Five-Minute Check (over Lesson 11–4)

CCSS

Then/Now

New Vocabulary

Key Concept: The Normal Distribution

Key Concept: The Empirical Rule

Example 1: Use the Empirical Rule to Analyze Data

Example 2: Real-World Example: Use the Empirical Rule to Analyze a Distribution

Key Concept: Formula for z-Values

Example 3:Use z-Values to Locate Position

Key Concept: Characteristics of the Standard Normal Distribution

Example 4:Real-World Example: Find Probabilities

Over Lesson 11–4

A. This experiment can be reduced to a binomial experiment because there are two possible outcomes. Success is yes, failure is no, a trial is asking a student, and the random variable is the number of yeses; n = 20, p = 0.5, q = 0.5.

B. This experiment cannot be reduced to a binomial experiment because there are more than two possible outcomes.

C. This experiment can be reduced to a binomial experiment. Success is yes, failure is no, a trial is asking a student, and the random variable is the number of yeses; n = 30, p = 0.75, q = 0.25.

D. This experiment can be reduced to a binomial experiment. Success is yes, failure is no, a trial is asking a student, and the random variable is the type of reply; n = the number of students in the class, p = 0.5, q = 0.5.

Determine whether the experiment is a binomial experiment or can be reduced to a binomial experiment. If so, describe a trial, determine the random variable, and state n, p, and q.You survey your class, asking what they plan to do for the holidays.

Over Lesson 11–4

A. This experiment cannot be reduced to a binomial experiment because there are more than two possible outcomes.

B. This experiment can be reduced to a binomial experiment because there are two possible outcomes. Success is one, failure is none, a trial is asking an adult, and the random variable is the number of ones; n = 30, p = 0.78, q = 0.22.

C. This experiment can be reduced to a binomial experiment. Success is yes, failure is no, a trial is asking an adult, and the random variable is the number of yeses; n = 30, p = 0.22, q = 0.78.

D. This experiment can be reduced to a binomial experiment. Success is yes, failure is no, a trial is asking an adult, and the random variable is the number of yeses; n = 30, p = 0.78, q = 0.22.

Determine whether the experiment is a binomial experiment or can be reduced to a binomial experiment. If so, describe a trial, determine the random variable, and state n, p, and q.A poll found that 78% of adults exercise at least one day a week. You ask 30 adults if they exercise at least one day a week.

Over Lesson 11–4

A. This experiment can be reduced to a binomial experiment. Success is red, failure is not red, a trial is asking a person, and the random variable is the number of reds; n = 100, p = 0.27, q = 0.73.

B. This experiment can be reduced to a binomial experiment. Success is red, failure is not red, a trial is asking a person, and the random variable is the number of reds; n = 100, p = 0.73, q = 0.27.

C. This experiment cannot be reduced to a binomial experiment because there are more than two possible outcomes.

D. This experiment can be reduced to a binomial experiment. Success is red, failure is blue, a trial is asking a person, and the random variable is the number of reds; n = 100, p = 0.27, q = 0.73.

Determine whether the experiment is a binomial experiment or can be reduced to a binomial experiment. If so, describe a trial, determine the random variable, and state n, p, and q.A study finds that 27% of people say that red is their favorite color. You ask 100 people if red is their favorite color.

Over Lesson 11–4

A. 25.2%

B. 48.7%

C. 51.3%

D. 74.8%

EQUIPMENT George’s garage door opener is malfunctioning and only works 65% of the time. What is the probability that the opener works at least 7 of the next 10 times George tries to use it?

Over Lesson 11–4

A. 3.8%

B. 12.4%

C. 23.2%

D. 39.4%

JUKEBOX Jason’s old jukebox contains 124 songs, 60 of which are from the 1980s. He programs the jukebox to play 8 random songs. What is the probability that 3 of those songs are from the 1980s?

Content Standards

S.ID.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. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.

Mathematical Practices

6 Attend to precision.

8 Look for and express regularity in repeated reasoning.

You constructed and analyzed discrete probability distributions.

• Use the Empirical Rule to analyze normally distributed variables.

• Apply the standard normal distribution and z-values.

• normal distribution

• Empirical Rule

• z-value

• standard normal distribution

Use the Empirical Rule to Analyze Data

A. A normal distribution has a mean of 45.1 and a standard deviation of 9.6. Find the values that represent the middle 99.7% of the distribution.

μ = 45.1 and σ = 9.6

The middle 99.7% of data in a normal distribution is the range from μ – 3σ to μ + 3σ.

45.1 – 3(9.6) = 16.3

45.1 + 3(9.6) = 73.9

Answer: Therefore, the range of values in the middle 99.7% is 16.3 < X < 73.9.

Use the Empirical Rule to Analyze Data

B. A normal distribution has a mean of 45.1 and a standard deviation of 9.6. What percent of the data will be greater than 54.7?

The value 54.7 is 1σ more than μ. Approximately 68% of the data fall between μ – σ and μ + σ, so the remaining data values represented by the two tails covers 32% of the distribution. We are only concerned with the upper tail, so 16% of the data will be greater than 54.7.

Answer: 16%

A. 0.3%

B. 2.5%

C. 5%

D. 97.5%

A normal distribution has a mean of 38.3 and a standard deviation of 5.9. What percent of the data will be less than 26.5?

Use the Empirical Rule to Analyze a Distribution

A. PACKAGING Students counted the number of candies in 100 small packages. They found that the number of candies per package was normally distributed, with a mean of 23 candies per package and a standard deviation of 1 piece of candy. About how many packages have between 22 and 24 candies?

22 and 24 are 1σ away from the mean. Therefore, about 68% of the data are between 22 and 24.

Since 100 × 68% = 68 we know that about 68 of the packages will contain 22 to 24 pieces.

Answer: about 68 packages

Use the Empirical Rule to Analyze a Distribution

B. PACKAGING Students counted the number of candies in 100 small packages. They found that the number of candies per package was normally distributed, with a mean of 23 candies per package and a standard deviation of 1 piece of candy. What is the probability that a package selected at random has more than 25 candies?

Values greater than 25 are more than 2σ from the mean. The values that are more than 2σ from the mean cover two tails and 5% of the distribution. We are only concerned with the upper tail, so 2.5% of the data will be greater than 25.

Answer: about 2.5%

A. 17%

B. 34%

C. 68%

D. 81.5%

DRIVER’S ED The number of students per driver’s education class is normally distributed, with a mean of 26 students per class and a standard deviation of 3 students. What is the probability that a driver’s education class selected at random will have between 23 and 32 students?

Use z-Values to Locate Position

Find σ if X = 28.3, μ = 24.6, and z = 0.63. Indicate the position of X in the distribution.

Formula for z-Values

X = 28.3, = 24.6, z = 0.63

Divide each side by 0.63.

Simplify.

0.63σ = 3.7 Multiply and subtract.

Use z-Values to Locate Position

Answer: σ is 5.87. Since z is 0.63, X is 0.63 standard deviations greater than the mean.

A. 26.082

B. 19.703

C. 18.698

D. 12.318

Find μ if X = 19.2, σ = 3.7, and z = –1.86.

Find Probabilities

HEALTH The cholesterol levels for adult males of a specific racial group are normally distributed with a mean of 158.3 and a standard deviation of 6.6. Find the probability. Then use a graphing calculator to sketch the corresponding area under the curve.

P(X > 150)

The question is asking for the percentage of adult males with a cholesterol level of at least 150. First, find the corresponding z-value for X = 150.

Find Probabilities

Formula for z-Values

Answer: 0.90

X = 150, = 158.3, z = 6.6

Simplify.

Using a graphing calculator, you can find the area between z = –1.26 and z = 4 to be about 0.90.

Find Probabilities

HEALTH The cholesterol levels for adult males of a specific racial group are normally distributed with a mean of 158.3 and a standard deviation of 6.6. Find the probability. Then use a graphing calculator to sketch the corresponding area under the curve.

P(145 < X < 165)

The question is asking for the percentage of adult males with a cholesterol level between 145 and 165. First, find the corresponding z-value for X = 145.

Find Probabilities

Formula for z-Values

Use 165 to find the other z-value.

X = 145, = 158.3, z = 6.6

Simplify.

Formula for z-Values

X = 145, = 158.3, z = 6.6

Simplify.

Find Probabilities

Answer: 0.82

Using a graphing calculator, you can find the area between z = –2.02 and z = 1.02 to be about 0.82.

A. 22%

B. 28%

C. 72%

D. 78%

INSECTS The lifespan of a specific insect is normally distributed with a mean of 12.3 days and a standard deviation of 3.9 days. Find P(X > 10).