A normal distribution has a mean of 416 and a
standard deviation of 55.
1.Find the range of values that represent the middle
99.7% of the distribution.
SOLUTION:
The middle 99.7% of data in a normal distribution is
the range from µ – 3σ to µ + 3σ. The standard
deviation is 55, so 3σ = 3 ∙55 or 165.
416 + 165 = 581 and 416 – 165 = 251
Therefore, the range of values in the middle 99.7% is
251 < X < 581.
3.CCSSTOOLSThe number of texts sent per day
by a sample of 811 teens is normally distributed with
a mean of 38 and a standard deviation of 7.
a. About how many teens sent between 24 and 38
texts?
b. What is the probability that a teen selected at
random sent less than 818 texts?
SOLUTION:
a. 24 is 2σ away from the mean, so the range of
values are between µ – 2σ and µ. This represents
half of 95% of the distribution, or 47.5%.
47.5% of 811 = 385.225
Therefore, about 386 teens sent between 24 and 38
texts.
b. 45 is σ away from the mean, so the range of
values are less than µ + σ. The range of values from
µ – σ to µ + σ represent 68% of the data, so the
range from µ to µ + σ represents half of these
values, or 34% of the data. 50% of the data are less
than µ, so the percentage of data values less than µ
+ σ is 50% + 34% or 84%. Therefore, the probability
that a teen selected at random sent less than 45 texts
is 84%.
Find the missing variable. Indicate the position
of X in the distribution.
5.z if μ = 13.3, X = 17.2, and σ = 1.9
SOLUTION:
The z-value that corresponds to X = 17.2 is
approximately 2.05 standard deviations more than the
mean.
z = Formula for z-values
z = X = 17.2, µ = 13.3, and
σ = 1.9
≈2.05 Simplify.
7.σ if μ = 21.1, X = 13.7, and z =-2.40
SOLUTION:
The z-value that corresponds to X = 13.7 is
approximately 2.4 standard deviations less than the
mean.
z = Formula for z-values
–2.40 = z = –2.40, X = 13.7,
and µ = 21.1
–2.40 = Subtract.
–2.4σ = –7.4 Multiply each side by σ.
σ = Divide each side by –
2.4.
σ ≈3.08 Simplify.
A normal distribution has a mean of 29.3 and a
standard deviation of 6.7.
9.Find the range of values that represent the outside
5% of the distribution.
SOLUTION:
The outside 5% of data in a normal distribution is
equal to 1 minus the middle 95%.
The middle 95% is the range from µ – 2σ to µ + 2σ.
The standard deviation is 6.7, so σ = 2 ∙6.7 or 13.4
29.3 – 13.4 = 15.9 and 29.3 + 13.4 = 42.7
The range of values in the middle 95% is 15.9 < X <
42.7. Therefore, the outside 5% is represented by X
< 15.9 or X > 42.7.
11.GYMS The number of visits to a gym per year by a
sample of 522 members is normally distributed with a
mean of 88 and a standard deviation of 19.
a. About how many members went to the gym at
least 50 times?
b. What is the probability that a member selected at
random went to the gym more than 145 times?
SOLUTION:
a. 50 is 2σ away from the mean of 88, so the range
of values are greater than µ – 2σ. µ – 2σ to µ
represents half of 95% of the distribution, or 47.5%.
50% of the distribution is greater than µ, so
everything greater than µ – 2σ is 47.5% + 50% or
97.5%.
97.5% of 522 = 508.95
Therefore, about 509 members went to the gym at
least 50 times.
b. 145 is 3σ away from the mean, so the range of
values are more than µ + 3σ. The range of values
from µ – 3σ to µ + 3σ represent 99.7% of the data,
so the range from µ to µ + 3σ represents half of
these values, or 49.85% of the data. The outer tail of
the distribution is 50% – 49.85% or 0.15% of the
data. Therefore, the probability that a member went
to the gym more than 145 times is 0.15%.
Find the missing variable. Indicate the position
of X in the distribution.
13.z if μ = 19.9, X = 18.7, and σ = 0.9
SOLUTION:
The z-value that corresponds to X = 18 is
approximately 1.33 standard deviations less than the
mean.
z = Formula for z-values
z = X = 18.7, µ = 19.9, and
σ = 0.9
≈–1.33 Simplify.
15.X if μ = 138.8, σ = 22.5, and z = 1.73
SOLUTION:
The z-value that corresponds to X = 177.725 is appro
standard deviations more than the mean.
z = Formula for z-val
1.73 = µ= 138.8, z = 1.7
38.925 = X – 138.8 Multiply each sid
177.725 = X Add 138.8 to eac
17.CAR BATTERIES The useful life of a certain car
battery is normally distributed with a mean of 113,627
miles and a standard deviation of 14,266 miles. The
company makes 20,000 batteries a month.
a. About how many batteries will last between
90,000 and 110,000 miles?
b. About how many batteries will last more than
125,000 miles?
c. What is the probability that if you buy a car battery
at random, it will last less than 100,000 miles?
SOLUTION:
a. Find the z-values associated with 90,000 and
110,000. The mean μis 113,627 and the standard
deviation σis 14,266.
Use a graphing calculator to find the area between
the z-values.
The value 0.35 is the percentage of batteries that will
last between 90,000 and 110,000 miles. The total
numberofbatteriesinthisgroupis0.35×20,000or
7000.
b. Find the z-value associated with 125,000.
We are looking for values greater than 125,000, so
we can use a graphing calculator to find the area
between z = 0.797 and z = 4.
The value 0.21 is the percentage of batteries that will
last between more than 125,000 miles. The total
numberofbatteriesinthisgroupis0.21×20,000or
4200.
c. Find the z-value associated with 100,000.
We are looking for values less than 100,000, so we
can use a graphing calculator to find the area
between z = −4 and z = −0.955.
The probability that if you buy a car battery at
random, it will last less than 100,000 miles is about
17.0%.
19.FINANCIAL LITERACY The insurance industry
uses various factors including age, type of car driven,
and driving record to determine an individual’s
insurance rate. Suppose insurance rates for a sample
population are normally distributed.
a. If the mean annual cost per person is $829 and the
standard deviation is $115, what is the range of rates
you would expect the middle 68% of the population
to pay annually?
b. If 900 people were sampled, how many would you
expect to pay more than $1000 annually?
c. Where on the distribution would you expect a
person with several traffic citations to lie? Explain
your reasoning.
d. How do you think auto insurance companies use
each factor to calculate an individual’s insurance
rate?
SOLUTION:
a. The middle 68% represents all data values within
one standard deviation of the mean. Add ±$115 to
$829. The range of rates is $714 to $944.
b. Find the z-value associated with 1000.
We are looking for values more than 1000, so we can
use a graphing calculator to find the area between z
= 1.487 and z = 4.
The probability that a customer selected at random
will pay more than $1000 is about 6.8%. Out of 900
people,about0.06847575·900or62peoplewillpay
more than $1000.
c. Sample answer: I would expect people with
several traffic citations to lie to the far right of the
distribution where insurance costs are highest,
because I think insurance companies would charge
them more.
d. Sample answer: I think auto insurance companies
would charge younger people more than older people
because they have not been driving as long. I think
they would charge more for expensive cars and
sports cars and less for cars that have good safety
ratings. I think they would charge a person less if
they have a good driving record and more if they
have had tickets and accidents.
21.CCSSCRITIQUEA set of normally distributed
tree diameters have mean 11.5 cm, standard
deviation 2.5, and range 3.6 to 19.8. Monica and
Hiroko are to find the range that represents the
middle 68% of the data. Is either of them correct?
Explain.
SOLUTION:
Hiroko; Monica’s solution would work with a uniform
distribution.
23.REASONING The term six sigma process comes
from the notion that if one has six standard deviations
between the mean of a process and the nearest
specification limit, there will be practically no items
that fail to meet the specifications. Is this a true
assumption? Explain.
SOLUTION:
Sample answer: True; according to the Empirical
Rule, 99% of the data lie within 3 standard deviations
of the mean. Therefore, only 1% will fall outside of
three sigma. An infinitesimally small amount will fall
outside of six-sigma.
25.OPEN ENDED Find a set of real-world data that
appears to be normally distributed. Calculate the
range of values that represent the middle 68%, the
middle 95%, and the middle 99.7% of the distribution.
SOLUTION:
Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
107.62. The middle 95% is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
27.The lifetimes of 10,000 light bulbs are normally
distributed. The mean lifetime is 300 days, and the
standard deviation is 40 days. How many light bulbs
will last between 260 and 340 days?
A 2500
B 3400
C 5000
D 6800
SOLUTION:
Given
The probability that a randomly selected value in the
distribution is between μ – σ and μ + σ, that is,
between 300 – 40 or 260 and 300 + 40 or 340.
Option D is the correct answer.
29.SHORT RESPONSE In the figure below, RT = TS
and QR = QT. What is the value of x?
SOLUTION:
Given RT = TS and QR = QT.
Since RT = TS, .
Since QR = QT, .
31.SNOW There is a 25% chance that it snows each da
the probability that it snows 3 out of the next 7 days.
SOLUTION:
The probability that it will snow in 3 of the next 7 day
P(X) = nCx
pXqn – X Binomial Pr
P(3) = 7C3(0.25)3(0.75)7– 3 n = 7, X = 3
≈0.173 Simplify.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
33.the amount of precipitation in a city per month
SOLUTION:
The amount of precipitation in a city is continuous
because it can take on any value.
Identify the type of function represented by
each graph.
35.
SOLUTION:
greatest integer
37.
SOLUTION:
constant
A normal distribution has a mean of 416 and a
standard deviation of 55.
1.Find the range of values that represent the middle
99.7% of the distribution.
SOLUTION:
The middle 99.7% of data in a normal distribution is
the range from µ – 3σ to µ + 3σ. The standard
deviation is 55, so 3σ = 3 ∙55 or 165.
416 + 165 = 581 and 416 – 165 = 251
Therefore, the range of values in the middle 99.7% is
251 < X < 581.
3.CCSSTOOLSThe number of texts sent per day
by a sample of 811 teens is normally distributed with
a mean of 38 and a standard deviation of 7.
a. About how many teens sent between 24 and 38
texts?
b. What is the probability that a teen selected at
random sent less than 818 texts?
SOLUTION:
a. 24 is 2σ away from the mean, so the range of
values are between µ – 2σ and µ. This represents
half of 95% of the distribution, or 47.5%.
47.5% of 811 = 385.225
Therefore, about 386 teens sent between 24 and 38
texts.
b. 45 is σ away from the mean, so the range of
values are less than µ + σ. The range of values from
µ – σ to µ + σ represent 68% of the data, so the
range from µ to µ + σ represents half of these
values, or 34% of the data. 50% of the data are less
than µ, so the percentage of data values less than µ
+ σ is 50% + 34% or 84%. Therefore, the probability
that a teen selected at random sent less than 45 texts
is 84%.
Find the missing variable. Indicate the position
of X in the distribution.
5.z if μ = 13.3, X = 17.2, and σ = 1.9
SOLUTION:
The z-value that corresponds to X = 17.2 is
approximately 2.05 standard deviations more than the
mean.
z = Formula for z-values
z = X = 17.2, µ = 13.3, and
σ = 1.9
≈2.05 Simplify.
7.σ if μ = 21.1, X = 13.7, and z =-2.40
SOLUTION:
The z-value that corresponds to X = 13.7 is
approximately 2.4 standard deviations less than the
mean.
z = Formula for z-values
–2.40 = z = –2.40, X = 13.7,
and µ = 21.1
–2.40 = Subtract.
–2.4σ = –7.4 Multiply each side by σ.
σ = Divide each side by –
2.4.
σ ≈3.08 Simplify.
A normal distribution has a mean of 29.3 and a
standard deviation of 6.7.
9.Find the range of values that represent the outside
5% of the distribution.
SOLUTION:
The outside 5% of data in a normal distribution is
equal to 1 minus the middle 95%.
The middle 95% is the range from µ – 2σ to µ + 2σ.
The standard deviation is 6.7, so σ = 2 ∙6.7 or 13.4
29.3 – 13.4 = 15.9 and 29.3 + 13.4 = 42.7
The range of values in the middle 95% is 15.9 < X <
42.7. Therefore, the outside 5% is represented by X
< 15.9 or X > 42.7.
11.GYMS The number of visits to a gym per year by a
sample of 522 members is normally distributed with a
mean of 88 and a standard deviation of 19.
a. About how many members went to the gym at
least 50 times?
b. What is the probability that a member selected at
random went to the gym more than 145 times?
SOLUTION:
a. 50 is 2σ away from the mean of 88, so the range
of values are greater than µ – 2σ. µ – 2σ to µ
represents half of 95% of the distribution, or 47.5%.
50% of the distribution is greater than µ, so
everything greater than µ – 2σ is 47.5% + 50% or
97.5%.
97.5% of 522 = 508.95
Therefore, about 509 members went to the gym at
least 50 times.
b. 145 is 3σ away from the mean, so the range of
values are more than µ + 3σ. The range of values
from µ – 3σ to µ + 3σ represent 99.7% of the data,
so the range from µ to µ + 3σ represents half of
these values, or 49.85% of the data. The outer tail of
the distribution is 50% – 49.85% or 0.15% of the
data. Therefore, the probability that a member went
to the gym more than 145 times is 0.15%.
Find the missing variable. Indicate the position
of X in the distribution.
13.z if μ = 19.9, X = 18.7, and σ = 0.9
SOLUTION:
The z-value that corresponds to X = 18 is
approximately 1.33 standard deviations less than the
mean.
z = Formula for z-values
z = X = 18.7, µ = 19.9, and
σ = 0.9
≈–1.33 Simplify.
15.X if μ = 138.8, σ = 22.5, and z = 1.73
SOLUTION:
The z-value that corresponds to X = 177.725 is appro
standard deviations more than the mean.
z = Formula for z-val
1.73 = µ= 138.8, z = 1.7
38.925 = X – 138.8 Multiply each sid
177.725 = X Add 138.8 to eac
17.CAR BATTERIES The useful life of a certain car
battery is normally distributed with a mean of 113,627
miles and a standard deviation of 14,266 miles. The
company makes 20,000 batteries a month.
a. About how many batteries will last between
90,000 and 110,000 miles?
b. About how many batteries will last more than
125,000 miles?
c. What is the probability that if you buy a car battery
at random, it will last less than 100,000 miles?
SOLUTION:
a. Find the z-values associated with 90,000 and
110,000. The mean μis 113,627 and the standard
deviation σis 14,266.
Use a graphing calculator to find the area between
the z-values.
The value 0.35 is the percentage of batteries that will
last between 90,000 and 110,000 miles. The total
numberofbatteriesinthisgroupis0.35×20,000or
7000.
b. Find the z-value associated with 125,000.
We are looking for values greater than 125,000, so
we can use a graphing calculator to find the area
between z = 0.797 and z = 4.
The value 0.21 is the percentage of batteries that will
last between more than 125,000 miles. The total
numberofbatteriesinthisgroupis0.21×20,000or
4200.
c. Find the z-value associated with 100,000.
We are looking for values less than 100,000, so we
can use a graphing calculator to find the area
between z = −4 and z = −0.955.
The probability that if you buy a car battery at
random, it will last less than 100,000 miles is about
17.0%.
19.FINANCIAL LITERACY The insurance industry
uses various factors including age, type of car driven,
and driving record to determine an individual’s
insurance rate. Suppose insurance rates for a sample
population are normally distributed.
a. If the mean annual cost per person is $829 and the
standard deviation is $115, what is the range of rates
you would expect the middle 68% of the population
to pay annually?
b. If 900 people were sampled, how many would you
expect to pay more than $1000 annually?
c. Where on the distribution would you expect a
person with several traffic citations to lie? Explain
your reasoning.
d. How do you think auto insurance companies use
each factor to calculate an individual’s insurance
rate?
SOLUTION:
a. The middle 68% represents all data values within
one standard deviation of the mean. Add ±$115 to
$829. The range of rates is $714 to $944.
b. Find the z-value associated with 1000.
We are looking for values more than 1000, so we can
use a graphing calculator to find the area between z
= 1.487 and z = 4.
The probability that a customer selected at random
will pay more than $1000 is about 6.8%. Out of 900
people,about0.06847575·900or62peoplewillpay
more than $1000.
c. Sample answer: I would expect people with
several traffic citations to lie to the far right of the
distribution where insurance costs are highest,
because I think insurance companies would charge
them more.
d. Sample answer: I think auto insurance companies
would charge younger people more than older people
because they have not been driving as long. I think
they would charge more for expensive cars and
sports cars and less for cars that have good safety
ratings. I think they would charge a person less if
they have a good driving record and more if they
have had tickets and accidents.
21.CCSSCRITIQUEA set of normally distributed
tree diameters have mean 11.5 cm, standard
deviation 2.5, and range 3.6 to 19.8. Monica and
Hiroko are to find the range that represents the
middle 68% of the data. Is either of them correct?
Explain.
SOLUTION:
Hiroko; Monica’s solution would work with a uniform
distribution.
23.REASONING The term six sigma process comes
from the notion that if one has six standard deviations
between the mean of a process and the nearest
specification limit, there will be practically no items
that fail to meet the specifications. Is this a true
assumption? Explain.
SOLUTION:
Sample answer: True; according to the Empirical
Rule, 99% of the data lie within 3 standard deviations
of the mean. Therefore, only 1% will fall outside of
three sigma. An infinitesimally small amount will fall
outside of six-sigma.
25.OPEN ENDED Find a set of real-world data that
appears to be normally distributed. Calculate the
range of values that represent the middle 68%, the
middle 95%, and the middle 99.7% of the distribution.
SOLUTION:
Sample answer: The scores per team in each game
of the first round of the 2010 NBA playoffs. The
mean is 96.56 and the standard deviation is 11.06.
The middle 68% of the distribution is 85.50 < X <
107.62. The middle 95% is 74.44 < X < 118.68. The
middle 99.7% is 63.38 < X < 129.74.
27.The lifetimes of 10,000 light bulbs are normally
distributed. The mean lifetime is 300 days, and the
standard deviation is 40 days. How many light bulbs
will last between 260 and 340 days?
A 2500
B 3400
C 5000
D 6800
SOLUTION:
Given
The probability that a randomly selected value in the
distribution is between μ – σ and μ + σ, that is,
between 300 – 40 or 260 and 300 + 40 or 340.
Option D is the correct answer.
29.SHORT RESPONSE In the figure below, RT = TS
and QR = QT. What is the value of x?
SOLUTION:
Given RT = TS and QR = QT.
Since RT = TS, .
Since QR = QT, .
31.SNOW There is a 25% chance that it snows each da
the probability that it snows 3 out of the next 7 days.
SOLUTION:
The probability that it will snow in 3 of the next 7 day
P(X) = nCx
pXqn – X Binomial Pr
P(3) = 7C3(0.25)3(0.75)7– 3 n = 7, X = 3
≈0.173 Simplify.
Identify the random variable in each
distribution, and classify it as discrete or
continuous. Explain your reasoning.
33.the amount of precipitation in a city per month
SOLUTION:
The amount of precipitation in a city is continuous
because it can take on any value.
Identify the type of function represented by
each graph.
35.
SOLUTION:
greatest integer
37.
SOLUTION:
constant
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10-5 The Normal Distribution
