Activity Assess 11-4 MODEL & DISCUSS Standard A meteorologist compares the high temperatures for two cities during the past 10 days. Deviation

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I CAN… quantify and analyze the spread of data. City A : 62° 64° 66° 65° 64° 64° 63° 65° 63° 64° VOCABULARY • normal distribution • standard deviation • variance

City B : 68° 63° 64° 62° 63° 62° 62° 63° 64° 63°

A. Create a data display for each city’s high temperatures. B. Use Structure What does the shape of each data display indicate about the data set and the measures of center?

ESSENTIAL QUESTION Why does the way in which data are spread out matter?

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 1,200 hours. The life spans, in hours, of a sample of Brand A light bulbs are shown. How close to the claim were the light bulbs in this sample?

1,150 1,231 1,305 1,080 1,125 1,295 1,127 1,184 1,099 1,123 1,204 1,345 1,173 1,126 1,220 1,245 1,283 1,225 1,185 1,275

A. 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.

USE STRUCTURE 8 How does the shape of the 6 distribution help you identify the mean? 4

Frequency 2 0 1,150–1,199 1,100–1,149 1,050–1,099 1,450–1,499 1,250–1,299 1,200–1,249 1,350–1,399 1,000–1,049 1,300–1,349 1,400–1,449 Life Span (h)

CONTINUED ON THE NEXT PAGE

LESSON 11-4 Standard Deviation 487

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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.

−1 standard deviation +1 standard deviation

34.1% 34.1% −2 standard deviation +2 standard deviation

−3 standard deviation 13.6% 13.6% +3 standard deviation 2.1% 2.1% 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% of data fall within three standard deviations of the mean. The standard deviation for this sample of lightbulbs is 75.5 hours. The mean of this data set is 1,200. 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 the normal distribution.

Try It! 1. What is the lifespan of light bulbs that are within 2 standard deviations of the mean? Within 3 standard deviations of the mean?

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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?

18 25 18 10 17 15 18 15

You can find the variability by solving for the standard deviation for a ______2 ∑(x – ¯x​ ​​) ​​ ​ sample, using the formula s = ​​ ______​ . To calculate it, follow these steps. √ n – 1 Step 1 Find the mean of the data by finding the sum of the data points and dividing by 8. The notation ¯x​​ ​ is used to indicate the mean. ¯​​ x​ = 17​ Step 2 Find the difference between each data value, x, and the mean, ¯x​​ ​​. Then square each difference. COMMON ERROR Remember to square the x 18 25 18 10 17 15 18 15 differences between each data x 17 17 17 17 17 17 17 17 value and the mean. Otherwise, the sum of the differences will be zero. x − x 1 8 1 −7 0 −2 1 −2 2 (x − x) 1 64 1 49 0 4 1 4

Step 3 Find the variance. 2 The variance of a total population, often noted σ​​ ​​ ​​, is the mean of the squares of the differences between each data value and the mean. When finding the variance of a sample, often noted ​​s​​ 2​, 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 by n – 1 instead of n, you get a slightly bigger number that is closer to the true population variance. 1 64 1 49 0 4 1 4 ​​s​​ 2​ ______​ + + + + + + + ​ = 7 In this sample, n = 8 so 2 ​​s​​ ​ ≈ 17.71​ we divide by n − 1, or 7. Step 4 Take the square root of the variance to find the standard deviation, s. ______​s ≈ √​ 17.71 ​​ Since only whole cars can be sold, it makes ​s ≈ 4.21​ sense to round the standard deviation to 4.

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 do the data show?

12 14 29 10 17 16 18 16

LESSON 11-4 Standard Deviation 489

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EXAMPLE 3 Find Standard Deviation of a Population

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

24 13 10 21 18 3 27 18 20 14 7 27

A. 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 ​​ ≈ 16.8 Find the variance. Use n as the denominator because the data represent STUDY TIP the full population. Use the Greek letter (sigma) for σ 2 2 ∑(x – ¯x​ ​​) ​​ ​ standard deviation when working σ​​ ​​ ​ = ​ ______​ with populations. Use s when n 51.36 14.69 46.69 17.36 1.36 191.36 103.36 1.36 10.03 8.03 96.69 103.36 ​​ ​​ 2​ ​​ ______+ + + + + + + + + + + ​ working with samples. σ ≈ 12 2 σ ​​ ​ ≈ 53.8 In this example, divide by n because you are working with the entire population of games.

Find the standard deviation by taking the square root of the variance. _____ σ ≈ √​ 53.8 ​​ σ ≈ 7.3 The mean number 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 percent of their games. B. The team played in two post-season 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 season? Use graphing technology to find the mean L1 L2 L3 1 and standard deviation. 24 13 10 Enter all 14 scores as a list in the graphing 21 18 calculator. 3 27 Use the STAT menu, to find the mean and L1(1)=24 standard deviation. When the team includes their post-season 1-Var Stats games, the mean number of points scored x = 15.92857143 decreases to about 15.9. The standard Σx = 223 Σx2 = 4291 deviation remains about 7.3. Sx = 7.539274457 σx = 7.265026651 n = 14

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

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APPLICATION 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 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. Brand A 12 10 mean: 1,200 h standard deviation: 8 75.5 h

6 –2s –s mean s 2s

Frequency 4 2 0 1,150–1,199 1,100–1,149 1,050–1,099 1,450–1,499 1,250–1,299 1,200–1,249 1,350–1,399 1,000–1,049 1,300–1,349 1,400–1,449 Hours

Brand B 12

10 mean: 1,200 h standard deviation: 8 90.25 h

6 –2s –s mean s 2s

Frequency 4 2 0 1,150–1,199 1,100–1,149 1,050–1,099 1,450–1,499 1,250–1,299 1,200–1,249 1,350–1,399 1,000–1,049 1,300–1,349 1,400–1,449 Hours

Both brands of light bulbs have an average life of about 1,200 hours. USE STRUCTURE The standard deviation for the data set for Brand A is less than the standard How does the shape of each graph help you compare the deviation for Brand B. This suggests that Brand A light bulbs show less standard deviations? 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 mean 1,250 hours and standard deviation 83 hours, to Brands A and B.

LESSON 11-4 Standard Deviation 491

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WORDS Standard deviation is a measure of spread, or variability. It indicates by 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 Standard Deviation of a Sample Standard Deviation of a Population ______2 2 ∑​(x − ¯x​ ​) ​​ ​ ______∑​(x − ¯x​ ​) ​​ ​ ​s = ​ ______​ ​ ​s = ​√ ​ ​ ​​ √ n − 1 n the sum of the squares of the deviation from the mean

DIAGRAM −1 standard deviation +1 standard deviation

34.1% 34.1% −2 standard deviation +2 standard deviation

−3 standard deviation 13.6% 13.6% +3 standard deviation 2.1% 2.1% mean

Do You UNDERSTAND? Do You KNOW HOW?

1. ESSENTIAL QUESTION Why does the Use the sample data for Exercises 5–7. way in which data are spread out matter? Sample A: 1, 2, 2, 5, 5, 5, 6, 6 2. Generalize What are the steps in finding Sample B: 5, 9, 9, 10, 10, 10, 11, 11 standard deviation? 5. What can you determine by using range to compare the spread of the two samples? 3. Error Analysis Marisol says that standard deviation is a measure of how much the 6. Find the standard deviation for each sample. values in a data set deviate from the 7. How can you use standard deviation to median. Is Marisol correct? Explain. compare the spread of each sample? 4. Use Structure If you add 10 to every 8. Based on the histogram, what data values are data value in a set, what happens to the within one standard deviation of the mean? mean, range, and standard deviation. Why? 6 Mean: 11.05 5 Standard Deviation: 2.40 4 3

Frequency 2 1 0 6–7 8–9 12–13 10–11 16–17 14–15 Score

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