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STUDY GUIDE FOR MAT 121 EXAM 3

Normal Distributions

Density Curve

A probability distribution is defined by a curve if

● The area under the whole curve is equal to 1

● The curve is above the horizontal scale

Area and probability If you want to learn more check out What are the strong acids?

Assume that X has a probability distribution defined by a curve

Uniform Distribution

A random variable x has a uniform distribution on an interval from a to b if all its values are equally likely (spread evenly). Its curve is a line segment from a to b at height

1__

b-a

Don't forget about the age old question of What are the phyton variable name rules?

Example

Suppose the temperature T of a piece of metal is always between 10 F and 20 F, and any value in between is equally likely. Find (P 12 <T < 17)

A: T has a uniform distribution. Its height = 1/ 20-10 = 1/10 = 0.10

Normal Distribution

A Normal Distribution has a bell-shaped symmetrical curve with a single central peak at he mean μ

Standard Normal Distribution

The Standard Normal Distribution is a normal distribution with mean μ = 0 and a standard σ = 1

deviation of and the horizontal scale uses z-scores instead of x-scores. We also discuss several other topics like How do we have social control?

Two Problems

● Finding the probability when a range of z-scores is given

● Finding z-scores when probabilities are given

Chart A2 pages 724 - 725

Table entry for z is the area under the standard normal curve to the left of z. Strategy Don't forget about the age old question of Poverty is caused by what?

Draw always the standard normal distribution curve, and shade the area that corresponds to the probability you want.

Table A2, Pages 724 - 725 gives the area to the left of a z-score

● To find it,

-Locate the ones and tenths digits of such z-score

-Locate its hundredths digits

-the left area is in the intersecting cell of the row at the ones and tenths digits and the column at its hundredths digits Don't forget about the age old question of Who is the father of containment?

● The area to the right = 1 - The area to the left

Example 1:

Find the probability P( z < 2.37)

We need distribution of z-scores If you want to learn more check out What was the name of the publication that darwin published?

Example 2:

Find the probability P (z > 1.82)

The area between a and b

To find the area between two positive z scores, find the area to the left of each one of them and then subtract the smaller area from the larger area

Example 3:

Find the probability P (0.21 < z < 0.75)

P(0.21 < z < 0.75) = left area to 0.75 - left area to 0.21

= 0.7734 - 0.5832

= 0.1902

By table A2, the left area

● To 0.21 is 0.5832

● To 0.75 is 0.7734

By calculator

● normal cdf (0.21, 0.75) = 0.1902

TI83/84: Finding probabilities

● Press 2nd and then Vars for DISTR menu

● Select normal cdf and then press enter

● Enter lower and upper bounds of z as follows

Normal cdf (lower bound, upper bound)

Press enter you get the desired probability

● Use -10 as lower bound if the area is unlimited from the left ● Use -10 as upper bound if the area is unlimited from the right

99.98% of values lie between

μ − 3σ

&

μ + 3σ

= 0 - 3 x 1 = -3 = 0 + 3 x 1 = 3

Finding z-scores when probabilities are given

If the area to the left is given , locate the closest value of it in the body of Table A-2 and identify the corresponding z-score

Area smaller than 50% look at the first page

Area bigger than 50% look at the second page

Keep in mind its left area, if its right area, you should convert it to left area!!!

Example 4

Find the z for which the left area is 0.9798

Example 5

Find the z for which the left area is 0.2546

Notation

Z is the z-score whose area to the right is . Here, are some useful values: α α

● Z0.1 = 1.28.

● Z0.05 = 1.64.

● Z0.025= 1.96.

● Z0.01 = 2.33

Caution

● Don’t confuse z-scores and areas

● Choose the correct (right/left) side of the graph.

● Areas (or probabilities) are positive or zero values, but they are never negative. Normal distributions

When working with a normal distribution that is not standard, we can use a simple conversion that allows us to standardize it so that the previous methods can be used Formula 6-2: z −score = x −µ

σ

(Round z-scores to 2 decimal places).

Procedure for Finding Probabilities

● Sketch a normal distribution curve, label the mean and the specific value of x, and then shade the region that represents the probability you want.

● Convert the specific value of x to a z-score using Formula 6-2.

● Solve the problem as in the Standard normal case

Example

μ = 10

Assume that x is normally distributed with a mean and a standard deviation

σ = 4.

Find P (x < 17)

A: x has a normal distribution

μ = 10 and σ = 4

P(X<17) = left area to z = 1.75 → 1.7 0.05 = 0.9599

Procedure for Finding values

● Sketch a normal distribution curve, label the mean and shade the region that represents the given probability.

● Find the z-score that corresponds to the shaded area.

● Convert it to x-value using the formula:

x = µ + z . σ.

Example

Suppose that men’s heights are normally distributed with mean 69 "in" and standard deviation of 2.8 "in", and that the Boeing 757-200 airliner has doors with a height of 72 "in".

● What proportion of men can fit through the doors without bending? In other words, find

P(x < 72),

where x stands for the height.

● What height would allow 98% of men to fit without bending?

TI83/84: Finding probabilities

● Press 2nd and then VARS for DISTR menu.

● Select normalcdf and then press enter.

● Enter lower and upper bounds of z as follows

normalcdf(lower bound, upper bound).

Press enter you get the desired probability.

● Use -10 as lower bound if the area is unlimited from the left. ● Use 10 as upper bound if the area is unlimited from the right.

TI83/84: Finding z-scores

● Press 2nd and then VARS for DISTR menu.

● Select invNorm and then press enter.

● Enter the given area to the left of the desired z-score as follows

invNorm(the given area to the left).

Press enter you get the desired z-score.

Sample Distributions and The Central Limit Theorem

We introduce the concept of a sampling distribution of a statistic to estimate a population parameter, and discuss the Central Limit Theorem.

The sampling distribution of the mean

If you take all samples fo the same size n from a population then their sample means vary

Their distribution is called the sampling distribution of the mean, and tells us how to spread out these samples means.

Example

The following table gives the final grades of a class of four students

Allan (A)

100

James (J)

90

Barbara (B)

70

Nicolas (N)

80

● Select a random student and find the probability that its final grade is more than 81 ● Find the average grade

● Select randomly two students, and find the probability that the average of their grades is more than 81.

● FInd the mean of all averages of two grades

● Select randomly three students, and find the probability that the average of their grades is more than 81

● Find the mean of all averages of three grades (find the average of all averages)

1., 2., 3., and 4

The mean is µ = 340/4 = 85.

Allan (A)

100

James (J)

90

Barbara (B)

70

Nicolas (N)

80

SUM = 340

P(x > 81) = 2/4= 0.5 where x represents a student's grade

Now, consider the following table of all possible averages of grades of groups of size 2.

Group of Size 2

Average grade (mean x̄)

A & J

95

A & B

85

A & N

90

J & B

80

J & N

85

B & N

75

SUM = 510

The mean of these averages is μ x̄ = 510/6 = 85

P (x̄ > 81) = 4/6 = 0.67

This table represents the sampling distribution of the mean of samples of size 2. 5., and 6

Now, consider the following table of all possible averages of grades of groups of size 3.

Group of size 3

Average grade (mean x̄)

A & J & B

86.67

A & J & N

90

A & B & N

83.33

J & B & N

80

SUM = 340

, μ is a notation - that stands for the mean of the distributions of x̄.

x̄

● The mean of these averages is μ x̄ = 340/ 4 = 85

● P (x̄ > 81) = 3/4 = 0.75

● This table represents the sampling distribution of the mean of samples of size 3. Properties

● The mean of all sample means coincides with the population mean

● The distribution of the sample means tends to be a normal distribution The Central Limit Theorem (CLT)

Assume that x is a random variable from a population with mean "µ" and standard deviation "σ".

If a sample of size n is selected from a such population and either n ≥ 30 or x is normally distributed, then the sampling distribution of the mean (of x̄) approaches a normal distribution for which

mean: μ x̄ = µ

And

standard deviation: σ x̄ = __σ__

√n

When you should use CLT

A CLT word problem will most likely contain the phrase "assume the variable is normally distributed", or "the given sample size n will be large".

You will be given:

● The population mean µ and standard deviation σ.

● A sample size n.

You may be asked: find the probability that the mean of a sample of size n is greater or less than a number. find the probability that the mean of a sample of size n is between two numbers. Be careful: When working with a sample mean x̄, its corresponding z-score is

Example

Assume that the final grades of the Fall 2013-MAT 121 are normally distributed with mean 80 and standard deviation 5.

(a) Find the probability that a student final grade is less than 76.

Info: Grades x have a normal distribution

µ = 80

σ = 5

Want:

P(x < 76) = ?

(b) Find the probability that the mean of the final grades of 7 students is less than 76.

Want: P (x̄ < 76) = ?

We need the distribution of x̄

It is a normal distribution, by CLT, with Mean

μ x̄ = μ = 80

σ

Standard Deviation x̄ = __σ__ = 5 = 1.89

√n √7

Example

Assume that weights of men are normally distributed with a mean of 172lb and standard deviation of 29lb.

● Find the probability that a randomly selected men has a weight less than 180lb. Info: Weights x have a normal distribution

μ = 172

σ = 29

Want: P (x< 180) = ?

● Find the probability that 20 randomly selected men have a mean weight less than 180lb.

P (x̄ < 180) = ? n = 20

CLT question

Remember that the A2 chart only gives you the LEFT areas, therefore if you look up the right area and need the left one, you need to convert it (by subtracting it from 1). ● If you use the calculator you do not need to convert

Note: knowing when to use the CLT and when it is just a normal problem for the final

Example

The waiting times at a Hospital are normally distributed with mean 10 minutes and standard deviation 5.5 minutes. What is the probability that the average waiting time of 37 randomly selected patients is more than 11.5 minutes at this hospital.

μ = 10

σ = 5.5

n = 37

P (x̄ > 11.5) = ?

X̄ has a normal distribution by CLT

μ x̄ = μ = 10

σ x̄ = __σ__ = 5.5 = 0.90

√n √37

Z of 11.5 = 11.5 - μ x̄ = 1.67

σ x̄

P( x̄ > 11.5) = Right area to z = 1.67

= 1 - left area to z = 1. 67

= 1 - 9525 = 0.0475

Normal CDF

Lower : 1.67

Upper : 10

μ : 0

σ : 1

Get same result 0.0475

Estimating a Population Proportion

Introduction

In this section, we learn:

1. How to construct a confidence interval from a sample to estimate a population proportion

2. How to find the necessary sample size to estimate a population proportion Definition and Notation

● A point estimate is a value that approximates a population parameter. ● If there are x individuals having a characteristic of interest in a sample of size n, then the sample proportion is

p̂ = x

n

It is the best point estimate of the population proportion p

Example 1

If 30 students from your section were asked about test 2, and 18 think that teh test was easy, then x = 18, n= 30 and

p̂ = 18 = 0.6

30

Example 2

If 883 Americans are surveyed and 280 respond that the economy is getting better, then x = 380, n = 883 and

p̂ = 380 = 0.43

883

Example 3

If 38% of a sample of 276 SU- students are members of fitness centers, then n = 276 0.38 = 38% = 0.38

Confidence Interval

● A confidence interval (CI) for a parameter is determined by using sample data and is an interval of numbers within which we expect the true value of the parameter to be in

● A confidence level (CL) is the probability 1−α (expressed as a percentage, typically 90%, or 95% or 99%) that the confidence interval contains the parameter. ● Keep in mind that α = 1−The confidence level

Requirements to construct a CI based on a sample

1 The sample is a simple random sample.

2 The conditions for binomial distribution are satisfied.

3 There are at least 5 successes and at least 5 failures.

Formula

A (1−α) confidence interval of a proportion p, based on a sample, is the interval given by p̂ −E < p < p̂ +E

where

● P̂ is the sample proportion.

● E is the margin error given by

● Here, z α 2 is the z-score for which the area to the right is α / 2

● n is the sample size.

● P̂ = 1−p̂.

Round the confidence interval limits for p to three significant digits

Most common choices for the confidence levels and the corresponding z-scores are listed in the following table.

Confidence Level (CL)

α

Za/2

99%

0.01

Z0.005 = 2.575

98%

0.02

Z0.001 = 2.33

95%

0.05

Z0.025 = 1.96

90%

0.1

Z0.05 = 1.645

Example 3

Randomly selected 30 students from your section were asked about test 2, and 18 of them think that the test was easy. Construct a 95% confidence interval of the true proportion p.

● n = 30

● p̂ = 18/30 = 0.6 and q̂ = 1- p̂ = 1 - 0.6 = 0.4

● For 95% confidence level, za/2 = Z 0.025 = 1.96

● The margin error

● The interval

-p̂ - E = 0.6 - 0.18 = 0.42

-p̂ + E = 0.6 + 0.18 = 0.78

CI : 0.42 < p < 0.78

Or

(0.42 , 0.78)

Example 4

In a sample of 519 SU students it was found that 285 use "Service Bus". Find a 99% confidence interval estimate of the proportion of SU students who use "Service Bus".

● n = 519

● p̂ = 285/518 = 0.55 and q̂ = 1- p̂ = 1 - 0.55 = 0.45

● For 99% confidence level, za/2/ = Z 0.005 = 2.575

● The error is

● The interval:

-p̂ - E = 0.55 - 0.056 = 0.494

-p̂ + E = 0.55 + 0.056 = 0.606

CI: 0.494 < p < 0.606

OR

(0.494, 0.606)

To estimate a proportion p with a known margin error E, you need a sample with size n given as follows:

1- If an estimate of p is known, then

2- If no estimate of p is known, then

Round the value of n UP to the next whole number

Example

We wish to estimate the proportion of all students from a college who own a car with a 95% confidence level and a margin of error E = 0.03.

(a) If you have no prior estimate of the sample proportion, how large should your sample size be?

(b) If the sample proportion is estimated by 0.65, how large should your sample size be?

As CL is 95%, za/2 = 1.96

a) Since p̂ is unknown ,

n= (za/2)2 0.25

E2

= 1.962 x 0.25 = 1067.11

0.032

Round it up to 1068

b) As p̂ = 0.65 and q̂ = 1 - p̂ = 1 - 0.65 = 0.35 are known

= 1.962 x 0.65 x 0.35

0.032

= 971.07

Round it up to 972.

CI:

P̂ - E = 0.65 - 0.03 = 0.62

P̂ + E = 0.65 + 0.03 = 0.68

If (0.62 - LL- , 0.68 -UL-) is CI for p then

p̂ = upper limit + lower limit = 0.68 + 0.62 = 0.65 = p̂ 2 2

p̂ = upper limit + lower limit = 0.68 - 0.62 = 0.03 = E 2 2

Finding p̂ and E from a CI If a CI is given, then

p̂ = upper limit + lower limit

2

and

E = upper limit − lower limit

2

Example

Based on a sample, a CI for a proportion is given by (0.3, 0.6)

Find the sample proportion and the margin error

If (0.03, 0.06) is a CI

For p

1. Find p̂ and E

2. Find n if the CL is 99%

Answer

1. p̂ = 0.3 + 0.6 = 0.65 = 0.45

2

E= 0.6 - 0.3 = 0.65 = 0.15

2

2.

Because CL is 99% → 2.575

= (2.575)2 0.45 x 0.55 = 72.94 round up to 73

152

T-distributions

A t-distribution has a degree of freedom (df), and its curve has a similar shape of the standard normal distribution. It is centered at 0 and symmetric about 0, and the total area under it is 1.

t-scores

To find a t-score, you must know the degrees of freedom df and ● the area to the right, if you are using Table A3.

The row at df and the column at the right area meet at the desired t-score. ● t is the t-score for which the right area is α. α

Example

Find the t-scores

t0.025 if df= 29

t0.005 if df = 36

Estimating a Population Mean (New PP Presentation)

Introduction

In this section:

● We use a sample mean x to construct a confidence interval (CI) to estimate a population mean µ. A such interval has the form

(x̄ − E , x̄+ E),

where E is a margin error.

● We show how to find the necessary sample size n to estimate a population mean. ● The sample mean x̄ is best point estimate of the population mean µ. ● Keep in mind that α = 1−The confidence level.

T-distributions

A t-distribution has a degree of freedom (df), and its curve has a similar shape of the standard normal distribution. It is centered at 0 and symmetric about 0, and the total area under it is 1.

t-scores

To find a t-score, you must know the degrees of freedom df and

● the area to the right, if you are using Table A3.

The row at df and the column at the right area meet at the desired t-score. ● t is the t-score for which the right area is α. α

Curve of t-distribution

Example

Find the t-scores

1) t0.025 if df= 29

2) t0.005 if df = 36

Confidence Interval

Requirements

● The sample is a simple random sample

● Either the population is normally distributed or the sample size n > 30 Formula

A confidence interval (CI) for the population mean μ based on a sample of size n is x̄ - E < μ < x̄ + E

Where

● x̄ is the sample mean

● E is the margin error that depends on whether σ is known or not

1. If σ is known, then

E = z α /2 σ

√n

Here z /2is the z-score for which the area to the right is α α /2

2. If σ is not know, then

E = t/2 s α

√n

Here, s is teh sample standard deviation and t/2is the t-score for which the area to the right is α α /2 and df= n-1

● Keep in mind that α = 1 - The confidence level

Population SD: σ

Sample SD: S

Most common choices for the confidence levels and the crresponding z-scores are listed in the following table

Confidence Level (CL)

α

z α /2

99%

0.01

Z0.005 = 2.575

98%

0.02

Z0.01 =2.33

95%

0.05

Z0.025 =1.96

90%

0.1

Z0.05 =1.645

Example

A research suggests that the population of weights of all women has a standard deviation σ = 30.86 lbe. If a simple random sample of weights of 40 women has a mean weight 146.22 lb, construct a 95% confidence interval of the mean weight of ALL WOMEN.

● Information:

-σ = 30.86 is given

- n = 40, the sample size (n > 30)

- x̄ = 146.22

- For The confidence level 95%, z α /2 = Z0.025 = 1.96

● CI for μ ? Is σ known? Yes! So, the form of such interval is given by

x̄ - E < μ < x̄ + E

Where

E = z α /2 σ = 1.96 = 30.86 = 9.56 and x̄ = 146.22

√n √40

● x̄ - E = 146.22 - 9.56 = 136.66 and x̄ + E = 146.22 + 9.56 = 155.78 136.66 < μ < 155.78

Example

40 overweight adults from a simple random sample participated in a test of a weight loss program. After 12 months, their mean weight loss was 2.1lb with a standard deviation of 4.8lb. Construct a 99% confidence interval estimate the mean weight loss of all overweight adults who follow a such program.

● Information:

- n = 40, the sample size (n > 30)

- x̄ = 2.1 lb and s = 4.8 lb

- The confidence level 99%, and α = 1 - 99% = 1% = 0.01 and thus, α/2 = 0.005 df = n - 1 = 4- - 1 = 39

● CI for μ ? Is σ known? No! So, the form of such interval is given by

x̄ - E < μ < x̄ + E

Where

E = t/2 s = t α 0.005 4.8 = 2.708 4.8 = 2.06 and x̄ = 2.1

√n √40 √40

● x̄ - E = 2.1 - 2.06 = 0.04

Example

Based on a simple random sample of 36 SU-students, the average weekly preparation time for their studies is 20.85 hours with a standard deviation of 13.41 hours. Construct a 98% confidence interval estimate the average weekly preparation time of all SU-students.

● Information:

- n = 36, the sample size (n > 30)

- x̄ = 20.85 hours, and s = 13.41 hours

-The confidence level 98%, and α = 1 - 98% = 2% = 0.02 and thus, α/2 = 0.01 ● CI for μ ? Is σ known? No! So, the form of such interval is given by

x̄ - E < μ < x̄ + E

Where

E = t/2 s = t α 0.01 13.41 = 2.438 13.41 = 5.45 and x̄ = 20.85 √n √36 √36

● x̄ - E = 20.85 - 25.45 = 15.4 and x̄ + E = 20.85 + 5.45 = 26.3. Finally,

15.4 < μ < 26.3

Sample Size

If both the margin error E and the population standard deviation σ are known, then

If n is not a whole number, then round it up to the next whole number.

Example

How many adults must be randomly selected to estimate the mean FICO (credit rating) score of working adults in the United States. We want 95% confidence that the sample mean is within 3 points of the population mean, and the population standard deviation is 68.

The corresponding z score to 95% confidence level is

Z0.025 = 1.96

From the text, we have

σ = 68 and E = 3

Thus,

Finding x and E from a confidence interval

If a CI is given, then

x̄ = upper limit + lower limit

2

E = upper limit − lower limit

2

Example

Based on a sample of randomly selected students from MAT121, a 95% confidence interval for the average final grade is given by

(80 , 85).

● Find the point estimate x and the margin error E.

● Find the sample size if standard deviation of all final grades is σ = 8.

-Always read if they want you to estimate the mean or proportion.

Exam Covers chapters 6&7 except sections 6.6 and 7.3

df= degrees of freedom (helpful for chart)

Examples

1) A Statistician plans to interview a simple random sample of american to estimate the proportion of all american smokers. He wishes to construct a 99% confidence interval for such a population with a margin of error E = 0.03. If the sample proportion is estimated by 0.37 (p̂), how large should the sample size be?

2) From a simple random sample of 425 americans, 289 of them believe that the minimum wage should be raised. COnstruct 95% confidence interval to estimate the proportion of all americans who believe that the minimum wage should be raised.