2= 0.03674$

$12.2-12.30833= -0.10833 \rightarrow -0.10833^2=0.01174$

$11.9-12.30833= -0.40833 \rightarrow -0.40833^2= 0.16673$

$12.6-12.30833=0.29167 \rightarrow 0.29167^2= 0.08507$

$12.35-12.30833=0.04167 \rightarrow 0.04167^2=0.00173$

Now we sum the squared deviations: $0.00007+0.03674+0.01174+0.16673+0.08507+0.00173=0.30208$ , and divide the total by the number of values:  $\frac{0.30208}{6}=0.050347$ to get the variance.

$The \ variance \ (\sigma^2 )=0.05347$

Finally, to get the standard deviation $(\sigma)$ , just take the square root of $\sigma^2$ .

$The \ standard \ deviation \ (\sigma) \ is \ \sqrt{0.05347}=0.23124$

2. Start by finding  $\mu: \frac{9.1+10.1+8.27+7.9+8.6+10.0}{6}=8.995$

Next, find the squared variation from the mean for each value:

• $9.1-8.995=0.105 \rightarrow 0.105^2=0.011025$
• $10.1-8.995=1.105 \rightarrow 1.105^2=1.221025$
• $8.27-8.995=-0.725 \rightarrow 0.525625^2=0.276281$
• $7.9-8.995=-1.095 \rightarrow -1.095^2=1.199025$
• $8.6-8.995=-0.395 \rightarrow 0.105^2=0.156025$
• $10.0-8.995=1.005 \rightarrow 0.105^2=1.010025$

Sum the squared deviations and divide by the number of values to get the variance:

$\sigma^2=\frac{3.873406}{6}=0.6455677$

Finally, take the square root of the variance to get the standard deviation:

$\sigma=\sqrt{0.6455677}=.8034723$

3. Follow the same series of steps to find the standard deviation of each set.

• $x=\left \{2,4,6,8,10 \right \} : \mu=6,\sigma^2=10,\sigma=3.16228$
• $y=\left \{3,5,7,9,11,13 \right \} : \mu=8, \sigma^2=14,\sigma=3.74166$

Set $y$ has the greater standard deviation

4. There are 13 values, with  $\mu= 16.46154$

• The sum of the squared deviations is: 15.2308, divide by 12 (since this is a sample!), to get the sample variance : $\frac{15.2308}{12}=1.26923$
• The square root of the sample variance is the sample standard deviation: $\sqrt{1.26923}=1.0824$

#### Practice

Find  $\mu, \sigma^2$ and $\sigma$ :

1. 265, 280.7, 293, 279, 314.2, 300, 289

2. 7200, 7020, 7165.9, 7100, 7196, 7112, 7116.1

3. 27, 20.3, 30.7, 40, 46, 36, 40, 33

4. 3607, 3600, 3600, 3631, 3600.6

5. 700, 700, 712, 736, 741, 716, 782

6. 3370, 3300.5, 3366, 3306.6, 3310, 3336, 3301.3

Calculate the sample standard deviation:

7. 34.4, 34, 34.7, 34.6, 34, 34.1, 31, 31.3

8. 989.22, 990.6, 992, 996.9, 981.1, 986, 975

9. 10, 16, 10.33, 10.63, 18, 17, 16.36, 10.46

10. 3240, 3260, 3250, 3280, 3280, 3300, 3310, 3270

### Vocabulary Language: English

68-95-99.7 Rule

68-95-99.7 Rule

When 68% of the data values would be located within 1 standard deviation of the mean, 95% of the data values would be located within 2 standard deviations of the mean, and 99.7% of the data values would be located within 3 standard deviations of the mean, statisticians refer to this as the 68-95-99.7 Rule.
inflection point

inflection point

An inflection point is a point in the domain where concavity changes from positive to negative or negative to positive.
standard deviation

standard deviation

The square root of the variance is the standard deviation. Standard deviation is one way to measure the spread of a set of data.
variance

variance

A measure of the spread of the data set equal to the mean of the squared variations of each data value from the mean of the data set.