Standard Deviation

Key Concepts

• Understand what is standard deviation.
• Understand what is variance.
• Understand which describes the data set completely.
• Understand the formula of standard deviation of a sample and population.

Model and Discuss: 

A meteorologist compares the low temperature for Jacksonville and Miami over the past 10 days. 

1. Create a data display for each city’s high temperature 

2. Use structure: What does the shape of each data display indicate about the data set and the measure of center? 

a. 

Step 1:  

The following is the dot plot for city A: 

Step 2:  

The following is the dot plot for city B: 

Solution: 

• City A’s data is symmetric, whereas for city B, it is not symmetric and is right-skewed. 

The mean and median temperature is the same for city A, equal to 64 degrees. 

• The mean and median temperatures are not the same and since it is right-skewed, the mean temperature would be higher than the temperature. 
• City A’s data is symmetric, whereas city B’s data is right-skewed. 

Example 1:   Interpret the variability of a data set 

The makers of a certain brand of light bulbs claim that the average life of the bulb is 1200 hours. The  

life spans of hours of a sample of brand A light bulbs are shown below. How close to claim were 

the light bulbs in this sample?  

Answer: 

1. What type of data display will provide the best view of the data? 

The numbers in the data set span a large range of numbers. To understand how these numbers relate to the claimed mean, create a histogram. 

2. What does the display reveal about the data? 

The data forms a bell-shaped curve. Data that is in this shape is said to have a normal distribution. 

When data are normally distributed, the most useful measure of spread is the standard deviation.  

Standard deviation is a measure that shows how data vary or deviate, from the mean. 

In a normal distribution, 

• About 68% of data fall within one standard deviation of the mean. 
• About 95% of data fall within two standard deviations of the mean. 

• About 99.7% data fall within three standard deviations of the mean. 

The standard deviation for this sample of light bulbs is 75.5 hours. 

The mean of this data set is 1200. Light bulbs with a life span between 1,124.5 hours and 1,275.5 hours  

lie within one standard deviation of the mean. In a large sample drawn randomly among such light  

bulbs, we would always expect about 68% of the life spans to be within one standard deviation  

of the mean, thanks to the predictive power of normal distribution. 

Try it 

1. What is the life span of light bulbs that are within 2 standard deviations of the mean? Within 3 standard deviations of the mean? 

Solution: 

Step 1: 

The mean (

x−x-

) and the standard deviation have been solved for and equal 1200 and 75.5 hours, 

respectively. The life span of light bulbs that are within 2 standard deviations of the mean lies  

between

x−x-

– 2x and

x−x- + 2s. The life span of light bulbs that are 2 standard deviations from the mean  

lies between

x−x-

– 3s and

x−x- + 3s. 

Step 2: 

x−x-

– 2s = 1200 – 2(75.5) 

= 1200 – 151   

=1049 

x−x-

+ 2s = 1200 + 2(75.5) 

= 1200 + 151   

=1351 

x−x-

– 3s = 1200 – 3(75.5) 

= 1200 – 226.5   

=973.5 

x−x-

– 3s = 1200 + 3(75.5) 

= 1200 +226.5   

=1426.5 

• The life span of light bulbs that are within 2 standard deviations of the mean lies between 1049 hours and 1351 hours  
• The life span of light bulbs that are within 2 standard deviations of the mean lies between 973.5 hours and 1426.5 hours 

Example 2: Calculate the standard deviation of a sample 

The table shows the number of cars sold by an auto sales associate over an eight-week 

period. How much variability do the data show? 

You can find the variability by solving for the standard deviation for a sample, using the formula 

s =

∑(x − x−)2n−1−−−−−−−−−−−−⎷∑(x − x-)2n−1

Step 1: 

Find the mean of the data by finding the sum of the data points and dividing by 8. the  

notation 

x−x-

  is used to indicate the mean  

x−x-

= 17 

Step 2: 

Find the difference between each data value, x, and the mean,

x−x-

.Then square each difference. 

Step 3:  

The variance of a total population, often

σ2𝜎2

, is the mean of the squares of the differences between  

each data and the mean. When finding the variance of a sample often noted

s2s2

dividing by n provides 

Too small an estimate of the variance of the population. This is because data points from the sample  

are likely to cluster more closely around the sample mean than the population mean.  If you divide n-1  

Instead of n, you get a slightly bigger number that is closer to the true population variance. 

Step 4: 

Take the square root of the variance to find the standard deviation, s. 

s ≈

17.71−−−−−√17.71

s ≈ 4.21  

The standard deviation is about 4 cars, and the mean is about 17 cars, so there is some variability in the data. The sales associate will sell between 13 and 21 cars about 68% of the time because those values are one standard deviation from the mean. 

Try it  

2. The table shows the number of cars sold by the auto sales associate over the next eight–week period. How much variability does the data show? 

Solution: 

Step 1:  

Find the mean,

x−x-

Mean of the data set is

∑xin∑xin

  , where

x′is xi′s are data values and n is the total number of values. 

Using this, we get a mean equal to 16.5. 

Step 2: 

Find the square of the difference between each data value and mean. 

Step 3: 

Find the variance of sample

s2s2

Variance,

s2s2

=

∑(x − x−)2n−1−−−−−−−−−−−−⎷∑(x − x-)2n−1

=

∑(x − x−)27−−−−−−−−−−−−⎷∑(x − x-)27

=

22872287

≈ 32.57 

Step 4: 

Find the standard deviation of sample s. 

s =

s2−−√s2

≈

32.57−−−−−√32.57

  ≈5.7 

The standard deviation is about 5.7 cars, and the mean is 16.5 hours, so there is some variability  

in the data. The sales associate will sell

x−x-

– s = 10.8 and

x−x- + s = 22.2 cars about 68%  

of the time in the next week 

Example 3: Find the standard deviation of a population 

The table displays the number of points scored by a football team during each of their regular season’s games. 

1. What are the mean and standard deviation for this data set? 

Find the mean of the data by finding the sum and dividing by 16. 

x−x-

= 16.8 

Find the variance. Use n as the denominator because the data represents the full population 

σ2𝜎2

=

∑(x − x−)2n−1∑(x − x-)2n−1

Find the standard deviation by taking the square root of  

the variance.  

σ ≈

53.8−−−−√53.8

σ ≈ 7.3 

The mean of points that the team scored was about 17. The standard deviation is about 7.3 

This means that the team scored between about 10 and 25 points in about 68% of them 

games. 

2. The team played in two post-sessions games, scoring 7 points in one and 14 points in the other.  

How do these games affect the overall mean and standard deviation of the number of points scored all sessions? 

• Use graphing technology to find the mean and standard deviation. 
• Enter all 14 scores as a list in the graphing calculator. 
• Use the STAT menu to find the mean and standard deviation. 

• When the team includes their post-session games, the mean number of points scored decreases to about 15.9. The standard deviation remains about 7.3 

Try it 

3. What was the range of points that the team scored in 95% of their regular-season games? 

Solution: 

Step 1: 

The range of points that are scored in 95% of games is 2 standard deviations of mean. 

We know from the example 3 that  

Standard deviation, s ≈ 7.3 and  

Mean,

x−x-

  ≈ 16.8 

Step 2: 

Finding 2 standard deviations from mean: 

x−x-

– 2s = 16.8 – 2(7.3) 

=16.8 – 14.6 

=2.2 

x−x-

+ 2s = 16.8 + 2(7.3) 

=16.8 + 14.6 

=31.4 

The range of points that are scored in 95% of games is between 2.2 and 31.4. 

Example 4: Compare data sets using standard deviation 

The histograms show the life spans of a sample of light bulbs from 2 companies. 

The first shows a sample from brand A and the second one from Brand B. The red lines indicate the 

mean and each standard deviation from the mean. Compare the distributions of life spans for the two light bulb brands. 

• Both brands of light bulbs have an average life of about 1200 hours. 

• The standard deviation for the data set for brand A is less than the standard deviation for brand B. This suggests that Brand A light bulbs show less variability in their life spans 
• Since their light bulbs are more consistently closer to the mean, Brand A light bulbs are more predictable in their life spans than Brand B 

Try it  

4. Compare Brand C, with a mean of 1250 hours and a standard deviation of 83 hours, to Brand A and B 

Solution: 

Comparing the means, the light bulbs of brand C perform better than Brand A and Brand B, on an average  

Comparing the standard deviations, the performance of brand C is more consistent than Brand B but less compared to Brand A 

Concept Summary  

Words 

Standard Deviation is a measure of spread or variability. It indicates how much the values in a data set deviate from the mean. It is the square root of the variance. The variance is the average of the squared deviations from the mean. When data are normally distributed (in a bell curve), the mean and the standard deviation describe the data set completely. 

Algebra 

The standard deviation of a sample 

s =

∑(x − x−)2n−1−−−−−−−−−−−−⎷∑(x − x-)2n−1

The standard deviation of the population 

s =

∑(x − x−)2n−−−−−−−−−−−−⎷∑(x − x-)2n

Diagram 

Check your knowledge 

• Find and use the mean and the standard deviation to compare the variability of each pair of sample data sets: 

1. 

Data set A: 6.9.1.2,3,4,4,4 

Data set B:10,5,5,2,3,7,4,8 

2. 

3. Find and use the mean and standard deviation to compare the variability of the populations represented in the dot plots. 

Data values in normally distributed data sets A and B are integers from 0 to 30 inclusive. Identify  

The range of values that satisfies each description. 

Data set A: mean12; standard deviation:2 

Data set B: mean: 18; Standard deviation :3  

Answers: 

1. 

Data set A: 6.9.1.2,3,4,4,4 

Data set B :10,5,5,2,3,7,4,8 

Solution: 

Step 1: Find the mean

x−x-

Mean of data set is

∑xn∑xn

, whereas x’s are data values and n is the total number of values. 

Using this, we get the means of sample A and B to be 4.25 and 5.5, respectively 

Step 2: For sample A 

Step 2: 

For sample B 

Step 3: 

Find the variance of samples,

s2As2A

and

s2Bs2B

Variance of sample A,

s2As2A

=

∑(x − x−)2n−1∑(x − x-)2n−1

=

∑(x − x−)27∑(x − x-)27

=

43.5743.57

≈6.21 

Variance of sample A,

s2Bs2B

=

∑(x − x−)2n−1∑(x − x-)2n−1

=

∑(x − x−)27∑(x − x-)27

=

507507

≈7.14 

Step 4: 

Find standard deviation of samples,

sA

  and

sBsB

sAsA

=

s2A−−−√s2A

=

6.21−−−−√6.21

≈2.49 

sBsB

=

s2B−−−√

=

7.14−−−−√

≈2.49 

Since standard deviation of sample B is greater, the variability of sample B. 

∴ Sample data set B has more variability. 

2. 

Solution: 

Step 1: 

Find mean,

x−x-

Mean of data set is

∑xn∑xn

, where x’s are data values and n is the total number of values. 

Using this, we get means of samples A and B to be 40.225 and 39.975, respectively. 

Find the square of the difference between each data values and mean. 

Step 2:                                                         

For Sample A:         

For Sample B: 

Step 3: 

Find the variance of samples ,

s2As2A

and

s2Bs2B

Variance of sample A,

s2A

= [Equation] 

= [Equation] 

≈

45259

 ≈ 502.8 

Variance of sample A,

s2A

= [Equation] 

= [Equation] 

≈

4199

≈ 46.5 

Step 4:  

Find the standard deviation of samples,

sA

  and

sB

sA

=

s2A−−−√

=

502.8−−−−−√

≈22.42 

sB

=

s2B−−−√

=

46.5−−−−√

≈6.82 

Since standard deviation is higher for sample data set A, variability is higher for sample data set A 

• Sample data set A has more variability  

3. 

Solution: 

Step 1: 

Find mean,

x−

Mean of data set is

∑xn

, where x’s are data values and n is the total number of values. 

Using this, we get the means of both populations to be 9. 

Step 2: 

Find the square of the difference between each data values and mean. 

For Sample A: 

For Sample B: 

Step 3: 

Find the variance of samples,

σ2A

and

σ2B

Variance of sample A,

σ2A

= [Equation] 

= [Equation] 

= 

8817

≈ 5.18 

Variance of sample B,

σ2B

= [Equation] 

= [Equation] 

≈

18417

≈ 10.82 

Step 4:  

Find standard deviation of samples,

sA

  and

sB

sA

=

s2A−−−√

=

1.58−−−−√

≈1.26 

sB

=

s2B−−−√

=

10.82−−−−−√

≈3.29 

The means are the same for both populations, but the variation is higher for data set B 

• Variability for Data set B.