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MATH 214 (NOTES)
Math 214
Al Nosedal
Department of Mathematics
Indiana University of Pennsylvania
MATH 214 (NOTES) – p. 1/11
CHAPTER 6
CONTINUOUS PROBABILITY DISTRIBUTIONS
MATH 214 (NOTES) – p. 2/11
Simple example
Random Experiment: Rolling a fair die 300 times.
Class
1≤x<2
2≤x<3
3≤x<4
4≤x<5
5≤x<6
6≤x<7
Expected Freq. Expected Relative Freq.
50
1/6
50
1/6
50
1/6
50
1/6
50
1/6
50
1/6
MATH 214 (NOTES) – p. 3/11
Histogram
50
40
30
freq
60
70
Histogram of frequencies
1
2
3
4
5
6
7
MATH 214 (NOTES) – p. 4/11
Histogram (Relative Freqs.)
0.18
0.14
0.10
freq
0.22
Histogram of
Relative Frequencies
1
2
3
4
5
6
7
MATH 214 (NOTES) – p. 5/11
Uniform Probability Distribution
Uniform Probability Density Function
1
f (x) =
b−a
f (x) = 0
for a ≤ x ≤ b
elsewhere
MATH 214 (NOTES) – p. 6/11
Expected Value and Variance
a+b
E(X) =
2
(b − a)2
V ar(X) =
12
MATH 214 (NOTES) – p. 7/11
Problem 2 (page 230)
The random variable x is known to be uniformly distributed
between 10 and 20.
a. Show the graph of the probability density function.
b. Compute P (x < 15).
c. Compute P (12 ≤ x ≤ 18).
d. Compute E(x).
e. Compute V ar(x).
MATH 214 (NOTES) – p. 8/11
Solution
0.10
0.08
0.06
f(x)
0.12
0.14
Graph of Probability
Density Function
10
12
14
16
18
20
MATH 214 (NOTES) – p. 9/11
Solution
b. P (x < 15) = (0.10)(5) = 0.50
c. P (12 ≤ x ≤ 18) = (0.10)(6) = 0.60
= 15
d. E(x) = 10+20
2
e. V (x) =
(20−10)2
12
= 8.33
MATH 214 (NOTES) – p. 10/11
Problem 3 (page 230)
Delta Airlines quotes a flight time of 2 hours, 5 minutes for
its flights from Cincinnati to Tampa. Suppose we believe that
actual flight times are uniformly distributed between 2 hours
and 2 hours, 20 minutes.
a. Show the graph of the probability density function for flight
time.
b. What is the probability that the flight will be no more than
5 minutes late?
c. What is the probability that the flight will be more than 10
minutes late?
d. What is the expected flight time?
MATH 214 (NOTES) – p. 11/11
Solution (problem 3 a)
0.05
0.04
0.03
f(x)
0.06
0.07
Graph of Probability
Density Function
120
125
130
135
140
minutes
MATH 214 (NOTES) – p. 12/11
Solution (problem 3 b)
10
20
= 0.5
0.05
0.04
0.03
f(x)
0.06
0.07
b. P (x ≤ 130) =
120
125
130
135
140
minutes
MATH 214 (NOTES) – p. 13/11
Solution (problem 3 c)
5
20
= 0.25
0.05
0.04
0.03
f(x)
0.06
0.07
c. P (x > 135) =
120
125
130
135
140
minutes
MATH 214 (NOTES) – p. 14/11
Solutions (without graphs)
10
= 0.5
b. P (x ≤ 130) = 20
5
= 0.25
c. P (x > 135) = 20
= 130 minutes
d. E(x) = 120+140
2
MATH 214 (NOTES) – p. 15/11
Standard Normal Probability Distribution
0.0
0.1
0.2
0.3
0.4
Normal Distribution
mean=0 and std.dev.=1
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 16/11
Standard Normal Probability Distribution
1 −z2
f (z) = √ e 2
2π
Note. Don’t worry about this formula, we are NOT going to
use it to do calculations.
MATH 214 (NOTES) – p. 17/11
Normal Probability Density Function
−(x−µ)2
1
f (x) = √ e 2σ2
σ 2π
where µ = mean, σ = standard deviation, e = 2.71828 and π
= 3.14159.
Note. Don’t worry about this formula, we are NOT going to
use it to do calculations.
MATH 214 (NOTES) – p. 18/11
0.00
0.05
0.10
0.15
0.20
Two Normal Distributions
−15
−10
−5
0
5
10
15
MATH 214 (NOTES) – p. 19/11
0.20
Two Different Standard Deviations
0.00
0.05
0.10
0.15
std.dev.=5
std.dev.=2
−15
−10
−5
0
5
10
15
MATH 214 (NOTES) – p. 20/11
Problem 11 (page 241)
Given that z is a standard normal random variable, compute
the following probabilities.
a. P (z ≤ −1).
b. P (z ≥ −1)
c. P (z ≥ −1.5)
d. P (−2.5 ≤ z)
e. P (−3 < z ≤ 0)
MATH 214 (NOTES) – p. 21/11
Problem 12 (page 241)
Given that z is a standard normal random variable, compute
the following probabilities.
a. P (0 ≤ z ≤ 0.83)
b. P (−1.57 ≤ z ≤ 0)
c. P (z ≥ 0.44)
d. P (z ≥ −0.23)
e. P (z ≤ 1.2)
f. P (z ≤ −0.71)
MATH 214 (NOTES) – p. 22/11
Solution (problem 12 a)
0.4
a. P (0 ≤ z ≤ 0.83) = 0.2967
0.0
0.1
0.2
0.3
0.7967−0.5 =
0.2967
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 23/11
Solution (problem 12 b)
0.5−0.0582 =
0.4418
0.0
0.1
0.2
0.3
0.4
b. P (−1.57 ≤ z ≤ 0) = 0.4418
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 24/11
Solution (problem 12 c)
0.4
c. P (z ≥ 0.44) = 0.3300
0.1
0.2
0.3
1−0.6700 =
0.3300
0.0
0.6700
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 25/11
Solution (problem 12 d)
0.4
d. P (z ≥ −0.23) = 0.5910
0.1
0.2
0.3
1−0.4090 =
0.5910
0.0
0.4090
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 26/11
Solution (problem 12 e)
0.1
0.2
0.3
0.4
e. P (z ≤ 1.2) = 0.8849
0.0
0.8849
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 27/11
Solution (problem 12 f)
0.2
0.3
0.4
f. P (z ≤ −0.71) = 0.2389
0.0
0.1
0.2389
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 28/11
Problem 14 (page 241)
Given that Z is a standard normal random variable, find z∗
for each situation.
a. The area to the left of z∗ is 0.9750.
b. The area between 0 and z∗ is 0.4750.
c. The area to the left of z∗ is 0.7291.
d. The area to the right of z∗ is 0.1314.
e. The area to the left of z∗ is 0.6700.
f. The area to the right of z∗ is 0.3300.
MATH 214 (NOTES) – p. 29/11
Problem 14 (solutions)
a. z∗ = 1.96 (TI-83: invNorm(0.975) = 1.9599 ).
b. z∗ = 1.96
c. z∗ = 0.61 (TI-83: invNorm(0.7291) = 0.6100 ).
d. z∗ = 1.12 (TI-83: invNorm(0.8686) = 1.1197 ).
e. z∗ = 0.44 (TI-83: invNorm(0.6700) = 0.4399 ).
f. z∗ = 0.44.
MATH 214 (NOTES) – p. 30/11
0.2
0.3
0.4
Problem 14 a
0.0
0.1
0.9750
−0.1
z*=1.96
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 31/11
0.1
0.2
0.3
0.4
Problem 14 b
0.4750
0.0
0.5
−0.1
z*=1.96
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 32/11
0.2
0.3
0.4
Problem 14 c
0.0
0.1
0.7291
−0.1
z*=0.61
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 33/11
0.2
0.3
0.4
Problem 14 d
0.1314
0.0
0.1
0.8686
−0.1
z*=1.12
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 34/11
0.1
0.2
0.3
0.4
Problem 14 e
0.0
0.67
−0.1
z*=0.44
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 35/11
0.1
0.2
0.3
0.4
Problem 14 f
0.33
0.0
0.67
−0.1
z*=0.44
−3
−2
−1
0
1
2
3
MATH 214 (NOTES) – p. 36/11
Problem 15 (page 242)
Given that z is a standard normal random variable, find z for
each situation.
a. The area to the left of z∗ is 0.2119.
b. The area between −z∗ and z∗ is 0.9030.
c. The area between −z∗ and z∗ is 0.2052.
d. The area to the left of z∗ is 0.9948.
e. The area to the right of z∗ is 0.6915.
MATH 214 (NOTES) – p. 37/11
Problem 15 (solutions)
a. z∗ = −0.80.
b. z∗ = 1.66.
c z∗ = 0.26.
d. z∗ = 2.56.
e. z∗ = −0.50.
MATH 214 (NOTES) – p. 38/11
Problem 17 (page 242)
For borrowers with good credit scores, the mean debt for
revolving and installment accounts is $ 15,015. Assume the
standard deviation is $ 3540 and that debt amounts are
normally distributed.
a. What is the probability that the debt for a randomly
selected borrower with good credit is more than $ 18,000?
b. What is the probability that the debt for a randomly
selected borrower with good credit is less than $ 10,000?
c. What is the probability that the debt for a randomly
selected borrower with good credit is between $ 12,000 and
$ 18,000?
d. What is the probability that the debt for a
randomly selected borrower with good credit
is no more than $ 14,000?
MATH 214 (NOTES) – p. 39/11
Problem 17 (solutions)
Let X = debt amount.
a. P (X > 18000) = P ( X−µ
σ >
18000−15015
)
3540
= P (Z > 0.8432) =
1 − P (Z < 0.8432) = 1 − 0.7995 = 0.2005
10000−15015
<
) = P (Z < −1.4166) w
b. P (X < 10000) = P ( X−µ
σ
3540
0.0773
c. P (12000 < X < 18000) =
X−µ
18000−15015
<
<
) = P (−0.8516 < Z < 0.8432) =
P ( 12000−15015
3540
σ
3540
P (Z < 0.8432) − P (Z < −0.8516) w 0.7995 − 0.1977 = 0.6018
14000−15015
d. P (X ≤ 14000) = P ( X−µ
≤
) = P (Z < −0.2867) w
σ
3540
0.3897
MATH 214 (NOTES) – p. 40/11
Problem 21 (page 242)
A person must score in the upper 2% of the population on
an IQ test to qualify for membership in MENSA, the
international high-IQ society. If IQ scores are normally
distributed with a mean of 100 and a standard deviation of
15, what score must a person have to qualify for MENSA?.
MATH 214 (NOTES) – p. 41/11
Problem 21 (solution)
1. State the problem. Let X = IQ score of a randomly
selected person. We want to find the IQ score x∗ with area
0.02 to its right under the Normal curve with mean µ = 100
and standard deviation σ = 15. That’s the same as finding
the IQ score x∗ with area 0.98 to its left. Because our table
gives the areas to the left of z-values, always state the
problem in terms of the area to the left of x∗ .
MATH 214 (NOTES) – p. 42/11
Solution (cont.)
2. Use the table. Look in the body of our table for the entry
closest to 0.98. It is 0.9798. This is the entry corresponding
to z∗ = 2.05. So z∗ = 2.05 is the standardized value with area
0.98 to its left.
3. Unstandardize to transform the solution from the Z back
to the original X scale. We know that the standardized value
of the unknown x∗ is z∗ = 2.05. So x∗ itself satisfies:
x∗ −100
= 2.05. Solving this equation for x∗ gives:
15
x∗ = 100 + (2.05)(15) = 130.75. We see that a person must
score at least 130.75 (131 perhaps) to place in the highest
2%.
MATH 214 (NOTES) – p. 43/11
Problem 25 (page 243)
According to the Sleep Foundation, the average night’s
sleep is 6.8 hours. Assume the standard deviation is 0.6
hours and that the probability distribution is normal.
a. What is the probability that a randomly selected person
sleeps more than 8 hours?
b. What is the probability that a randomly selected person
sleeps 6 hours or less?
c. Doctors suggest getting between 7 and 9 hours of sleep
each night. What percentage of the population gets this
much sleep?
MATH 214 (NOTES) – p. 44/11
Problem 25 (solution)
Let X = number of hours a randomly selected person sleeps
per night.
8−6.8
a. P (X > 8) = P ( X−µ
>
σ
0.6 ) = P (Z > 2) = 1 − 0.9772 = 0.0228
6−6.8
≤
b.P (X ≤ 6) = P ( X−µ
σ
0.6 ) = P (Z ≤ −1.33) = 0.0918
X−µ
9−6.8
c.P (7 ≤ X ≤ 9) = P ( 7−6.8
≤
≤
0.6
σ
0.6 )
= P (0.33 ≤ Z < 3.66) ' 1 − 0.6293 = 0.3707
MATH 214 (NOTES) – p. 45/11
Standard Normal Distribution (again)
The Standard Normal Distribution is the Normal Distribution
N (0, 1) with mean 0 and standard deviation 1.
If a variable X has any Normal Distribution N (µ, σ) with mean
µ and standard deviation σ , then the standardized variable
X −µ
Z=
σ
has the Standard Normal Distribution.
MATH 214 (NOTES) – p. 46/11
Finding Normal Probabilities
1. State the problem in terms of the observed variable X .
2. Standardize X to restate the problem in terms of a
Standard Normal variable Z . Draw a picture to show the
area under the Standard Normal curve.
3. Find the required area under the Standard Normal curve,
using a table (or a calculator) and the fact that the total area
under the curve is 1.
MATH 214 (NOTES) – p. 47/11
Exponential Probability Distribution
EXPONENTIAL PROBABILITY DENSITY FUNCTION
1 −x
f (x) = e µ
µ
for x ≥ 0 and µ ≥ 0
(where µ = expected value or mean)
MATH 214 (NOTES) – p. 48/11
Exponential Distribution (cont.)
CUMULATIVE PROBABILITIES.
P (X ≤ k) = 1 − e−k/µ
MATH 214 (NOTES) – p. 49/11
Problem 32 (page 249)
Consider the following exponential probability density
function.
1 −x/8
f (x) = e
8
a. Find P (X ≤ 6).
b. Find P (X ≤ 4).
c. Find P (X ≥ 6).
d. Find P (4 ≤ X ≤ 6).
MATH 214 (NOTES) – p. 50/11
Solution
a. P (X ≤ 6) = 1 − e−6/8 = 0.5276
b. P (X ≤ 4) = 1 − e−4/8 = 0.3934
c. P (X ≥ 6) = 1 − P (X < 6) = 1 − 0.5276 = 0.4724
d. P (4 ≤ X ≤ 6) = P (X ≤ 6) − P (X < 4)
= 0.5276 − 0.3934 = 0.1342
MATH 214 (NOTES) – p. 51/11
Problem 34 (page 249)
The time required to pass through security screening at the
airport can be annoying to travelers. The mean wait time
during peak periods at Cincinnati/Northern Kentucky
International Airport is 12.1 minutes. Assume the time to
pass through security screening follows an exponential
distribution.
a. What is the probability it will take less than 10 minutes to
pass through security screening during a peak period?
MATH 214 (NOTES) – p. 52/11
Problem 34 (cont.)
b. What is the probability it will take more than 20 minutes to
pass through security screening during a peak period?
c. What is the probability it will take between 10 and 20
minutes to pass through security screening during a peak
period?
d. It is 8:00 a.m. (a peak period) and you just entered the
security line. To catch your plane you must be at the gate
within 30 minutes. If it takes 12 minutes from the time you
clear security until you reach your gate, what is the
probability you will miss your flight?
MATH 214 (NOTES) – p. 53/11
Solution
Let T be the time required to pass through security
screening. T has an exponential distribution with µ = 12.1
a. P (T < 10) = 1 − e−10/12.1 = 0.5623
b. P (T > 20) = 1 − P (T ≤ 20) = 1 − (1 − e−20/12.1 ) =
1 − 0.8085 = 0.1915
c. P (10 < T < 20) = P (T < 20) − P (T < 10) = 0.8085 − 0.5623 =
0.2462
d. P (missing your flight) = P (T > 18) = 1 − P (T ≤ 18) =
1 − (1 − e−18/12.1 ) = 1 − 0.7740 = 0.226
MATH 214 (NOTES) – p. 54/11
CHAPTER 7
SAMPLING AND SAMPLING DISTRIBUTIONS
MATH 214 (NOTES) – p. 55/11
Toy example
Consider a population of 5 families with annual incomes
shown below. We want to choose a simple random sample
of size 2 from this population. How can this be done? And
how do the sample mean of the chosen families compare to
population mean?
Incomes: 30,000 35,000 40,000 45,000 50,000.
MATH 214 (NOTES) – p. 56/11
Solution
30,000+40,000
x̄1 = 30,000+35,000
=
32,
500
x̄
=
= 35, 000
2
2
2
30,000+50,000
x̄3 = 30,000+45,000
=
37,
500
x̄
=
= 40, 000
4
2
2
35,000+45,000
=
37,
500
x̄
=
= 40, 000
x̄5 = 35,000+40,000
6
2
2
40,000+45,000
x̄7 = 35,000+50,000
=
42,
500
x̄
=
= 42, 500
8
2
2
45,000+50,000
=
45,
000
x̄
=
= 47, 500
x̄9 = 40,000+50,000
10
2
2
µ = 40, 000 (population mean).
MATH 214 (NOTES) – p. 57/11
1.0
0.5
0.0
frequency
1.5
2.0
Graph
32500
37500
42500
47500
income
MATH 214 (NOTES) – p. 58/11
Population, Sample
The entire group of individuals that we want information
about is called the population.
A sample is a part of the population that we actually
examine in order to gather information.
MATH 214 (NOTES) – p. 59/11
Simple Random Sample
A simple random sample (SRS) of size n consists of n
individuals from the population chosen in such a way that
every set of n individuals has an equal chance to be the
sample actually selected.
MATH 214 (NOTES) – p. 60/11
Example
A couple plans to have three children. There are 8 possible
arrangements of girls and boys. For example, GGB means
the first two children are girls and the third child is a boy. All
8 arrangements are (approximately) equally likely.
a) Write down all 8 arrangements of the sexes of three
children. What is the probability of any one of these
arrangements?
MATH 214 (NOTES) – p. 61/11
Example (cont.)
b) Let X be the number of girls the couple has. What is
the probability that X = 2 ?
c) Starting from your work in a), find the distribution of X .
That is, what values can X take, and what are the
probabilities for each value?
MATH 214 (NOTES) – p. 62/11
Binomial Probability Function
n!
f (x) =
px (1 − p)n−x
x!(n − x)!
f (x)= the probability of x successes in n trials.
n= the number of trials.
p= the probability of a success on any one trial.
1 − p= the probability of a failure on any one trial.
MATH 214 (NOTES) – p. 63/11
More about Binomial Distributions
E(X) = µ = np
V ar(X) = σ 2 = np(1 − p)
MATH 214 (NOTES) – p. 64/11
Problem
We are interested in estimating the average number of cars
per household in a little town call Statstown. Let X represent
the number of cars in a house picked at random. God knows
that X has a Binomial distribution with n = 4 and p = 0.5.
Suppose that we can only afford a sample of size 4 and that
we are going to use this sample to estimate that population
average.
MATH 214 (NOTES) – p. 65/11
Problem (cont.)
What we are going to do next is called a simulation. First,
we will draw a lot of random samples coming from a
Binomial Distribution with n = 4 and p = 0.5. Then we will
make a histogram for all the x̄’s corresponding to our
samples. We are going to do this to see what the histogram
of x̄ looks like. This will give us an idea of what to expect in a
similar situation.
MATH 214 (NOTES) – p. 66/11
Sampling Distribution of x̄
0.6
0.4
0.2
0.0
Density
0.8
1.0
Histogram of means
0
1
2
3
4
means
MATH 214 (NOTES) – p. 67/11
Central Limit Theorem
Draw a random sample of size n from any population with
mean µ and finite standard deviation σ . When n is large, the
sampling distribution of the sample mean x̄ is approximately
Normal:
(1)
x̄
σ
is approximately N (µ, √ )
n
MATH 214 (NOTES) – p. 68/11
Example
The number of accidents per week at a hazardous
intersection varies with mean 2.2 and standard deviation
1.4. This distribution takes only whole-number values, so it
is certainly not Normal.
a) Let x̄ be the mean number of accidents per week at
the intersection during a year (52 weeks). What is the
approximate distribution of x̄ according to the central
limit theorem?
MATH 214 (NOTES) – p. 69/11
Example (cont.)
b) What is the approximate probability that x̄ is less than
2?
c) What is the approximate probability that there are
fewer than 100 accidents at the intersection in a year?
(Hint: Restate this event in terms of x̄)
MATH 214 (NOTES) – p. 70/11
Solution
a) By the Central Limit Theorem, x̄ is roughly Normal with
= 0.1941.
mean µ = 2.2 and standard deviation √σn = √1.4
52
b) P (x̄ < 2) = P ( x̄−µ
<
√σ
n
2−2.2
0.1941 )
= P (Z < −1.0303) = 0.1515.
c) Let xi be the number of accidents during week i.
P52
P52
i=1 xi
P (Total < 100) = P ( i=1 xi < 100) = P ( 52 <
P (x̄ < 1.9230) = P (Z < −1.4270) = 0.0768
100
52 )
=
MATH 214 (NOTES) – p. 71/11
Problem 18 (page 278)
A population has a mean of 200 and a standard deviation of
50. A simple random sample of size 100 will be taken and
the sample mean x̄ will be used to estimate the population
mean.
a. What is the expected value of x̄.
b. What is the standard deviation of x̄.
c. Show the sample distribution of x̄.
MATH 214 (NOTES) – p. 72/11
Solution
a. E(x̄) = µ = 200.
=5
b. σx̄ = √σn = √50
100
c. By the Central Limit Theorem, x̄ is roughly Normal with
mean µ = 200 and standard deviation √σn = 5.
MATH 214 (NOTES) – p. 73/11
Problem 19 (page 278)
A population has a mean of 200 and a standard deviation of
50. Suppose a simple random sample of size 100 is
selected and x̄ is used to estimate µ.
a. What is the probability that the sample mean will be within
±5 of the population mean?
b. What is the probability that the sample mean will be within
±10 of the population mean?
MATH 214 (NOTES) – p. 74/11
Solution
By the Central Limit Theorem, x̄ is roughly Normal with
mean µ = 200 and standard deviation √σn = 5.
a. P (195 ≤ x̄ ≤ 205) = P ( 195−200
≤
5
x̄−µ
√σ
n
≤
205−200
)
5
=
P (−1 ≤ Z ≤ 1) = P (Z ≤ 1) − P (Z ≤ −1) = 0.8413 − 0.1587 =
0.6826
x̄−µ
210−200
≤
b. P (190 ≤ x̄ ≤ 210) = P ( 190−200
≤
)=
σ
√
5
5
n
P (−2 ≤ Z ≤ 2) = P (Z ≤ 2) − P (Z ≤ −2) = 0.9772 − 0.0228 =
0.9544
MATH 214 (NOTES) – p. 75/11
Problem 24
The mean tuition cost at state universities throughout the
United States is $ 4260 per year . Use this value as the
population mean and assume that the population standard
deviation is σ = $900. Suppose that a random sample of 50
state universities will be selected.
a. Show the sampling distribution of x̄ where x̄ is the sample
mean tuition cost for the 50 state universities.
b. What is the probability that the simple random sample will
provide a sample mean within $250 of the population mean?
c. What is the probability that the simple random sample will
provide a sample mean within $100 of the population mean?
MATH 214 (NOTES) – p. 76/11
Solution
a. By the Central Limit Theorem, x̄ has roughly a N (µ, √σn ).
In this case, x̄ has roughly a N (4260, √900
), that is, x̄ has
50
roughly a N (4260, 127.28)
4510−4260
b. P (4010 ≤ x̄ ≤ 4510) = P ( 4010−4260
≤
Z
≤
127.28
127.28 ) =
P (−1.96 ≤ Z ≤ 1.96) = 0.9750 − 0.025 = 0.95
4360−4260
≤
Z
≤
c. P (4160 ≤ x̄ ≤ 4360) = P ( 4160−4260
127.28
127.28 ) =
P (−0.79 ≤ Z ≤ 0.79) = 0.7852 − 0.2148 = 0.5704
MATH 214 (NOTES) – p. 77/11
Problem 25
The College Board College Testing Program reported a
population mean SAT score of µ = 1020. Assume that the
population standard deviation is σ = 100.
a. What is the probability that a random sample of 75
students will provide a sample mean SAT score within 10 of
the population mean?
b. What is the probability a random sample of 75 students
will provide a sample mean SAT score within 20 of the
population mean?
MATH 214 (NOTES) – p. 78/11
CHAPTER 8
INTERVAL ESTIMATION
MATH 214 (NOTES) – p. 79/11
M & M’s
If you buy an 8 pack of m & m’s fun size you will find the
following information:
Nutrition Facts
Serving Size 3 packs (45 g)
Which implies that: 1 pack = 15 g, right?. Since we are
buying an 8 pack the net weight should be (8)(15) = 120 g,
right?. However, the net weight on the label is 117.1 g, How
can we explain that?.
MATH 214 (NOTES) – p. 80/11
Interval Estimate of a Population Mean:σ kn
σ
x̄ ± z∗ ( √ )
n
where z∗ is a number coming from a Standard Normal that
depends on the confidence level required.
MATH 214 (NOTES) – p. 81/11
Problem 1
A simple random sample of 40 items resulted in a sample
mean of 25. The population standard deviation is σ = 5.
a. What is the standard error of the mean, σx̄ ?
b. At 95% confidence, what is the margin of error?
MATH 214 (NOTES) – p. 82/11
Solution
a. standard error of the mean = σx̄ = √σn = √540 = 0.7905
b. margin of error = z∗ √σn = (1.96)(0.7905) = 1.5493
MATH 214 (NOTES) – p. 83/11
Problem 2 (page 306)
A simple random sample of 50 items from a population with
σ = 6 resulted in a sample mean of 32.
a. Provide a 90% confidence interval for the population
mean?
b. Provide a 95% confidence interval for the population
mean?
c. Provide a 99% confidence interval for the population
mean?
MATH 214 (NOTES) – p. 84/11
Solution
a. x̄ = 32, σ = 6, n = 50 and z∗ = 1.65
(x̄ − z∗ √σn , x̄ − z∗ √σn )
(32 − 1.65( √650 ), 32 − 1.65( √650 ))
(30.5999, 33.4000) 90% Confidence Interval for µ.
b. (32 − 1.96( √650 ), 32 − 1.96( √650 ))
(30.3368, 33.6631) 95% Confidence Interval for µ.
MATH 214 (NOTES) – p. 85/11
Problem 5 (page 306)
In an effort to estimate the mean amount spent per
customer for dinner at a major Atlanta restaurant, data were
collected for a sample of 49 customers. Assume a
population standard deviation of $ 5.
a. At 95% confidence, what is the margin of error?
b. If the sample mean is $ 24.80, what is the 95%
confidence interval for the population mean?
MATH 214 (NOTES) – p. 86/11
Solution
In this case, µ = mean amount spent per customer for dinner
for all customers at a major Atlanta restaurant.
a. margin of error = z∗ ( √σn ) = 1.96( √549 ) = 1.4
b. (x̄ − z∗ √σn , x̄ − z∗ √σn )
(24.80 − 1.4, 24.80 + 1.4))
(23.40, 26.20) 95 % Confidence Interval for µ.
MATH 214 (NOTES) – p. 87/11
Problem 10 (page 306)
Playbill magazine reported that the mean annual household
income of its readers is $ 119,155. Assume this estimate of
the mean annual household income is based on a sample of
80 households, and based on past studies, the population
standard deviation is known to be σ = $ 30,000.
a. Develop a 90 % confidence interval estimate of the
population mean.
b. Develop a 95 % confidence interval estimate of the
population mean.
c. Develop a 99 % confidence interval estimate of the
population mean.
MATH 214 (NOTES) – p. 88/11
Solution
a. (x̄ − z∗ √σn , x̄ − z∗ √σn )
30,000
√
√
),
119,
155
−
1.65(
(119, 155 − 1.65( 30,000
))
80
80
(113, 620.73; 124, 689.26) 90% Confidence Interval for µ.
30,000
√
√
),
119,
155
−
1.96(
))
b. (119, 155 − 1.96( 30,000
80
80
(112, 580.96; 125, 729.03) 95% Confidence Interval for µ.
30,000
√
√
c. (119, 155 − 2.58( 30,000
))
),
119,
155
−
2.58(
80
80
(110501.41; 127808.58) 99% Confidence Interval for µ.
MATH 214 (NOTES) – p. 89/11
Problem 11 (page 314)
For a t distribution with 16 degrees of freedom, find the area,
or probability, in each region.
a. To the right of 2.120.
b. To the left of 1.337.
c. To the left of -1.746.
d. To the right of 2.583.
e. Between -2.120 and 2.120.
f. Between -1.746 and 1.746.
MATH 214 (NOTES) – p. 90/11
Solution
(Go to page 920)
Let T represent a t distribution with 16 degrees of freedom.
a. P (T > 2.120) = 0.025.
b. P (T < 1.337) = 0.10.
c. P (T < −1.746) = P (T > 1.746) = 0.05. (Because t
distributions are symmetric).
d. P (T > 2.583) = 0.01.
e. P (−2.120 < T < 2.120) = 0.95.
f. P (−1.746 < T < 1.746) = 0.90.
MATH 214 (NOTES) – p. 91/11
Problem 12 (page 314)
Find the t value(s) for each of the following cases.
a. Upper tail area of 0.025 with 12 degrees of freedom.
b. Lower tail area of 0.05 with 50 degrees of freedom.
c. Upper tail area of 0.01 with 30 degrees of freedom.
d. Where 90 % of the area falls between these two t values
with 25 degrees of freedom?
e. Where 95 % of the area falls between these two t values
with 45 degrees of freedom?
MATH 214 (NOTES) – p. 92/11
Solution
a. t∗ = 2.179. (TI-84: invT(0.975,12)= 2.1788).
b. t∗ = −1.676 (TI-84: invT(0.05,50)= -1.6759).
c. t∗ = 2.457 (TI-84: invT(0.99,30)= 2.4572).
d. t∗ = 2.014 (TI-84: invT(0.975,45)= 2.014).
e. t∗ = 1.708 (TI-84: invT(0.99,30)= 2.4572).
MATH 214 (NOTES) – p. 93/11
Problem 13 (page 314)
The following sample data are from a normal population: 10,
8, 12, 15, 13, 11, 6, 5.
a. What is the point estimate of the population mean?
b. What is the point estimate of the population standard
deviation?
c. With 95 % confidence, what is the margin of error for the
estimation of the population mean?
d. What is the 95 % confidence interval for the population
mean?
MATH 214 (NOTES) – p. 94/11
Solution
a. x̄ = 10.
b. s = 3.4641.
√ ) = 2.8965.
c. margin of error = t∗ √sn = 2.365( 3.4641
8
d. (x̄ − t∗ ( √sn ), x̄ + t∗ ( √sn ))
(7.1039, 12.896) (using a TI-84).
MATH 214 (NOTES) – p. 95/11
Problem 14 (page 314)
A simple random sample with n = 54 provided a sample
mean of 22.5 and a sample standard deviation of 4.4.
a. Develop a 90% confidence interval for the population
mean.
b. Develop a 95% confidence interval for the population
mean.
c. Develop a 99% confidence interval for the population
mean.
MATH 214 (NOTES) – p. 96/11
Solution
a. (x̄ − t∗ ( √sn ), x̄ + t∗ ( √sn ))
4.4
√
(22.5 − 1.674( √4.4
),
22.5
+
1.674(
))
54
54
(21.498, 23.502). (using a TI-84)
4.4
√
b. (22.5 − 2.006( √4.4
))
),
22.5
+
2.006(
54
54
(21.299, 23.701) (using a TI-84)
4.4
√
c. (22.5 − 2.672( √4.4
),
22.5
+
2.672(
))
54
54
(20.9, 24.1) (using a TI-84)
MATH 214 (NOTES) – p. 97/11
Problem 15 (page 315)
Sales personnel for Skillings Distributors submit weekly
reports listing the customer contacts made during the week.
A sample of 65 weekly reports showed a sample mean of
19.5 customer contacts per week. The sample standard
deviation was 5.2. Provide a 90% and 95% confidence
intervals for the population mean number of weekly
customer contacts for the sales personnel.
MATH 214 (NOTES) – p. 98/11
Solution
90 % Confidence.
(x̄ − t∗ ( √sn ), x̄ + t∗ ( √sn ))
5.2
√
),
19.5
+
1.669(
))
(19.5 − 1.669( √5.2
65
65
(18.42, 20.58)
95 % Confidence.
(x̄ − t∗ ( √sn ), x̄ + t∗ ( √sn ))
5.2
√
(19.5 − 1.998( √5.2
),
19.5
+
1.998(
))
65
65
(18.21, 20.79)
MATH 214 (NOTES) – p. 99/11
Problem 23 (page 318)
How large a sample should be selected to provide a 95%
confidence interval with a margin of error of 10? Assume
that the population standard deviation is 40.
MATH 214 (NOTES) – p. 100/11
Problem 24 (page 318)
The range for a set of data is estimated to be 36.
a. What is the planning value for the population standard
deviation?
b. At 95% confidence, how large a sample would provide a
margin of error of 3?
c. At 95% confidence, how large a sample would provide a
margin of error of 2?
MATH 214 (NOTES) – p. 101/11
Problem 25 (page 318)
Scheer Industries is considering a new computer-assisted
program to train maintenance employees to do machine
repairs. In order to fully evaluate the program, the director of
manufacturing requested an estimate of the population
mean time required for maintenance employees to complete
the computer-assisted training. Use 6.84 days as a planning
value for the population standard deviation.
a. Assuming 95% confidence, what sample size would be
required to obtain a margin of error of 1.5 days?
b. If the precision statement was made with 90%
confidence, what sample size would be required to obtain a
margin of error of 2 days?
MATH 214 (NOTES) – p. 102/11
Problem 28 (page 318)
Smith Travel Research provides information on the one-night
cost of hotel rooms through-out the United States. Use $ 2
as the desired margin of error and $ 22.50 as the planning
value for the population standard deviation to find the
sample size recommended in a), b), and c).
a. A 90% confidence interval estimate of the population
mean cost of hotel rooms.
b. A 95% confidence interval estimate of the population
mean cost of hotel rooms.
c. A 99% confidence interval estimate of the population
mean cost of hotel rooms.
MATH 214 (NOTES) – p. 103/11
Problem
A market research consultant hired by the Pepsi-Cola Co. is
interested in determining the proportion of IUP students who
favor Pepsi-Cola over Coke Classic. A random sample of
100 students shows that 40 students favor Pepsi over Coke.
Use this information to construct a 95% confidence interval
for the proportion of all students in this market who prefer
Pepsi.
MATH 214 (NOTES) – p. 104/11
Bernoulli Distribution
xi =
(
1 i-th person prefers Pepsi
0 i-th person prefers Coke
µ = E(xi ) = p
σ 2 = V (xi ) = p(1 − p)
Pn
xi
Let p̂ be our estimate of p. Note that p̂ = n = x̄. If n is
"large", by the Central Limit Theorem, we know that:
x̄ is roughly N (µ, √σn ), that is,
i=1
q
p̂ is roughly N p, p(1−p)
n
MATH 214 (NOTES) – p. 105/11
Problem 31 (page 322)
A simple random sample of 400 individuals provides 100
Yes responses.
a. What is the point estimate of the proportion of the
population that would provide Yes responses?
b. What is the point estimate of the standard error of the
proportion, σp̂ ?
c. Compute the 95% confidence interval for the population
proportion.
MATH 214 (NOTES) – p. 106/11
Problem 32 (page 323)
A simple random sample of 800 elements generates a
sample proportion p̂ = 0.70.
a. Provide a 90% confidence interval for the population
proportion.
b. Provide a 95% confidence interval for the population
proportion.
MATH 214 (NOTES) – p. 107/11
Problem 35 (page 323)
A survey of 611 office workers investigated telephone
answering practices, including how often each office worker
was able to answer incoming telephone calls and how often
incoming telephone calls went directly to voice mail. A total
of 281 office workers indicated that they never need voice
mail and are able to take every telephone call.
a. What is the point estimate of the proportion of the
population of office workers who are able to take every
telephone call?
b. At 90% confidence, what is the margin of error?
c. What is the 90% confidence interval for the proportion of
the population of office workers who are able to
take every telephone call?
MATH 214 (NOTES) – p. 108/11
Problem 33 (page 323)
In a survey, the planning value for the population proportion
is p∗ = 0.35. How large a sample should be taken to provide
a 95% confidence interval with a margin of error of 0.05?
MATH 214 (NOTES) – p. 109/11
Problem 34 (page 323)
At 95% confidence, how large a sample should be taken to
obtain a margin of error of 0.03 for the estimation of a
population proportion? Assume that past data are not
available for developing a planning value for p∗ .
MATH 214 (NOTES) – p. 110/11
TO BE CONTINUED...
MATH 214 (NOTES) – p. 111/11