MATH 214 (NOTES)math.unm.edu/~alvaro/math214-p2.pdf · MATH 214 (NOTES) Math 214 Al Nosedal Department of Mathematics Indiana University of Pennsylvania ... A person must score in

MATH 214 (NOTES)Math 214

Al Nosedal

Department of Mathematics

Indiana University of Pennsylvania

MATH 214 (NOTES) – p. 1/111

CHAPTER 6CONTINUOUS PROBABILITY DISTRIBUTIONS

MATH 214 (NOTES) – p. 2/111

Simple example

Random Experiment: Rolling a fair die 300 times.

Class Expected Freq. Expected Relative Freq.

1 ≤ x < 2 50 1/62 ≤ x < 3 50 1/63 ≤ x < 4 50 1/64 ≤ x < 5 50 1/65 ≤ x < 6 50 1/66 ≤ x < 7 50 1/6

MATH 214 (NOTES) – p. 3/111

Histogram

1 2 3 4 5 6 7

3040

5060

70

freq

Histogram of frequencies

MATH 214 (NOTES) – p. 4/111

Histogram (Relative Freqs.)

1 2 3 4 5 6 7

0.10

0.14

0.18

0.22

freq

Histogram of Relative Frequencies

MATH 214 (NOTES) – p. 5/111

Uniform Probability Distribution

Uniform Probability Density Function

f(x) =1

b − afor a ≤ x ≤ b

f(x) = 0 elsewhere

MATH 214 (NOTES) – p. 6/111

Expected Value and Variance

E(X) =a + b

2

V ar(X) =(b − a)2

12

MATH 214 (NOTES) – p. 7/111

Problem 2 (page 230)

The random variable x is known to be uniformly distributedbetween 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. 8/111

Solution

10 12 14 16 18 20

0.06

0.08

0.10

0.12

0.14

f(x)

Graph of ProbabilityDensity Function

MATH 214 (NOTES) – p. 9/111

Solution

b. P (x < 15) = (0.10)(5) = 0.50

c. P (12 ≤ x ≤ 18) = (0.10)(6) = 0.60

d. E(x) = 10+202 = 15

e. V (x) = (20−10)2

12 = 8.33

MATH 214 (NOTES) – p. 10/111


Delta Airlines quotes a flight time of 2 hours, 5 minutes forits flights from Cincinnati to Tampa. Suppose we believe thatactual flight times are uniformly distributed between 2 hoursand 2 hours, 20 minutes.a. Show the graph of the probability density function for flighttime.b. What is the probability that the flight will be no more than5 minutes late?c. What is the probability that the flight will be more than 10minutes late?d. What is the expected flight time?

MATH 214 (NOTES) – p. 11/111

Solution (problem 3 a)

120 125 130 135 140

0.03

0.04

0.05

0.06

0.07

minutes

f(x)

Graph of ProbabilityDensity Function

MATH 214 (NOTES) – p. 12/111

Solution (problem 3 b)

b. P (x ≤ 130) = 1020 = 0.5

120 125 130 135 140

0.03

0.04

0.05

0.06

0.07

minutes

f(x)

MATH 214 (NOTES) – p. 13/111

Solution (problem 3 c)

c. P (x > 135) = 520 = 0.25

120 125 130 135 140

0.03

0.04

0.05

0.06

0.07

minutes

f(x)

MATH 214 (NOTES) – p. 14/111

Solutions (without graphs)

b. P (x ≤ 130) = 1020 = 0.5

c. P (x > 135) = 520 = 0.25

d. E(x) = 120+1402 = 130 minutes

MATH 214 (NOTES) – p. 15/111

Standard Normal Probability Distribution

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

Normal Distributionmean=0 and std.dev.=1

MATH 214 (NOTES) – p. 16/111

Standard Normal Probability Distribution

f(z) =1√2π

e−z2

2

Note. Don’t worry about this formula, we are NOT going touse it to do calculations.

MATH 214 (NOTES) – p. 17/111

Normal Probability Density Function

f(x) =1

σ√

2πe

−(x−µ)2

2σ2

where µ = mean, σ = standard deviation, e = 2.71828 and π

= 3.14159.Note. Don’t worry about this formula, we are NOT going touse it to do calculations.

MATH 214 (NOTES) – p. 18/111

Two Normal Distributions

−15 −10 −5 0 5 10 15

0.00

0.05

0.10

0.15

0.20

MATH 214 (NOTES) – p. 19/111

Two Different Standard Deviations

−15 −10 −5 0 5 10 15

0.00

0.05

0.10

0.15

0.20

std.dev.=5std.dev.=2

MATH 214 (NOTES) – p. 20/111


Given that z is a standard normal random variable, computethe 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. 21/111


Given that z is a standard normal random variable, computethe 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. 22/111

Solution (problem 12 a)

a. P (0 ≤ z ≤ 0.83) = 0.2967

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.7967−0.5 =0.2967

MATH 214 (NOTES) – p. 23/111

Solution (problem 12 b)

b. P (−1.57 ≤ z ≤ 0) = 0.4418

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.5−0.0582 = 0.4418

MATH 214 (NOTES) – p. 24/111

Solution (problem 12 c)

c. P (z ≥ 0.44) = 0.3300

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.6700

1−0.6700 =0.3300

MATH 214 (NOTES) – p. 25/111

Solution (problem 12 d)

d. P (z ≥ −0.23) = 0.5910

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.4090

1−0.4090 =0.5910

MATH 214 (NOTES) – p. 26/111

Solution (problem 12 e)

e. P (z ≤ 1.2) = 0.8849

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.8849

MATH 214 (NOTES) – p. 27/111

Solution (problem 12 f)

f. P (z ≤ −0.71) = 0.2389

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

0.2389

MATH 214 (NOTES) – p. 28/111


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. 29/111

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. 30/111

Problem 14 a

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.9750

z*=1.96

MATH 214 (NOTES) – p. 31/111

Problem 14 b

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.47500.5

z*=1.96

MATH 214 (NOTES) – p. 32/111

Problem 14 c

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.7291

z*=0.61

MATH 214 (NOTES) – p. 33/111

Problem 14 d

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.8686

z*=1.12

0.1314

MATH 214 (NOTES) – p. 34/111

Problem 14 e

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.67

z*=0.44

MATH 214 (NOTES) – p. 35/111

Problem 14 f

−3 −2 −1 0 1 2 3

−0.

10.

00.

10.

20.

30.

4

0.67 0.33

z*=0.44

MATH 214 (NOTES) – p. 36/111


Given that z is a standard normal random variable, find z foreach 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. 37/111


a. z∗ = −0.80.b. z∗ = 1.66.c z∗ = 0.26.d. z∗ = 2.56.e. z∗ = −0.50.

MATH 214 (NOTES) – p. 38/111


For borrowers with good credit scores, the mean debt forrevolving and installment accounts is $ 15,015. Assume thestandard deviation is $ 3540 and that debt amounts arenormally distributed.a. What is the probability that the debt for a randomlyselected borrower with good credit is more than $ 18,000?b. What is the probability that the debt for a randomlyselected borrower with good credit is less than $ 10,000?c. What is the probability that the debt for a randomlyselected borrower with good credit is between $ 12,000 and$ 18,000?d. What is the probability that the debt for arandomly selected borrower with good creditis no more than $ 14,000?

MATH 214 (NOTES) – p. 39/111


Let X = debt amount.a. P (X > 18000) = P (X−µ

σ > 18000−150153540 ) = P (Z > 0.8432) =

1 − P (Z < 0.8432) = 1 − 0.7995 = 0.2005

b. P (X < 10000) = P (X−µσ < 10000−15015

3540 ) = P (Z < −1.4166) w

0.0773

c. P (12000 < X < 18000) =

P (12000−150153540 < X−µ

σ < 18000−150153540 ) = P (−0.8516 < Z < 0.8432) =

P (Z < 0.8432) − P (Z < −0.8516) w 0.7995 − 0.1977 = 0.6018

d. P (X ≤ 14000) = P (X−µσ ≤ 14000−15015

3540 ) = P (Z < −0.2867) w

0.3897

MATH 214 (NOTES) – p. 40/111


A person must score in the upper 2% of the population onan IQ test to qualify for membership in MENSA, theinternational high-IQ society. If IQ scores are normallydistributed with a mean of 100 and a standard deviation of15, what score must a person have to qualify for MENSA?.

MATH 214 (NOTES) – p. 41/111

Problem 21 (solution)

1. State the problem. Let X = IQ score of a randomlyselected person. We want to find the IQ score x∗ with area0.02 to its right under the Normal curve with mean µ = 100

and standard deviation σ = 15. That’s the same as findingthe IQ score x∗ with area 0.98 to its left. Because our tablegives the areas to the left of z-values, always state theproblem in terms of the area to the left of x∗.

MATH 214 (NOTES) – p. 42/111

Solution (cont.)

2. Use the table. Look in the body of our table for the entryclosest to 0.98. It is 0.9798. This is the entry correspondingto z∗ = 2.05. So z∗ = 2.05 is the standardized value with area0.98 to its left.3. Unstandardize to transform the solution from the Z backto the original X scale. We know that the standardized valueof the unknown x∗ is z∗ = 2.05. So x∗ itself satisfies:x∗−100

15 = 2.05. Solving this equation for x∗ gives:x∗ = 100 + (2.05)(15) = 130.75. We see that a person mustscore at least 130.75 (131 perhaps) to place in the highest2%.

MATH 214 (NOTES) – p. 43/111


According to the Sleep Foundation, the average night’ssleep is 6.8 hours. Assume the standard deviation is 0.6hours and that the probability distribution is normal.a. What is the probability that a randomly selected personsleeps more than 8 hours?b. What is the probability that a randomly selected personsleeps 6 hours or less?c. Doctors suggest getting between 7 and 9 hours of sleepeach night. What percentage of the population gets thismuch sleep?

MATH 214 (NOTES) – p. 44/111

Problem 25 (solution)

Let X = number of hours a randomly selected person sleepsper night.a. P (X > 8) = P (X−µ

σ > 8−6.80.6 ) = P (Z > 2) = 1− 0.9772 = 0.0228

b.P (X ≤ 6) = P (X−µσ ≤ 6−6.8

0.6 ) = P (Z ≤ −1.33) = 0.0918

c.P (7 ≤ X ≤ 9) = P (7−6.80.6 ≤ X−µ

σ ≤ 9−6.80.6 )

= P (0.33 ≤ Z < 3.66) ' 1 − 0.6293 = 0.3707

MATH 214 (NOTES) – p. 45/111

Standard Normal Distribution (again)

The Standard Normal Distribution is the Normal DistributionN(0, 1) with mean 0 and standard deviation 1.If a variable X has any Normal Distribution N(µ, σ) with meanµ and standard deviation σ, then the standardized variable

Z =X − µ

σ

has the Standard Normal Distribution.

MATH 214 (NOTES) – p. 46/111

Finding Normal Probabilities

1. State the problem in terms of the observed variable X.2. Standardize X to restate the problem in terms of aStandard Normal variable Z. Draw a picture to show thearea 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 areaunder the curve is 1.

MATH 214 (NOTES) – p. 47/111

Exponential Probability Distribution

EXPONENTIAL PROBABILITY DENSITY FUNCTION

f(x) =1

µe

−x

µ

for x ≥ 0 and µ ≥ 0

(where µ = expected value or mean)

MATH 214 (NOTES) – p. 48/111

Exponential Distribution (cont.)

CUMULATIVE PROBABILITIES.

P (X ≤ k) = 1 − e−k/µ

MATH 214 (NOTES) – p. 49/111


Consider the following exponential probability densityfunction.

f(x) =1

8e−x/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. 50/111

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. 51/111


The time required to pass through security screening at theairport can be annoying to travelers. The mean wait timeduring peak periods at Cincinnati/Northern KentuckyInternational Airport is 12.1 minutes. Assume the time topass through security screening follows an exponentialdistribution.a. What is the probability it will take less than 10 minutes topass through security screening during a peak period?

MATH 214 (NOTES) – p. 52/111

Problem 34 (cont.)

b. What is the probability it will take more than 20 minutes topass through security screening during a peak period?c. What is the probability it will take between 10 and 20minutes to pass through security screening during a peakperiod?d. It is 8:00 a.m. (a peak period) and you just entered thesecurity line. To catch your plane you must be at the gatewithin 30 minutes. If it takes 12 minutes from the time youclear security until you reach your gate, what is theprobability you will miss your flight?

MATH 214 (NOTES) – p. 53/111

Solution

Let T be the time required to pass through securityscreening. 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. 54/111

CHAPTER 7SAMPLING AND SAMPLING DISTRIBUTIONS

MATH 214 (NOTES) – p. 55/111

Toy example

Consider a population of 5 families with annual incomesshown below. We want to choose a simple random sampleof size 2 from this population. How can this be done? Andhow do the sample mean of the chosen families compare topopulation mean?Incomes: 30,000 35,000 40,000 45,000 50,000.

MATH 214 (NOTES) – p. 56/111

Solution

x̄1 = 30,000+35,0002 = 32, 500 x̄2 = 30,000+40,000

2 = 35, 000

x̄3 = 30,000+45,0002 = 37, 500 x̄4 = 30,000+50,000

2 = 40, 000

x̄5 = 35,000+40,0002 = 37, 500 x̄6 = 35,000+45,000

2 = 40, 000

x̄7 = 35,000+50,0002 = 42, 500 x̄8 = 40,000+45,000

2 = 42, 500

x̄9 = 40,000+50,0002 = 45, 000 x̄10 = 45,000+50,000

2 = 47, 500

µ = 40, 000 (population mean).

MATH 214 (NOTES) – p. 57/111

Graph

0.0

0.5

1.0

1.5

2.0

income

freq

uenc

y

32500 37500 42500 47500

MATH 214 (NOTES) – p. 58/111

Population, Sample

The entire group of individuals that we want informationabout is called the population.

A sample is a part of the population that we actuallyexamine in order to gather information.

MATH 214 (NOTES) – p. 59/111

Simple Random Sample

A simple random sample (SRS) of size n consists of nindividuals from the population chosen in such a way thatevery set of n individuals has an equal chance to be thesample actually selected.

MATH 214 (NOTES) – p. 60/111

Example

A couple plans to have three children. There are 8 possiblearrangements of girls and boys. For example, GGB meansthe first two children are girls and the third child is a boy. All8 arrangements are (approximately) equally likely.

a) Write down all 8 arrangements of the sexes of threechildren. What is the probability of any one of thesearrangements?

MATH 214 (NOTES) – p. 61/111

Example (cont.)

b) Let X be the number of girls the couple has. What isthe 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 theprobabilities for each value?

MATH 214 (NOTES) – p. 62/111

Binomial Probability Function

f(x) =n!

x!(n − x)!px(1 − p)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. 63/111

More about Binomial Distributions

E(X) = µ = np

V ar(X) = σ2 = np(1 − p)

MATH 214 (NOTES) – p. 64/111

Problem

We are interested in estimating the average number of carsper household in a little town call Statstown. Let X representthe number of cars in a house picked at random. God knowsthat X has a Binomial distribution with n = 4 and p = 0.5.Suppose that we can only afford a sample of size 4 and thatwe are going to use this sample to estimate that populationaverage.

MATH 214 (NOTES) – p. 65/111

Problem (cont.)

What we are going to do next is called a simulation. First,we will draw a lot of random samples coming from aBinomial Distribution with n = 4 and p = 0.5. Then we willmake a histogram for all the x̄’s corresponding to oursamples. We are going to do this to see what the histogramof x̄ looks like. This will give us an idea of what to expect in asimilar situation.

MATH 214 (NOTES) – p. 66/111

Sampling Distribution of x̄

Histogram of means

means

Den

sity

0 1 2 3 4

0.0

0.2

0.4

0.6

0.8

1.0

MATH 214 (NOTES) – p. 67/111

Central Limit Theorem

Draw a random sample of size n from any population withmean µ and finite standard deviation σ. When n is large, thesampling distribution of the sample mean x̄ is approximatelyNormal:

x̄ is approximately N(µ,σ√n

)(1)

MATH 214 (NOTES) – p. 68/111

Example

The number of accidents per week at a hazardousintersection varies with mean 2.2 and standard deviation1.4. This distribution takes only whole-number values, so itis certainly not Normal.

a) Let x̄ be the mean number of accidents per week atthe intersection during a year (52 weeks). What is theapproximate distribution of x̄ according to the centrallimit theorem?

MATH 214 (NOTES) – p. 69/111

Example (cont.)

b) What is the approximate probability that x̄ is less than2?

c) What is the approximate probability that there arefewer than 100 accidents at the intersection in a year?(Hint: Restate this event in terms of x̄)

MATH 214 (NOTES) – p. 70/111

Solution

a) By the Central Limit Theorem, x̄ is roughly Normal withmean µ = 2.2 and standard deviation σ√

n= 1.4√

52= 0.1941.

b) P (x̄ < 2) = P ( x̄−µσ√n

< 2−2.20.1941) = P (Z < −1.0303) = 0.1515.

c) Let xi be the number of accidents during week i.

P (Total < 100) = P (∑52

i=1 xi < 100) = P (∑52

i=1 xi

52 < 10052 ) =

P (x̄ < 1.9230) = P (Z < −1.4270) = 0.0768

MATH 214 (NOTES) – p. 71/111


A population has a mean of 200 and a standard deviation of50. A simple random sample of size 100 will be taken andthe sample mean x̄ will be used to estimate the populationmean.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. 72/111

Solution

a. E(x̄) = µ = 200.b. σx̄ = σ√

n= 50√

100= 5

c. By the Central Limit Theorem, x̄ is roughly Normal withmean µ = 200 and standard deviation σ√

n= 5.

MATH 214 (NOTES) – p. 73/111


A population has a mean of 200 and a standard deviation of50. Suppose a simple random sample of size 100 isselected 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. 74/111

Solution

By the Central Limit Theorem, x̄ is roughly Normal withmean µ = 200 and standard deviation σ√

n= 5.

a. P (195 ≤ x̄ ≤ 205) = P (195−2005 ≤ x̄−µ

σ√n

≤ 205−2005 ) =

P (−1 ≤ Z ≤ 1) = P (Z ≤ 1) − P (Z ≤ −1) = 0.8413 − 0.1587 =

0.6826

b. P (190 ≤ x̄ ≤ 210) = P (190−2005 ≤ x̄−µ

σ√n

≤ 210−2005 ) =

P (−2 ≤ Z ≤ 2) = P (Z ≤ 2) − P (Z ≤ −2) = 0.9772 − 0.0228 =

0.9544

MATH 214 (NOTES) – p. 75/111

Problem 24

The mean tuition cost at state universities throughout theUnited States is $ 4260 per year . Use this value as thepopulation mean and assume that the population standarddeviation is σ = $900. Suppose that a random sample of 50state universities will be selected.a. Show the sampling distribution of x̄ where x̄ is the samplemean tuition cost for the 50 state universities.b. What is the probability that the simple random sample willprovide a sample mean within $250 of the population mean?c. What is the probability that the simple random sample willprovide a sample mean within $100 of the population mean?

MATH 214 (NOTES) – p. 76/111

Solution

a. By the Central Limit Theorem, x̄ has roughly a N(µ, σ√n).

In this case, x̄ has roughly a N(4260, 900√50

), that is, x̄ has

roughly a N(4260, 127.28)

b. P (4010 ≤ x̄ ≤ 4510) = P (4010−4260127.28 ≤ Z ≤ 4510−4260

127.28 ) =

P (−1.96 ≤ Z ≤ 1.96) = 0.9750 − 0.025 = 0.95

c. P (4160 ≤ x̄ ≤ 4360) = P (4160−4260127.28 ≤ Z ≤ 4360−4260

127.28 ) =

P (−0.79 ≤ Z ≤ 0.79) = 0.7852 − 0.2148 = 0.5704

MATH 214 (NOTES) – p. 77/111

Problem 25

The College Board College Testing Program reported apopulation mean SAT score of µ = 1020. Assume that thepopulation standard deviation is σ = 100.a. What is the probability that a random sample of 75students will provide a sample mean SAT score within 10 ofthe population mean?b. What is the probability a random sample of 75 studentswill provide a sample mean SAT score within 20 of thepopulation mean?

MATH 214 (NOTES) – p. 78/111

CHAPTER 8INTERVAL ESTIMATION

MATH 214 (NOTES) – p. 79/111

M & M’s

If you buy an 8 pack of m & m’s fun size you will find thefollowing information:Nutrition FactsServing Size 3 packs (45 g)Which implies that: 1 pack = 15 g, right?. Since we arebuying an 8 pack the net weight should be (8)(15) = 120 g,right?. However, the net weight on the label is 117.1 g, Howcan we explain that?.

MATH 214 (NOTES) – p. 80/111

Interval Estimate of a Population Mean:σ kno

x̄ ± z∗(σ√n

)

where z∗ is a number coming from a Standard Normal thatdepends on the confidence level required.

MATH 214 (NOTES) – p. 81/111

Problem 1

A simple random sample of 40 items resulted in a samplemean 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. 82/111

Solution

a. standard error of the mean = σx̄ = σ√n

= 5√40

= 0.7905

b. margin of error = z∗σ√n

= (1.96)(0.7905) = 1.5493

MATH 214 (NOTES) – p. 83/111


A simple random sample of 50 items from a population withσ= 6 resulted in a sample mean of 32.a. Provide a 90% confidence interval for the populationmean?b. Provide a 95% confidence interval for the populationmean?c. Provide a 99% confidence interval for the populationmean?

MATH 214 (NOTES) – p. 84/111

Solution

a. x̄ = 32, σ = 6, n = 50 and z∗ = 1.65

(x̄ − z∗σ√n, x̄ − z∗

σ√n)

(32 − 1.65( 6√50

), 32 − 1.65( 6√50

))

(30.5999, 33.4000) 90% Confidence Interval for µ.b. (32 − 1.96( 6√

50), 32 − 1.96( 6√

50))

(30.3368, 33.6631) 95% Confidence Interval for µ.

MATH 214 (NOTES) – p. 85/111


In an effort to estimate the mean amount spent percustomer for dinner at a major Atlanta restaurant, data werecollected for a sample of 49 customers. Assume apopulation 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. 86/111

Solution

In this case, µ = mean amount spent per customer for dinnerfor all customers at a major Atlanta restaurant.a. margin of error = z∗(

σ√n) = 1.96( 5√

49) = 1.4

b. (x̄ − z∗σ√n, x̄ − z∗

σ√n)

(24.80 − 1.4, 24.80 + 1.4))

(23.40, 26.20) 95 % Confidence Interval for µ.

MATH 214 (NOTES) – p. 87/111


Playbill magazine reported that the mean annual householdincome of its readers is $ 119,155. Assume this estimate ofthe mean annual household income is based on a sample of80 households, and based on past studies, the populationstandard deviation is known to be σ = $ 30,000.a. Develop a 90 % confidence interval estimate of thepopulation mean.b. Develop a 95 % confidence interval estimate of thepopulation mean.c. Develop a 99 % confidence interval estimate of thepopulation mean.

MATH 214 (NOTES) – p. 88/111

Solution

a. (x̄ − z∗σ√n, x̄ − z∗

σ√n)

(119, 155 − 1.65(30,000√80

), 119, 155 − 1.65(30,000√80

))

(113, 620.73; 124, 689.26) 90% Confidence Interval for µ.b. (119, 155 − 1.96(30,000√

80), 119, 155 − 1.96(30,000√

80))

(112, 580.96; 125, 729.03) 95% Confidence Interval for µ.c. (119, 155 − 2.58(30,000√

80), 119, 155 − 2.58(30,000√

80))

(110501.41; 127808.58) 99% Confidence Interval for µ.

MATH 214 (NOTES) – p. 89/111


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. 90/111

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 tdistributions 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. 91/111


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 valueswith 25 degrees of freedom?e. Where 95 % of the area falls between these two t valueswith 45 degrees of freedom?

MATH 214 (NOTES) – p. 92/111

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. 93/111


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 standarddeviation?c. With 95 % confidence, what is the margin of error for theestimation of the population mean?d. What is the 95 % confidence interval for the populationmean?

MATH 214 (NOTES) – p. 94/111

Solution

a. x̄ = 10.b. s = 3.4641.c. margin of error = t∗

s√n

= 2.365(3.4641√8

) = 2.8965.

d. (x̄ − t∗(s√n), x̄ + t∗(

s√n))

(7.1039, 12.896) (using a TI-84).

MATH 214 (NOTES) – p. 95/111


A simple random sample with n = 54 provided a samplemean of 22.5 and a sample standard deviation of 4.4.a. Develop a 90% confidence interval for the populationmean.b. Develop a 95% confidence interval for the populationmean.c. Develop a 99% confidence interval for the populationmean.

MATH 214 (NOTES) – p. 96/111

Solution

a. (x̄ − t∗(s√n), x̄ + t∗(

s√n))

(22.5 − 1.674( 4.4√54

), 22.5 + 1.674( 4.4√54

))

(21.498, 23.502). (using a TI-84)b. (22.5 − 2.006( 4.4√

54), 22.5 + 2.006( 4.4√

54))

(21.299, 23.701) (using a TI-84)c. (22.5 − 2.672( 4.4√

54), 22.5 + 2.672( 4.4√

54))

(20.9, 24.1) (using a TI-84)

MATH 214 (NOTES) – p. 97/111


Sales personnel for Skillings Distributors submit weeklyreports listing the customer contacts made during the week.A sample of 65 weekly reports showed a sample mean of19.5 customer contacts per week. The sample standarddeviation was 5.2. Provide a 90% and 95% confidenceintervals for the population mean number of weeklycustomer contacts for the sales personnel.

MATH 214 (NOTES) – p. 98/111

Solution

90 % Confidence.(x̄ − t∗(

s√n), x̄ + t∗(

s√n))

(19.5 − 1.669( 5.2√65

), 19.5 + 1.669( 5.2√65

))

(18.42, 20.58)

95 % Confidence.(x̄ − t∗(

s√n), x̄ + t∗(

s√n))

(19.5 − 1.998( 5.2√65

), 19.5 + 1.998( 5.2√65

))

(18.21, 20.79)

MATH 214 (NOTES) – p. 99/111


How large a sample should be selected to provide a 95%confidence interval with a margin of error of 10? Assumethat the population standard deviation is 40.

MATH 214 (NOTES) – p. 100/111


The range for a set of data is estimated to be 36.a. What is the planning value for the population standarddeviation?b. At 95% confidence, how large a sample would provide amargin of error of 3?c. At 95% confidence, how large a sample would provide amargin of error of 2?

MATH 214 (NOTES) – p. 101/111


Scheer Industries is considering a new computer-assistedprogram to train maintenance employees to do machinerepairs. In order to fully evaluate the program, the director ofmanufacturing requested an estimate of the populationmean time required for maintenance employees to completethe computer-assisted training. Use 6.84 days as a planningvalue for the population standard deviation.a. Assuming 95% confidence, what sample size would berequired 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 amargin of error of 2 days?

MATH 214 (NOTES) – p. 102/111


Smith Travel Research provides information on the one-nightcost of hotel rooms through-out the United States. Use $ 2as the desired margin of error and $ 22.50 as the planningvalue for the population standard deviation to find thesample size recommended in a), b), and c).a. A 90% confidence interval estimate of the populationmean cost of hotel rooms.b. A 95% confidence interval estimate of the populationmean cost of hotel rooms.c. A 99% confidence interval estimate of the populationmean cost of hotel rooms.

MATH 214 (NOTES) – p. 103/111

Problem

A market research consultant hired by the Pepsi-Cola Co. isinterested in determining the proportion of IUP students whofavor Pepsi-Cola over Coke Classic. A random sample of100 students shows that 40 students favor Pepsi over Coke.Use this information to construct a 95% confidence intervalfor the proportion of all students in this market who preferPepsi.

MATH 214 (NOTES) – p. 104/111

Bernoulli Distribution

xi =

{

1 i-th person prefers Pepsi0 i-th person prefers Coke

µ = E(xi) = p

σ2 = V (xi) = p(1 − p)

Let p̂ be our estimate of p. Note that p̂ =∑n

i=1 xi

n = x̄. If n is"large", by the Central Limit Theorem, we know that:x̄ is roughly N(µ, σ√

n), that is,

p̂ is roughly N

(

p,

√

p(1−p)n

)

MATH 214 (NOTES) – p. 105/111


A simple random sample of 400 individuals provides 100Yes responses.a. What is the point estimate of the proportion of thepopulation that would provide Yes responses?b. What is the point estimate of the standard error of theproportion, σp̂?c. Compute the 95% confidence interval for the populationproportion.

MATH 214 (NOTES) – p. 106/111


A simple random sample of 800 elements generates asample proportion p̂ = 0.70.a. Provide a 90% confidence interval for the populationproportion.b. Provide a 95% confidence interval for the populationproportion.

MATH 214 (NOTES) – p. 107/111


A survey of 611 office workers investigated telephoneanswering practices, including how often each office workerwas able to answer incoming telephone calls and how oftenincoming telephone calls went directly to voice mail. A totalof 281 office workers indicated that they never need voicemail and are able to take every telephone call.a. What is the point estimate of the proportion of thepopulation of office workers who are able to take everytelephone call?b. At 90% confidence, what is the margin of error?c. What is the 90% confidence interval for the proportion ofthe population of office workers who are able totake every telephone call?

MATH 214 (NOTES) – p. 108/111


In a survey, the planning value for the population proportionis p∗ = 0.35. How large a sample should be taken to providea 95% confidence interval with a margin of error of 0.05?

MATH 214 (NOTES) – p. 109/111


At 95% confidence, how large a sample should be taken toobtain a margin of error of 0.03 for the estimation of apopulation proportion? Assume that past data are notavailable for developing a planning value for p∗.

MATH 214 (NOTES) – p. 110/111

TO BE CONTINUED...

MATH 214 (NOTES) – p. 111/111

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