common core state standards cc-20 normal distributions ...teachers.dadeschools.net/rbaglos/cc-20...

86 Chapter 11 Probability and Statistics

24. Open Ended The government uses a variety of methods to estimate how the general public is feeling about the economy. A researcher wants to conduct a study to determine whether people who live in his state are representative of the latest government results. What type of study should the researcher use? Explain.

25. In a recent telephone survey, respondents were asked questions to determine whether they supported the new law that required every passenger to wear a seat belt while in a moving vehicle. The first question was, “According to the National Highway Traffic Safety Administration, wearing seat belts could prevent 45% of the fatalities suffered in car accidents. Do you think that everyone should wear safety belts?” Does this question introduce a bias into the survey? Explain.

ChallengeC

HSM15_A2Hon_SE_CC_19_TrKit.indd 86 8/5/13 6:51 PM

Lesson H-15 Normal Distributions 87

Objective To use a normal distribution

A discrete probability distribution has a finite number of possible events, or values.

The events for a continuous probability distribution can be any value in an interval of real numbers. If a data set is large, the distribution of its discrete values approximates a continuous distribution.

Essential Understanding Many common statistics (such as human height, weight, or blood pressure) gathered from samples in the natural world tend to have a normal distribution about their mean.

A normal distribution has data that vary randomly from the mean. The graph of a normal distribution is a normal curve.

Even and odd functions are defined as follows.Even function: f (x) = f (−x)Odd function: −f (x) = f (−x)Which is the graph of an even function? Of an odd function? Justify your answers.

x

y

O x

y

O

y

O

Normal Distributions

Try it out. Suppose (2, 2) is a point on f(x). If f(x) is even, what other point is on the graph of f(x)?

Lesson Vocabulary

•discrete probability distribution

•continuous probability distribution

•normal distribution

LessonVocabulary

Key Concept Normal Distribution

�3deviations

�1deviation

mean

�2deviations

�1deviation

�3deviations

�2deviations

2.35% 2.35%13.5% 13.5%

34% 34%

A normal distribution has a symmetric bell shape centered on the mean.

In a normal distribution,• 68% of data fall within one standard

deviation of the mean• 95% of data fall within two standard

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

deviations of the mean

CC-20

MATHEMATICAL PRACTICES

MACC.912.S-ID.1.4 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate . . .

MP 1, MP 3, MP 4

Common Core State Standards

HSM15_A2Hon_SE_CC_20_TrKit.indd 87 8/5/13 6:51 PMCC-20 Normal Distributions 87

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CC-20 87

Big idea ProbabilityEssEntial UndErstanding

Normal distributions model many common natural phenomena, such as human height, weight, and blood pressure.

Math BackgroundNormal distributions occur often in real life such as standardized test scores and heights of adults. The normal distribution curve has the following characteristics:•Thegraphhasitsmaximumatthecenter.•Thegraphissymmetricaboutthemean.•Themean,mode,andmedianareequal.•About68%ofthevaluesfallwithinonestandarddeviationofthemean,34%fall within one standard deviation to the rightofthemeanand34%fallwithinone standard deviation to the left of the mean.

•About95%ofthevaluesfallwithintwostandarddeviationsofthemean,47.5%

fall within two standard deviations to the rightofthemeanand47.5%fallwithintwo standard deviations to the left of the mean.

•About99.7%ofthevaluesfallwithinthree standard deviations of the mean, 49.85%fallwithinthreestandarddeviations to the right of the mean and49.85%fallwithinthreestandarddeviations to the left of the mean.

Mathematical PracticesConstruct viable arguments and critique the reasoning of others. With knowledge of variance and standard deviation, students will construct arguments about the distribution of a set of data.

1 Interactive LearningSolve It!PUrPOsE To distinguish between the graphs of even and odd functions using the definitionsPrOCEss Students may•testorderedpairsfromthegraphstoseewhether

the graph satisfies the definitions.•considerhoweachdefinitionaffectstheordered

pair given by (x, f (x)).

FaCilitatE

Q When the x-value of an ordered pair of an even function changes sign, what happens to the y-value? [It stays the same.]

Q When the x-value of an ordered pair of an odd function changes sign, what happens to the y-value? [It changes sign.]

Q How are even functions symmetric? Odd functions? [Even functions are symmetric about the y-axis. Odd functions are symmetric about the origin.]

ansWEr SeeSolveItinAnswersonnextpage.COnnECt tHE MatH In the Solve It students explorethedefinitionsandsymmetryofgraphsofeven and odd functions. In the lesson, students will explorethegraphsofnormaldistributionsthataresymmetric about the mean.

2 Guided InstructionTake Note

Q Ifanormaldistributionhadameanof50andastandarddeviationof11,68percentofthedatapoints would have to fall between which two values? [39 and 61]

Preparing to TeachCC-20

Problem 1

Got It?

88 Honors Supplement

Sometimes data are not normally distributed. A data set could have a distribution that is skewed, an asymmetric curve where one end stretches out further than the other end. When a data set is skewed, the data do not vary predictably from the mean. This means that the data do not fall within the standard deviations of the mean like normally distributed data, and so it is inappropriate to use mean and standard deviation to estimate percentages for skewed data.

Analyzing Normally Distributed Data

Zoology The bar graph gives the weights of a population of female brown bears. The red curve shows how the weights are normally distributed about the mean, 115 kg. Approximately what percent of female brown bears weigh between 100 and 129 kg?

Estimate and add the percents for the intervals 1009109, 1109119, and 1209129.

23 + 42 + 23 = 88

About 88% of female brown bears weigh between 100 and 129 kg.

1. a. Approximately what percent of female brown bears in Problem 1 weigh less than 120 kg?

b. The standard deviation in the weights of female brown bears is about 10 kg. Approximately what percent of female brown bears have weights that are within 1.5 standard deviations of the mean?

When data are normally distributed, you can sketch the graph of the distribution using the fact that a normal curve has a symmetric bell shape.

Positively Skewed Normally Distributed Negatively Skewed

50

40

30

20

10

0� 79 � 15080�

8990�

99100�

109120�

129110�

119130�

139140�

149Weight (kilograms)

Female Brown Bear Weights

Perc

ent

of B

ears

How do you find this percent?The percents for each bar are based on the same sample population of bears. You can add the percents.

STEM


–1 standard deviation

18.5 in.

12.9 in.

15.7 in.

+1 standard deviation

mean body length

Problem 2

Got It?


Sketching a Normal Curve

Zoology For a population of male European eels, the mean body length and one positive and negative standard deviation is shown below. Sketch a normal curve showing the eel lengths at one, two, and three standard deviations from the mean.

2. For a population of female European eels, the mean body length is 21.1 in. The standard deviation is 4.7 in. Sketch a normal curve showing eel lengths at one, two, and three standard deviations from the mean.

When you show a probability distribution as a bar graph, the height of the bar indicates probability. For a normal distribution, however, the area between the curve and an interval on the x-axis represents probability.

• Multiplythestandarddeviationby1,2,and3.

• Drawverticallinesatthemean { these values.

•Sketchthenormalcurve.

Lengthsthatareone,two,andthree standard deviations from the mean

The mean and the standard deviation of the population

�3deviations

�1deviation

mean

�2deviations

�1deviation

�3deviations

�2deviations

7.3 10.1 12.9 15.7

Length (inches)

Distribution of Body Lengths for Male European Eels

Freq

uenc

y of

Mea

sure

18.5 21.3 24.1

2.8 2.82.8 2.8 2.8 2.8

How high do you draw the curve?Unlessyouactuallylabeltheverticalscale, it doesn’t matter.

STEM

HSM15_A2Hon_SE_CC_20_TrKit.indd 89 8/5/13 6:51 PM88 Common Core


88 Common Core

Problem 1

Got It?

Q Whydoyouexpecttheweightsofthefemalebrown bear to be normally distributed? [Although there may be a few heavy or light bears, most bears are similar in size, so the weights of most of the bears should be close to the average weight of the female brown bears.]

Q Aboutwhatpercentofthefemalebrownbearsweighlessthan100kgormorethan129kg?Describe a way to find this value. [Samples: Since 88% of the bears weigh between 100 kg and 129 kg, 100 ∙ 88 ∙ 12% of the bears weigh less than 100 kg or more than 129 kg. You could also add the percentages for the bars for 80–89, 90–99, 130–139, and 140–149 to get 1 ∙ 5 ∙ 5 ∙ 1 ∙ 12%]

Q What percent of female brown bears should weigh lessthan115kg?Explain. [Since the weight is normally distributed, half of the bears, or 50%, should weigh less than the mean and half should weigh more than the mean.]

Q For which intervals on the graph will you add the respective percentages to find the percent for 1a? [80–89, 90–99, 100–109, and 110–119]

Q Whatweightvaluesare1.5standarddeviationsfrom the mean? How do you find these values? [130 kg and 100 kg; 115 ∙ 1.5(10) ∙ 130 and 115 ∙ 1.5(10) ∙ 100]

AnswersSolve It!The first graph is of an even function because the y-values are the same for x and -x. The second graph is of an odd function because the y-value for -x is equaltotheoppositeofthey-value for x. The third graph is not of a function.

Got It? 1. a. 71% b. 88% 2.

3deviations

1deviation

mean

2deviations

1deviation

3deviations

2deviations

7.0 11.7 16.4 21.1

Length (inches)

Distribution of FemaleEuropean Eels

Freq

uenc

y of

Mea

sure

25.8 30.5 35.2

4.7 4.7 4.7 4.7 4.7 4.7

Problem 1

Got It?


Sometimes data are not normally distributed. A data set could have a distribution that is skewed, an asymmetric curve where one end stretches out further than the other end. When a data set is skewed, the data do not vary predictably from the mean. This means that the data do not fall within the standard deviations of the mean like normally distributed data, and so it is inappropriate to use mean and standard deviation to estimate percentages for skewed data.

Analyzing Normally Distributed Data

Zoology The bar graph gives the weights of a population of female brown bears. The red curve shows how the weights are normally distributed about the mean, 115 kg. Approximately what percent of female brown bears weigh between 100 and 129 kg?

Estimate and add the percents for the intervals 1009109, 1109119, and 1209129.

23 + 42 + 23 = 88

About 88% of female brown bears weigh between 100 and 129 kg.

1. a. Approximately what percent of female brown bears in Problem 1 weigh less than 120 kg?

b. The standard deviation in the weights of female brown bears is about 10 kg. Approximately what percent of female brown bears have weights that are within 1.5 standard deviations of the mean?

When data are normally distributed, you can sketch the graph of the distribution using the fact that a normal curve has a symmetric bell shape.

Positively Skewed Normally Distributed Negatively Skewed

50

40

30

20

10

0� 79 � 15080�

8990�

99100�

109120�

129110�

119130�

139140�

149Weight (kilograms)

Female Brown Bear Weights

Perc

ent

of B

ears

How do you find this percent?The percents for each bar are based on the same sample population of bears. You can add the percents.

STEM


–1 standard deviation

18.5 in.

12.9 in.

15.7 in.

+1 standard deviation

mean body length

Problem 2

Got It?


Sketching a Normal Curve

Zoology For a population of male European eels, the mean body length and one positive and negative standard deviation is shown below. Sketch a normal curve showing the eel lengths at one, two, and three standard deviations from the mean.

2. For a population of female European eels, the mean body length is 21.1 in. The standard deviation is 4.7 in. Sketch a normal curve showing eel lengths at one, two, and three standard deviations from the mean.

When you show a probability distribution as a bar graph, the height of the bar indicates probability. For a normal distribution, however, the area between the curve and an interval on the x-axis represents probability.

• Multiplythestandarddeviationby1,2,and3.

• Drawverticallinesatthemean { these values.

•Sketchthenormalcurve.

Lengthsthatareone,two,andthree standard deviations from the mean

The mean and the standard deviation of the population

�3deviations

�1deviation

mean

�2deviations

�1deviation

�3deviations

�2deviations

7.3 10.1 12.9 15.7

Length (inches)

Distribution of Body Lengths for Male European Eels

Freq

uenc

y of

Mea

sure

18.5 21.3 24.1

2.8 2.82.8 2.8 2.8 2.8

How high do you draw the curve?Unlessyouactuallylabeltheverticalscale, it doesn’t matter.

STEM



CC-20 89

Problem 2

EXtEnsiOn

Got It?

Q Do you have to sketch the graph very far beyond threestandarddeviations?Explain. [No; less than 0.3 percent of the data points fall outside three standard deviations.]

Q What effect does the standard deviation have on theshapeofthebellcurve?Abouthowtallshouldthe bell curve be at {1 standard deviation from the mean compared to the height of the bell curve at the mean? [no effect; a little more than half as tall]

Q What should the height of the bell curve be at {3 standard deviations from the mean? [close to zero]

Q Why are the intervals between the vertical lines spaced the same distance apart? [The distances between the lines represent the same value of one standard deviation.]

Q What will be the values along the bottom of your graph? [7, 11.7, 16.4, 21.1, 25.8, 30.5, and 35.2]

Q How high should your graph be? [The vertical axis is not scaled, so the height does not matter.]

Q Aroundwhatvalueshouldthegraphbesymmetric? [the mean of 21.1 in]

Additional Problems 1. You track the number of words in your

textmessagesforamonthandsketchthebar graph shown. The number of words is normallydistributedaboutameanof10.Aboutwhatpercentofyourtextmessagesarebetween9and11wordslong?

ansWEr 56%

2. ForanEnglishclass,theaveragescoreonaresearchprojectwas82andthestandard deviation of the normally distributedscoreswas5.Sketchanormalcurve showing the project scores and three standard deviations from the mean.ansWEr

67 72 77 82 87 92 97

3. UsingthenormalcurvefromAdditionalProblem2,whatpercentofstudentsscoredbetween72and82points?ansWEr 47.5%

Perc

ent

ofM

essa

ges

Number of Words in Text Messages

10

0

20

30

5 6 7 8 9 10 12 13 14 1511

Problem 3

Got It?


Analyzing a Normal Distribution

The heights of adult American males are approximately normally distributed with mean 69.5 in. and standard deviation 2.5 in.

A What percent of adult American males are between 67 in. and 74.5 in. tall?

Draw a normal curve. Label the mean. Divide the graph into sections of standard-deviation widths. Label the percentages for each section.

Distribution of Heights—Adult American Males

P ( 67 6 height 6 74.5) = 0.34 + 0.34 + 0.135 = 0.815

About 82% of adult American males are between 67 in. and 74.5 in. tall.

B In a group of 2000 adult American males, about how many would you expect to be taller than 6 ft (or 72 in.)?

Because the graph is symmetric about the mean, the right half of the distribution contains 50% of the data. If you subtract everything between 69.5 in. and 72 in. from the right half, only the part of the distribution that is greater than 72 in. remains.


P (height 7 72) = 0.50 - 0.34 = 0.16

You would expect about 16% of the 2000 adult American males to be taller than 72 in. You would expect about 0.16 # 2000 = 320 to be over 6 ft tall.

3. The scores on the Algebra 2 final are approximately normally distributed with a mean of 150 and a standard deviation of 15.

a. What percentage of the students who took the test scored above 180? b. If 250 students took the final exam, approximately how many scored above 135? c. Reasoning If 13.6% of the students received a B on the final, how can you

describe their scores? Explain.

62 64.5 69.567 72 74.5 77

13.5% 13.5%

34% 34%2.35% 2.35%

50%

62 64.5 69.567 72 74.5 77

13.5% 13.5%

34% 34%2.35% 2.35%


How do you divide the graph of the distribution?Drawverticallinesatintervals that are one standard deviation wide,onbothsides of the mean.



Lesson CheckDo you know HOW? 1. Use the graph from Problem 1. What is the

approximate percent of female brown bears weighing at least 100 kg?

2. Draw a curve to represent a normally distributed experiment that has a mean of 180 and a standard deviation of 15. Label the x-axis and indicate the probabilities.

3. The scores on an exam are normally distributed, with a mean of 85 and a standard deviation of 5. What percent of the scores are between 85 and 95?

Do you UNDERSTAND? 4. Vocabulary Why is a normal distribution “normal”?

5. Compare and Contrast How do the mean and median compare in a normal distribution?

6. Reasoning What is the effect on a normal distribution if each data value increases by 10? Justify your answer.

Practice and Problem-Solving Exercises

Biology The heights of men in a survey are normally distributed about the mean. Use the graph for Exercises 7–10.

7. About what percent of men aged 25 to 34 are 69–71 in. tall?

8. About what percent of men aged 25 to 34 are less than 70 in. tall?

9. Suppose the survey included data on 100 men. About how many would you expect to be 69–71 in. tall?

10. The mean of the data is 70, and the standard deviation is 2.5. Approximately what percent of men are within one standard deviation of the mean height?

Sketch a normal curve for each distribution. Label the x-axis values at one, two, and three standard deviations from the mean.

11. mean = 45, standard deviation = 5 12. mean = 45, standard deviation = 10

13. mean = 45, standard deviation = 2 14. mean = 45, standard deviation = 3.5

A set of data has a normal distribution with a mean of 50 and a standard deviation of 8. Find the percent of data within each interval.

15. from 42 to 58 16. greater than 34 17. less than 50

PracticeA See Problem 1.

64

20

10

066 68 70 72 7674

Height (in.)

Perc

ent

of M

en

Heights of Men Ages 25–34

See Problem 2.

See Problem 3.





90 Common Core

AnswersGot It? (continued) 3. a. 2.5% b. 210students c. The students that received a B had scores

between165and180.

Problem 3

EXtEnsiOn

Got It?

Q Is there another way to find out what percent of malesyouwouldexpecttobetallerthan72in.? [You could add 13.5 ∙ 2.35 ∙ 0.15 to get 16%.]

Q Since the area between the curve and the x-axisrepresents probability for a normal distribution, whatistheareaundertheentirecurve?Explain. [The area under the entire normal curve is one. One hundred percent of the data is represented by the entire curve.]

Q Whatpercentofthescoresarebelow150?Howdo you know? [Since the graph is symmetric, 50% of the data lie above the mean and 50% lie below the mean of 150.]

Q For 3b, what is the most convenient way to divide the graph of the normal curve to find the percent ofstudentsscoringabove135? [Divide the graph into the 34% between 135 and 150 and the 50% scoring above the mean of 150.]

Q For 3c, is there more than one area under the normalcurvethatcontains13.5%ofthedata?WhichismostlikelyaB?Explain. [A grade of B is most likely the area between the first standard deviation and the second deviation. Although an area of 13.5% could be found in other places along the normal curve, a B is a higher grade, and is above the mean but is not the highest grade.]

Problem 3

Got It?


Analyzing a Normal Distribution

The heights of adult American males are approximately normally distributed with mean 69.5 in. and standard deviation 2.5 in.

A What percent of adult American males are between 67 in. and 74.5 in. tall?

Draw a normal curve. Label the mean. Divide the graph into sections of standard-deviation widths. Label the percentages for each section.


P ( 67 6 height 6 74.5) = 0.34 + 0.34 + 0.135 = 0.815

About 82% of adult American males are between 67 in. and 74.5 in. tall.

B In a group of 2000 adult American males, about how many would you expect to be taller than 6 ft (or 72 in.)?

Because the graph is symmetric about the mean, the right half of the distribution contains 50% of the data. If you subtract everything between 69.5 in. and 72 in. from the right half, only the part of the distribution that is greater than 72 in. remains.


P (height 7 72) = 0.50 - 0.34 = 0.16

You would expect about 16% of the 2000 adult American males to be taller than 72 in. You would expect about 0.16 # 2000 = 320 to be over 6 ft tall.

3. The scores on the Algebra 2 final are approximately normally distributed with a mean of 150 and a standard deviation of 15.

a. What percentage of the students who took the test scored above 180? b. If 250 students took the final exam, approximately how many scored above 135? c. Reasoning If 13.6% of the students received a B on the final, how can you

describe their scores? Explain.

How do you divide the graph of the distribution?Drawverticallinesatintervals that are one standard deviation wide,onbothsides of the mean.



Lesson CheckDo you know HOW? 1. Use the graph from Problem 1. What is the

approximate percent of female brown bears weighing at least 100 kg?

2. Draw a curve to represent a normally distributed experiment that has a mean of 180 and a standard deviation of 15. Label the x-axis and indicate the probabilities.

3. The scores on an exam are normally distributed, with a mean of 85 and a standard deviation of 5. What percent of the scores are between 85 and 95?

Do you UNDERSTAND? 4. Vocabulary Why is a normal distribution “normal”?

5. Compare and Contrast How do the mean and median compare in a normal distribution?

6. Reasoning What is the effect on a normal distribution if each data value increases by 10? Justify your answer.

Practice and Problem-Solving Exercises

Biology The heights of men in a survey are normally distributed about the mean. Use the graph for Exercises 7–10.

7. About what percent of men aged 25 to 34 are 69–71 in. tall?

8. About what percent of men aged 25 to 34 are less than 70 in. tall?

9. Suppose the survey included data on 100 men. About how many would you expect to be 69–71 in. tall?

10. The mean of the data is 70, and the standard deviation is 2.5. Approximately what percent of men are within one standard deviation of the mean height?

Sketch a normal curve for each distribution. Label the x-axis values at one, two, and three standard deviations from the mean.

11. mean = 45, standard deviation = 5 12. mean = 45, standard deviation = 10

13. mean = 45, standard deviation = 2 14. mean = 45, standard deviation = 3.5

A set of data has a normal distribution with a mean of 50 and a standard deviation of 8. Find the percent of data within each interval.

15. from 42 to 58 16. greater than 34 17. less than 50

PracticeA See Problem 1.

64

20

10

066 68 70 72 7674

Height (in.)

Perc

ent

of M

en

Heights of Men Ages 25–34

See Problem 2.

See Problem 3.





CC-20 91

Lesson Check 1. 94% 2.

3. 47.5% 4. Normal distribution means that most of

theexamplesinadatasetareclosetothe mean; the distribution of the data iswithin1,2,or3standarddeviationsof the mean.

5. Themeanandmedianareequivalentin a normal distribution.

3deviations

1deviation

mean

2deviations

1deviation

3deviations

2deviations

135 150 165 180

Freq

uenc

y of

Mea

sure

195 210 225

15 152.35% 13.5%

15

34%

15

34%

15 1513.5% 2.35%

3 Lesson CheckDo you know HOW?•IfstudentshavedifficultyfindingthepercentinExercise1,askwhichbarsofthebargraphareincludedbythecondition“atleast100kg.”

•IfstudentscannotdrawthecurveinExercise2,remind them that the curve is symmetric about the mean and covers {3 standard deviations.

•IfstudentshavedifficultysolvingExercise3,askthem to draw and label the normal curve that fits the conditions.

Do you UNDERSTAND?•Ifstudentshavedifficultycomparingthemean

and median of a normal distribution for Exercise5,askthemwhatpercentofthedatafalls above and below each of the median and mean. Point out that the normal distribution is symmetric about the mean.

•Ifstudentshavedifficultyreasoningthechangeinthe graph of the distribution with an increase in themeanforExercise6,thenaskwhatkindsofvalues increase the mean and where those values occur in a graph.

Close

Q How can the graph of a normal distribution of data help you understand the data? [Knowing that data fits a normal distribution allows you to calculate what percentage of data falls within various standard deviations of the mean.]

6. meanincreasesby10:thebellcurveistranslated10unitstothert.

7–17. Seenextpage.


18. Think About a Plan The numbers of paper clips per box in a truckload of boxes are normally distributed, with a mean of 100 and a standard deviation of 5. Find the probability that a box will not contain between 95 and 105 clips.

• How should you label the vertical lines on the graph of the normal distribution?

• Which parts of the graph are not between 95 and 105 clips?

19. a. From the table at the right, select the set of values that appears to be distributed normally.

b. Using the set you chose in part (a), make a histogram of the values. c. Sketch a normal curve over your graph.

20. Writing In a class of 25, one student receives a score of 100 on a test. The grades are distributed normally, with a mean of 78 and a standard deviation of 5. Do you think the student’s score is an outlier? Explain.

21. Sports To qualify as a contestant in a race, a runner has to be in the fastest 16% of all applicants. The running times are normally distributed, with a mean of 63 min and a standard deviation of 4 min. To the nearest minute, what is the qualifying time for the race?

22. Agriculture To win a prize at the county fair, the diameter of a tomato must be greater than 4 in. The diameters of a crop of tomatoes grown in a special soil are normally distributed, with a mean of 3.2 in. and a standard deviation of 0.4 in. What is the probability that a tomato grown in the special soil will be a winner?

A normal distribution has a mean of 100 and a standard deviation of 10. Find the probability that a value selected at random is in the given interval.

23. from 80 to 100 24. from 70 to 130 25. from 90 to 120

26. at least 100 27. at most 110 28. at least 80

29. Weather The table at the right shows the number of tornadoes that were recorded in the U.S. in 2008.

a. Draw a histogram to represent the data. b. Does the histogram approximate a normal curve? Explain. c. Is it appropriate to use a normal curve to estimate the percent of

tornados that occur during certain months of the year? Explain.

30. Error Analysis In a set of data, the value 332 is 3 standard deviations from the mean and the value 248 is 1 standard deviation from the mean. A classmate claims that there is only one possible mean and standard deviation for this data set. Do you agree? Explain.

31. Reasoning Jake and Elena took the same standardized test, but are in different classes. They both received a score of 87. In Jake’s group, the mean was 80 and the standard deviation was 6. In Elena’s group, the mean was 76 and the standard deviation was 4. Did either student score in the top 10% of his or her group? Explain.

ApplyB

Set 1 Set 2

1

10

5

19

2

7

1

7

2

10

6

9

5

7

7

7

4

11

7

7

7

9

7

7

Set 3

5

6

9

1

1

5

11

1

10

4

2

8

Month Tornadoes

1

2

3

4

5

6

7

8

9

10

11

12

84

147

129

189

461

294

93

101

111

21

20

40

STEM



32. Manufacturing Tubs of yogurt weigh 1.0 lb each, with a standard deviation of 0.06 lb. At a quality control checkpoint, 12 of the tubs taken as samples weighed less than 0.88 lb. Assume that the weights of the samples were normally distributed. How many tubs of yogurt were taken as samples?

33. Reasoning Describe how you can use a normal distribution to approximate a binomial distribution. Draw a binomial histogram and a normal curve to help with your explanation.

ChallengeC



92 Common Core

AnswersPractice and Problem-Solving Exercises (continued) 7. ≈43% 8. ≈39% 9. ≈43 men 10. ≈66% 11.

12.

13.

14.

30 35 40 45 50 55 60

15 25 35 45 55 65 75

39 41 43 45 47 49 51

34.5 38 41.5 45 48.5 52 55.5

15. 68% 16. 97.5% 17. 50% 18. 32% 19. a. Set2 b–c.

20. Yes; 99%ofallgradesareexpectedto be within 3 standard deviations of the mean, and this score is 4.4 standard deviations above the mean.

21.59min 22. 2.5% 23. 47.5% 24. 99.7% 25. 81.5% 26. 50% 27. 84% 28. 97.5% 29. a.

b. No; the curve is skewed to the left. c. No, because the data are skewed it

is not appropriate to use the mean and standard deviations shown on a normal curve to estimate percents of populations.

30. No;themeancouldbe206withstandarddeviationof42,orthemeancouldbe269withastandarddeviationof21.

31. Yes;Elenascoredwithinthetop10%ofhergroup.Herscoreis2.75standard deviations above the mean, which places her in the top 1%. Jake did not score in the top 10%. Hisscoreis1.16standarddeviationsabovethemean,oratthe88thpercentile.

4 5 6 7 8 9 1011

1 2 3 4 5 6 7 8 9 10 11 12

50100150200250300350400450500

00

Tornad

oes

Month

4 PracticeassignMEnt gUidEBasic: 7–17all,18–22even,29,31

Average: 7–17odd,18–31

Advanced: 7–17odd,18–33

Mathematical Practices are supported by exerciseswithredheadings.HerearethePracticessupported in this lesson:

MP1:MakeSenseofProblemsEx.18MP3:ConstructArgumentsEx.6,33MP3:CommunicateEx.6,20MP3:CompareArgumentsEx.5,31MP3:CritiquetheReasoningofOthersEx.30

Applicationsexerciseshaveblueheadings. Exercises7–10,21,22supportMP4:Model.

STEMexercisesfocusonscienceorengineeringapplications.

HOMEWOrK QUiCK CHECKTo check students’ understanding of key skills and concepts,gooverExercises9,13,18,22,and31.


18. Think About a Plan The numbers of paper clips per box in a truckload of boxes are normally distributed, with a mean of 100 and a standard deviation of 5. Find the probability that a box will not contain between 95 and 105 clips.

• How should you label the vertical lines on the graph of the normal distribution?

• Which parts of the graph are not between 95 and 105 clips?

19. a. From the table at the right, select the set of values that appears to be distributed normally.

b. Using the set you chose in part (a), make a histogram of the values. c. Sketch a normal curve over your graph.

20. Writing In a class of 25, one student receives a score of 100 on a test. The grades are distributed normally, with a mean of 78 and a standard deviation of 5. Do you think the student’s score is an outlier? Explain.

21. Sports To qualify as a contestant in a race, a runner has to be in the fastest 16% of all applicants. The running times are normally distributed, with a mean of 63 min and a standard deviation of 4 min. To the nearest minute, what is the qualifying time for the race?

22. Agriculture To win a prize at the county fair, the diameter of a tomato must be greater than 4 in. The diameters of a crop of tomatoes grown in a special soil are normally distributed, with a mean of 3.2 in. and a standard deviation of 0.4 in. What is the probability that a tomato grown in the special soil will be a winner?

A normal distribution has a mean of 100 and a standard deviation of 10. Find the probability that a value selected at random is in the given interval.

23. from 80 to 100 24. from 70 to 130 25. from 90 to 120

26. at least 100 27. at most 110 28. at least 80

29. Weather The table at the right shows the number of tornadoes that were recorded in the U.S. in 2008.

a. Draw a histogram to represent the data. b. Does the histogram approximate a normal curve? Explain. c. Is it appropriate to use a normal curve to estimate the percent of

tornados that occur during certain months of the year? Explain.

30. Error Analysis In a set of data, the value 332 is 3 standard deviations from the mean and the value 248 is 1 standard deviation from the mean. A classmate claims that there is only one possible mean and standard deviation for this data set. Do you agree? Explain.

31. Reasoning Jake and Elena took the same standardized test, but are in different classes. They both received a score of 87. In Jake’s group, the mean was 80 and the standard deviation was 6. In Elena’s group, the mean was 76 and the standard deviation was 4. Did either student score in the top 10% of his or her group? Explain.

ApplyB

Set 1 Set 2

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10

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19

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7

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7

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10

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7

7

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Set 3

5

6

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5

11

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10

4

2

8

Month Tornadoes

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2

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6

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84

147

129

189

461

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93

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STEM



32. Manufacturing Tubs of yogurt weigh 1.0 lb each, with a standard deviation of 0.06 lb. At a quality control checkpoint, 12 of the tubs taken as samples weighed less than 0.88 lb. Assume that the weights of the samples were normally distributed. How many tubs of yogurt were taken as samples?

33. Reasoning Describe how you can use a normal distribution to approximate a binomial distribution. Draw a binomial histogram and a normal curve to help with your explanation.

ChallengeC



CC-20 93

32. 480tubs 33. Abinomialdistributionhasafinite

no. of possible probability events which sum to 1 and are a subset of a larger normal distribution. For exampleusing,n = 6, p = 0.5, the binomial distribution probabilities are P (0) ≈ 0.0156, P (1) ≈ 0.0938, P (2) ≈ 0.2344, P (3) ≈ 0.3125, P (4) ≈ 0.2344, P (5) ≈ 0.0938, P (6) ≈ 0.0156.

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ability

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common core state standards cc-20 normal distributions ...teachers.dadeschools.net/rbaglos/cc-20...

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