Understanding Basic Statistics Fourth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Seven Normal Curves and Sampling.

Understanding Basic StatisticsFourth Edition

By Brase and BrasePrepared by: Lynn Smith

Gloucester County College

Chapter Seven

Normal Curves and Sampling Distributions

Copyright © Houghton Mifflin Company. All rights reserved. 7 | 2

Properties of The Normal Distribution

The curve is bell-shaped with the highest point over the mean, μ.


The curve is symmetrical about a vertical line through μ.



The curve approaches the horizontal axis but never touches or crosses it.



The transition points between cupping upward and downward occur

above μ + σ and μ – σ .



The Empirical Rule

• Applies to any symmetrical and bell-shaped distribution


The Empirical Rule

Approximately 68% of the data values lie within one standard deviation of the mean.


Approximately 95% of the data values lie within two standard deviations of the mean.

The Empirical Rule


Almost all (approximately 99.7%) of the data values will be within three standard deviations

of the mean.

The Empirical Rule


Application of the Empirical Rule

The life of a particular type of light bulb is normally distributed with a mean of

1100 hours and a standard deviation of 100 hours.

What is the probability that a light bulb of this type will last between 1000 and

1200 hours?Approximately 68%


Standard Score

• The z value or z score tells the number of standard deviations between the original measurement and the mean.

• The z value is in standard units.


Formula for z score

x

z


x Values and Corresponding z Values


Calculating z scores

The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. Convert 21 minutes to a z score.

00.22

2521xz


Mean delivery time = 25 minutes Standard deviation = 2 minutes

Convert 29.7 minutes to a z score.

35.22

257.29xz

Calculating z scores


Raw Score

• A raw score is the result of converting from standard units (z scores) back to original measurements, x values.

• Formula: x = z +


Mean delivery time = 25 minutes Standard deviation = 2 minutes

Interpret a z score of 1.60.

2.2825)2(6.1zx The delivery time is 28.2 minutes.

Interpreting z-scores


Standard Normal Distribution:

= 0

= 1

Any x values are converted to z

scores.


Importance of the Standard Normal Distribution:

Areas will be equal.

Any Normal Curve:


Areas of a Standard Normal Distribution

• Appendix

• Table 3

• Pages A6 - A7


Use of the Normal Probability Table

Appendix Table 3 is a left-tail style table.

Entries give the cumulative areas to the left of a specified z.


To Find the area to the Left of a Given z score

• Find the row associated with the sign, units and tenths portion of z in the left column of Table 3.

• Move across the selected row to the column headed by the hundredths digit of the given z.


Find the area to the left of z = – 2.84


To find the area to the left of z = – 2.84


To find the area to the left of z = – 2.84


The area to the left of z = – 2.84 is .0023


Use Table 3 of the Appendix directly.

To Find the Area to the Left of a Given Negative z Value:


Use Table 3 of the Appendix directly.

To Find the Area to the Left of a Given Positive z Value:


Subtract the area to the left of z from 1.0000.

To Find the Area to the Right of a Given z Value:


Use the symmetry of the normal distribution.

Area to the right of z

= area to left of –z.

Alternate Way To Find the Area to the Right of a Given Positive z Value:


Subtract area to left of z1 from area to left of z2 . (When z2 > z1.)

To Find the Area Between Two z Values


Convention for Using Table 3

• Treat any area to the left of a z value smaller than 3.49 as 0.000

• Treat any area to the left of a z value greater than 3.49 as 1.000


a. P( z < 1.64 ) = __________

b. P( z < - 2.71 ) = __________

.9495

.0034




c. P(0 < z < 1.24) = ______

d. P(0 < z < 1.60) = _______

e. P( 2.37 < z < 0) = ______

.3925

.4452

.4911



f. P( 3 < z < 3 ) = ________

g. P( 2.34 < z < 1.57 ) = _____

h. P( 1.24 < z < 1.88 ) = _______

.9974

.9322

.0774



i. P( 2.44 < z < 0.73 ) = _______

j . P( z > 2.39 ) = ________

k. P( z > 1.43 ) = __________

.0084

.2254

.9236


To Work with Any Normal Distributions

• Convert x values to z values using the formula:

x

z

Use Table 3 of the Appendix to find corresponding areas and probabilities.


Rounding

• Round z values to the hundredths positions before using Table 3.

• Leave area results with four digits to the right of the decimal point.


Application of the Normal Curve

The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be:

a. between 25 and 27 minutes. a. __________

b. less than 30 minutes. b. __________

c. less than 22.7 minutes. c. __________

.3413

.9938

.1251


Inverse Normal Probability Distribution

• Finding z or x values that correspond to a given area under the normal curve


Look up area A in body of Table 3 and use corresponding z value.

Inverse Normal Left Tail Case


Look up the number 1 – A in body of Table 3 and use corresponding z value.

Inverse Normal Right Tail Case:


Look up the number (1 – A)/2 in body of Table 3 and use corresponding ± z value.

Inverse Normal Center Case:


Using Table 3 for Inverse Normal Distribution

• Use the nearest area value rather than interpolating.

• When the area is exactly halfway between two area values, use the z value exactly halfway between the z values of the corresponding table areas.


When the area is exactly halfway between two area values

• When the z value corresponding to an area is smaller than 2, use the z value corresponding to the smaller area.

• When the z value corresponding to an area is larger than 2, use the z value corresponding to the larger area.


Find the indicated z score:



z = _______– 2.57


zz = _______ = _______2.33



Find the indicated z scores:

z = ____–z = _____–1.23 1.23



± z =__________ ± 2.58


± z = ________± 1.96



Application of Determining z Scores

The Verbal SAT test has a mean score of 500 and a standard deviation of 100.

Scores are normally distributed. A major university determines that it will

accept only students whose Verbal SAT scores are in the top 4%. What is the minimum score that a student must

earn to be accepted?


The cut-off score is 1.75 standard deviations above the mean.



The cut-off score is 500 + 1.75(100) = 675.

Mean = 500

standard deviation = 100



Introduction to Sampling Distributions


Review of Statistical Terms

• Population

• Sample

• Parameter

• Statistic


Population

• the set of all measurements or counts

• (either existing or conceptual)

• under consideration


Sample

• a subset of measurements from a population


Parameter

• a numerical descriptive measure of a population


Statistic

• a numerical descriptive measure of a sample


We use a statistic to make inferences about a population parameter.


Some Common Statistics and Corresponding Parameters


Principal Types of Inferences

• Estimation: estimate the value of a population parameter

• Testing: formulate a decision about the value of a population parameter

• Regression: Make predictions or forecasts about the value of a statistical variable


Sampling Distribution

• a probability distribution for the sample statistic

• based on all possible random samples of the same size from the same population


Example of a Sampling Distribution

• Select samples with two elements each (in sequence with replacement) from the set

• {1, 2, 3, 4, 5, 6}.


Constructing a Sampling Distribution of the Mean for Samples of Size n = 2

List all samples (with 2 items in each) and compute the mean of each sample.

sample: mean: sample: mean

{1,1} 1.0 {1,6} 3.5{1,2} 1.5 {2,1} 1.5{1,3} 2.0 {2,2} 4{1,4} 2.5 … ...{1,5} 3.0

How many different samples are there? 36


Sampling Distribution of the Mean

p

1.0 1/361.5 2/362.0 3/362.5 4/363.0 5/363.5 6/364.0 5/364.5 4/36 5.0 3/365.5 2/36 6.0 1/36

x


Let x be a random variable with a normal distribution with mean and standard deviation . Let x be the

sample mean corresponding to random samples of size n taken from

the distribution .


Theorem 7.1:

• The x distribution is a normal distribution.

• The mean of the x distribution is (the same mean as the original distribution).

• The standard deviation of the x distribution is (the standard deviation of the original distribution, divided by the square root of the sample size).

n


We can use this theorem to draw conclusions about means of samples

taken from normal distributions.

If the original distribution is normal, then the sampling distribution will be

normal.


The Mean of the Sampling Distribution

x


x

The mean of the sampling distribution is equal to the mean of the original

distribution.


x

The Standard Deviation of the Sampling Distribution


The standard deviation of the sampling distribution is equal to the standard

deviation of the original distribution divided by the square root of the sample size.

nx


To Calculate z Scores

n

xxz

x

x


The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2

minutes.

1. Find the probability that a trip between the cities takes more than 15 minutes.

2. Find the probability that mean time of nine trips between the cities is more than 15 minutes.


• Find the probability that a trip between the cities takes more than 15 minutes.

3264.06736.01)45.0z(P

45.02.2

1415z

Mean = 14 minutes, standard deviation = 2.2 minutes


• Find the probability that the mean time of nine trips between the cities is more than 15 minutes.

73.09

2.2

n

14

x

x




• Find the probability that mean time of nine trips between the cities is more than 15 minutes.

0853.09147.01)37.1(

37.173.0

1415

zP

z


Standard Error of the Mean

• The standard error of the mean is the standard deviation of the sampling distribution

•

n

mean the of

error standard the on,distributi sampling x theFor

x


What if the Original Distribution Is Not Normal?

• Use the Central Limit Theorem.


Central Limit Theorem

If x has any distribution with mean and standard deviation , then the sample mean based on a random sample of

size n will have a distribution that approaches the normal distribution

(with mean and standard deviation divided by the square root of n) as n

increases without limit.


How large should the sample size be to permit the application of the Central Limit

Theorem?

• In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.



• For most x distributions, if we use a sample size of 30 or larger, the x distribution will be approximately normal.



• The mean of the sampling distribution is the same as the mean of the original distribution.

• The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.


Central Limit Theorem Formula

x


nx



n/

xxz

x

x



Application of the Central Limit Theorem

Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be:

a. more than 530 pounds?b. less than 500 pounds?c. between 495 and 515 pounds?


Are we authorized to use the Normal Distribution?

• Yes, we are attempting to draw conclusions about means of large samples.


Applying the Central Limit Theorem

What is the probability that the mean weight will be more than 530 pounds?Consider the distribution of sample means:

P( x > 530): z = 530 – 510 = 20 = 2.22 9 9

P(z > 2.22) = _______.0132

9100/90,510 xx



What is the probability that their mean weight will be less than 500 pounds?

P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9

P(z < – 1.11) = _______.1335



What is the probability that their mean weight will be between 495 and 515 pounds?

P(495 < x < 515) :

for 495: z = 495 – 510 = - 15 = - 1.67 9 9

for 515: z = 515 – 510 = 5 = 0.56 9 9

P(–1.67 < z < 0.56) = ______.6648


Normal Approximation of The Binomial Distribution:

• If n (the number of trials) is sufficiently large, a binomial random variable has a distribution that is approximately normal.


Define “sufficiently large”

The sample size, n, is considered to be "sufficiently large" if

np and nq

are both greater than 5.


Mean and Standard Deviation: Binomial Distribution

qpnandpn


Experiment: tossing a coin 20 times

Problem: Find the probability of getting exactly 10 heads.

Distribution of the number of heads appearing should look like:

10 200


Using the Binomial Probability Formula

n =

x =

p =

q = 1 – p =

P(10) = 0.176197052

20

10

0.5

0.5


Normal Approximation of the Binomial Distribution

First calculate the mean and standard deviation:

= np = 20 (.5) = 10

24.25)5(.)5(.20)p1(pn


The Continuity Correction

Continuity Correction allows us to approximate a discrete probability distribution with a

continuous distribution.



• We are using the area under the curve to approximate the area of the rectangle.



• If r is the left-point of an interval, subtract 0.5 to obtain the corresponding normal variable.

• x = r 0.5• If r is the right-point of an interval, add

0.5 to obtain the corresponding normal variable.

• x = r + 0.5



• Continuity Correction: to compute the probability of getting exactly 10 heads, find the probability of getting between 9.5 and 10.5 heads.


Using the Normal Distribution

P(9.5 < x < 10.5 ) = ?

For x = 9.5: z = – 0.22

For x = 10.5: z = 0.22


P(9.5 < x < 10.5 ) =

P( – 0.22 < z < 0.22 ) =

.5871 – .4129 = .1742

Using the Normal Distribution


Application of Normal Distribution

If 22% of all patients with high blood pressure have side effects from a certain medication,

and 100 patients are treated, find the probability that at least 30 of them will have

side effects.Using the Binomial Probability Formula we would need to compute:

P(30) + P(31) + ... + P(100) or 1 - P( x < 29)


Using the Normal Approximation to the Binomial Distribution

Is n sufficiently large?

Check: np =

nq =


Is n sufficiently large?

np = 22

nq = 78

Both are greater than five.

Using the Normal Approximation to the Binomial Distribution


Find the mean and standard deviation

= 100(.22) = 22

and =

14.416.17

)78)(.22(.100


Applying the Normal Distribution

To find the probability that at least 30 of them will have side effects, find P(x 29.5)


z = 29.5 – 22 = 1.81 4.14

Find P(z 1.81)

The probability that at least 30 of the patients will have

side effects is 0.0351.

Applying the Normal Distribution


Reminders:

• Use the normal distribution to approximate the binomial only if both np and nq are greater than 5.

• Always use the continuity correction when using the normal distribution to approximate the binomial distribution.

Understanding Basic Statistics Fourth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Seven Normal Curves and Sampling.

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