Math 118 Ppt 12.4

Section 12.4 The Normal Distribution 1. 2. 3. 4. 5. 6. 7. 8. 9.

Objectives Recognize characteristics of normal distributions. Understand the 68-95-99.7 Rule. Find scores at a specified number of standard deviations from the mean. Use the 68-95-99.7 Rule Convert a data item to a z-score. Understand percentiles and quartiles. Solve applied problems involving normal distributions. Use and interpret margins of error. Recognize distributions that are not normal.

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Normal Distribution • Also called the bell curve or Gaussian distribution. • Normal distribution is bell shaped and symmetric about a vertical line through its center. • Mean, median and mode are all equal and located at the center of the distribution.

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Normal Distribution continued

The shape of the normal distribution depends on the mean and the standard deviation. These three graphs have the same mean but different standard deviations. As the standard deviation increases, the distribution becomes more spread out. 12/03/09

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Standard Deviation and the 68-95-99.7 Rule 1. Approximately 68% of the data items fall within 1 standard deviation of the mean (in both directions). 2. Approximately 95% of the data items fall within 2 standard deviations of the mean. 3. Approximately 99.7% of the data items fall within 3 standard deviations of the mean. 12/03/09

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Example 1 Finding Scores at a Specified Standard Deviation From the Mean • Male adult heights in North America are approximately normally distributed with a mean of 70 inches and a standard deviation of 4 inches. Find the height that is 2 standard deviations above the mean. • Solution. • Height = mean + 2∙standard deviation = 70 + 2∙4 = 70 + 8 = 78

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Example 2 Using the 68-95-99.7 Rule • Use the distribution of male adult heights in the figure to find the percentage of men in North America with heights between 66 inches and 74 inches. • Solution. The 68-95-99.7 Rule states That approximately 68% of the data items fall within 1 standard deviation, 4, of the mean, 70.

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Mean – 1∙standard deviation = 70 - 1∙ 4= 66. Mean + 1∙standard deviation = 70 + 1∙ 4 = 74. 68% of males have heights between 66 and 74 inches.

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Computing z-Scores in Normal Distributions. • Describes how many standard deviations a data item in a normal distribution lies above or below the mean. Computing z-scores:

data item - mean z-score = standard deviation • Data items above the mean have positive z-scores. • Data items below the mean have negative z-scores. • The z-score for the mean is 0. 12/03/09

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Example 3 Computing z-Scores • The mean weight of newborn infants is 7 pounds and the standard deviation is 0.8 pound. The weights of newborn infants are normally distributed. Find the z-score for a weight of 9 pounds.

• Solution: The mean is 7 and the standard deviation is 0.8. The z-score written z9, is: data item - mean z9 = standard deviation 9−7 2 = = 0.8 0.8 = 2.5

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Example 4 Understanding Scores • Intelligence quotients (IQs) on the Stanford – Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. What is the IQ corresponding to a z-score of -1.5? • Solution: The negative sign in -1.5 tells us that the IQ is 1½ standard deviations below the mean. Score = mean – 1.5 ∙ standard deviation = 100 – 1.5 (16) = 100 – 24 = 76. The IQ corresponding to a z-score of -1.5 is 76. 12/03/09

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Percentiles and Quartiles • Percentiles: If n% of the items in a distribution are less than a particular data item, we say that the data item is in the nth percentile of the distribution. • Quartiles: Divide data sets into four equal parts. The 25th percentile is the first quartile. 25% of the data fall below the first quartile. The 50th percentile is the second quartile. The 75th percentile is the third quartile. 12/03/09

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A Percentile Interpretation for z-Scores • Using the z-score and a table, you can find the percentage of data items that are less than any data item in a normal distribution. • In a normal distribution, the mean, median, and mode all have a corresponding z-score of 0 and are the 50th percentile. Thus, 50% of the data items are greater than or equal to the mean, median and mode.

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Example 5 Finding the Percentage of Data Items Less Than a Given Data Item • According to the Department of Health and Education, cholesterol levels are normally distributed. For men between 18 and 24 years, the mean is 178.2 and the standard deviation is 40.7. What percentage of men in this age range have a cholesterol level less than 239.15? • Solution Compute the z-score for a 239.15 cholesterol level. z239.15 12/03/09

data item - mean 239.15 − 178.1 61.05 = = = = 1.5 standard deviation 40.7 40.7 Section 12.4

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Example 5 continued • We must find the percentage of men with a cholesterol level less than z = 1.5. The table gives this percentage as a percentile. Finding 1.5 in the z-score column gives a percentile of 93.32. thus , 93.32% of men between 18 and 24 have a cholesterol level less than 239.15

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z-score

Percentile

1.4

91.92

1.5

93.32

1.6

94.52

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Finding the Percentage of Data Items Between Two Given Data Items. 1. Convert each given data item to a z-score. 2. Use the table to find the percentile corresponding to each z-score in step 1. 3. Subtract the lesser percentile from the greater percentile and attach a % sign.

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Example 6 • The amount of time that self-employed Americans work each week is normally distributed with a mean of 44.6 hours and a standard deviation of 14.4 hours. What percentage of self-employed individuals in the United States work between 37.4 and 80.6 hours per week? Solution: Step 1. Convert each given data item to a z-score.

z37.4

data item - mean 37.4 − 44.6 - 7.2 = = = = −0.5 standard deviation 14.4 14.4

z80.6

data item - mean 80.6 − 44.6 36 = = = = 2.5 standard deviation 14.4 14.4

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Example 6 continued Step 2. Use the Table to find the percentile corresponding to these z-scores. The percentile corresponding to -0.50 is 30.85. This means that 30.85 percent of self-employed Americans work fewer than 37.4 hours per week. The Table also shows that the percentile that corresponds to a z-score of 2.5 is 99.38. That means that 99.38% of self- employed Americans work fewer than 80.6 hours per week. 12/03/09

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Example 6 continued Step 3. Subtract the lesser percentile from the greater percentile and attach a % sign. 99.38 – 30.85 = 68.53.

= 68.53 %

Thus, 68.53% of self-employed Americans work between 37.4 and 80.6 hours per week.

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Summary of Computing Percentage of Data Items for Normal Distributions Description of Percentage

Graph

Computation of Percentage

Percentage of data items less than a given data item with z = b

Use the table percentile for z = b and add a % sign

Percentage of data items greater than a given data item with z=a

Subtract the table percentile for z = a from 100 and add a % sign.

Percentage of data items between two given data items with z = a and z = b

Subtract the table percentile for z = a from the table percentile for z = b and add a % sign.

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Polls and Margins of Error • Statisticians use properties of the normal distribution to estimate the probability that a result obtained from a single sample reflects what is truly happening. • If n is the sample size, there is a 95% probability that it lies within 1 of the true population statistic. n • ±

1 is called the margin of error. n

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Example 7 Using and Interpreting Margin of Error In a random sample of 1172 children ages 6 through 14, 17% of the children said getting bossed around is a bad thing about being a kid. a. Verify the margin of error. Solution: The sample size is n = 1172. The margin of error is 1 1 ± =± ≈ ±2.9% n 1172 b. Write a statement about the percentage of children who feel that getting bossed around is a bad thing about being a kid. Solution: There is a 95% probability that the true population percentage lies between 17% -2.9% = 14.1% and 17% + 2.9% = 19.9% 12/03/09

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Other Kinds of Distributions • This graph represents the population distribution of weekly earnings in the United States. There is no upper limit on weekly earnings. The relatively few people with very high weekly incomes pull the mean income to a value greater than the median. • The most frequent income, the mode, occurs towards the low end of the data items. • This is called a skewed distribution because a large number of data items are piled up at one end or the other with a “tail” at the other end. • This graph is skewed to the right. 12/03/09

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