Transcript Over Lesson 11–4

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
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.
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.
Over Lesson 11–4
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.
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.
Over Lesson 11–4
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.
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.
Over Lesson 11–4
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?
A. 25.2%
B. 48.7%
C. 51.3%
D. 74.8%
Over Lesson 11–4
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?
A. 3.8%
B. 12.4%
C. 23.2%
D. 39.4%
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 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?
A. 0.3%
B. 2.5%
C. 5%
D. 97.5%
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%
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?
A. 17%
B. 34%
C. 68%
D. 81.5%
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
0.63σ = 3.7
Multiply and subtract.
Divide each side by 0.63.
Simplify.
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.
Find μ if X = 19.2, σ = 3.7, and z = –1.86.
A. 26.082
B. 19.703
C. 18.698
D. 12.318
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
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.
Answer: 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
X = 145,  = 158.3, z = 6.6
Simplify.
Use 165 to find the other z-value.
Formula for z-Values
X = 145,  = 158.3, z = 6.6
Simplify.
Find Probabilities
Using a graphing calculator, you can find the area
between z = –2.02 and z = 1.02 to be about 0.82.
Answer: 0.82
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).
A. 22%
B. 28%
C. 72%
D. 78%