Elementary Statistics Larson Farber 5 Normal Probability Distributions.

Elementary Statistics

Larson Farber

5 Normal Probability Distributions

Properties of a Normal Distribution

• The mean, median, and mode are equal

• Bell shaped and is symmetric about the mean

• The total area that lies under the curve is one or 100%

x

• As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never

touches it.

• The points at which the curvature changes are called inflection points. The graph curves downward between the

inflection points and curves upward past the inflection points to the left and to the right.

x

Inflection pointInflection point

Properties of a Normal Distribution

Means and Standard Deviations

2012 15 1810 11 13 14 16 17 19 21 229

12 15 1810 11 13 14 16 17 19 20

Curves with different means, different standard deviations

Curves with different means, same standard deviation

Empirical Rule

About 95% of the area lies within 2 standard

deviations

About 99.7% of the area lies within 3 standard deviations of the mean

About 68% of the area lies within 1 standard deviation of the mean

68%

4.2 4.5 4.8 5.13.93.63.3

Determining Intervals

An instruction manual claims that the assembly time for a product is normally distributed with a mean of 4.2 hours

and standard deviation 0.3 hour. Determine the interval in which 95% of the assembly times fall.

x

4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8. 95% of the assembly times will be between 3.6 and 4.8 hrs.

95% of the data will fall within 2 standard deviations of the mean.

The Standard Normal

Distribution

The Standard ScoreThe standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean.

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of:(a) 161 (b) 148 (c) 152

(a) (b) (c)

The Standard Normal Distribution

The standard normal distribution has a mean of 0 and a standard deviation of 1.

Using z-scores any normal distribution can be transformed into the standard normal distribution.

–4 –3 –2 –1 0 1 2 3 4 z

Cumulative Areas

0 1 2 3–1–2–3 z

The total area

under the curve

is one.

The cumulative area for z = 0 is 0.5000, indicating that the probability of getting a z value of 0 or less is .5

Find the cumulative area for a z-score of –1.25.

0 1 2 3–1–2–3 z

Cumulative Areas - Finding

0.1056

•Table in Appendix•Table on fold-out card•Excel•Internet calculator

Using the table …Find the cumulative area that corresponds to a z-score of -0.24.

Solution:Find -0.2 in the left hand column.

Move across the row to the column under 0.04

Using the table …Find the cumulative area that corresponds to a z-score of -0.24.

Solution:Find -0.2 in the left hand column.

The area to the left of z = -0.24 is 0.4052.Move across the row to the column under 0.04

Using Excel

x value of interest

mean

standard deviation

Online: Rossman-Chance

http://www.rossmanchance.com/applets/NormalCalcs/NormalCalculations.html

Online: SurfStat

http://surfstat.anu.edu.au/surfstat-home/tables/normal.php

Find the cumulative area for a z-score of –1.25.

0 1 2 3–1–2–3 z

Cumulative Areas

0.1056

Read down the z column on the left to z = –1.25 and across to the column under .05. The value in the cell is 0.1056, the

cumulative area.

The probability that z is at most –1.25 is 0.1056.

“Less than”To find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score.

0 1 2 3–1–2–3 z

Read down the z-column to –1.4 and across to .05. The cumulative area is 0.0735.

Find P(z < –1.45).

P (z < –1.45) = 0.0735

“Greater than”To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1.

0 1 2 3–1–2–3 z

P(z > –1.24) = 0.8925

Find P(z > –1.24).

The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925.

0.10750.8925

“Between”To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from

the larger.

Find P(–1.25 < z < 1.17).

1. P(z < 1.17) = 0.8790 2. P(z < –1.25) = 0.1056

3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734

0 1 2 3–1–2–3 z

0 1 2 3-1 -2-3 z

Summary

0 1 2 3-1-2-3 zTo find the probability is greater than a given value, subtract the cumulative area in the table from 1.

0 1 2 3-1-2-3 z

To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger.

To find the probability that z is lessthan a given value, read thecorresponding cumulative area.

Normal Distributions

Finding Probabilities

Probability and Normal Distributions

• If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.

μ = 500σ = 100

600μ =500x

Remember that the total area under the curve is 1.0 (equal to 100%).


• If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.

P(x < 600) = Area μ = 500σ = 100

600μ =500x

Remember that the total area under the curve is 1.0 (equal to 100%).


Normal Distribution

600μ =500

P(x < 600)

μ = 500 σ = 100

x

600 5001

100

xz


P(x < 500) = P(z < 1)

Normal Distribution

600μ =500

P(x < 600)

μ = 500 σ = 100

x

Standard Normal Distribution

600 5001

100

xz

1μ = 0

μ = 0 σ = 1

z

P(z < 1)

Same Area

Example

A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed.

Solution

P(x < 2) = P(z < -0.80) = 0.2119

Normal Distribution

2 2.4

P(x < 2)

μ = 2.4 σ = 0.5

x

2 2.40.80

0.5

xz

Standard Normal Distribution

-0.80 0

μ = 0 σ = 1

z

P(z < -0.80)

0.2119

Example:

A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes.

Solution: Finding Probabilities for Normal Distributions

P(24 < x < 54) = P(-1.75 < z < 0.75) = 0.7734 – 0.0401 = 0.7333

1

24 45 1 7512

xz - - - .

24 45

P(24 < x < 54)

x

Normal Distribution μ = 45 σ = 12

0.0401

54

2

54 45 0 7512

xz - - .

-1.75z

Standard Normal Distribution μ = 0 σ = 1

0

P(-1.75 < z < 0.75)

0.75

0.7734

Example:

Find the probability that the shopper will be in the store more than 39 minutes. (Recall μ = 45 minutes and σ = 12 minutes)

Solution: Finding Probabilities for Normal Distributions

P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915

39 45 0 5012

- - - .xz

39 45

P(x > 39)

x

Normal Distribution μ = 45 σ = 12

Standard Normal Distribution μ = 0 σ = 1

0.3085

0

P(z > -0.50)

z

-0.50

Normal Distributions

Finding Values

Section 5.4

z

From Areas to z-Scores

Locate 0.9803 in the area portion of the table. Read the values at the beginning of the corresponding row and at

the top of the column. The z-score is 2.06.

Find the z-score corresponding to a cumulative area of 0.9803.

z = 2.06 correspondsroughly to the

98th percentile.

–4 –3 –2 –1 0 1 2 3 4

0.9803

Finding z-Scores from Areas

Find the z-score corresponding to the 90th percentile.

z0

.90

The closest table area is .8997. The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28.

A z-score of 1.28 corresponds to the 90th percentile.

Find the z-score with an area of .60 falling to its right.

.60.40

0 zz

With .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is 0.2 and column heading is .05. The z-score is 0.25.

A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile


Find the z-score such that 45% of the area under the curve falls between –z and z.

0 z–z

The area remaining in the tails is .55. Half this area isin each tail, so since .55/2 = .275 is the cumulative area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is 0.60. The positive z score is 0.60.

.45.275.275


From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard z-score of: (a) 2.33 (b) –1.75 (c) 0

To find the data value, x when given a standard score, z:

From z-Scores to Raw Scores

The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard z-score of: (a) 2.33 (b) –1.75 (c) 0

(a) x = 152 + (2.33)(7) = 168.31

(b) x = 152 + (–1.75)(7) = 139.75

(c) x = 152 + (0)(7) = 152

To find the data value, x when given a standard score, z:

Finding Percentiles or Cut-off ValuesMonthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills?

10%90%

Find the cumulative area in the table that is closest to 0.9000 (the 90th percentile.) The area 0.8997 corresponds to a z-score of 1.28.

x = 100 + 1.28(12) = 115.36.

$115.36 is the smallestvalue for the top 10%.

z

To find the corresponding x-value, use

Sampling Distributions & The Central Limit Theorem

Sample

Sampling DistributionsA sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means.

Sample

The sampling distribution consists of the values of the sample means,

SampleSample

Sample

Sample

Sampling Distribution of x-bar

Sample 11x

Sample 55x

Sample 22x

3x

4x

Population with μ, σ

The sampling distribution consists of the values of the sample means,

1 2 3 4 5, , , , ,...x x x x x

x

the sample means will have a normal distribution

The Central Limit Theorem

and standard deviation

If a sample n >= 30 is taken from a population withany type distribution that has a mean =and standard deviation =

the distribution of means of sample size n, will be normal with a mean

standard deviation

The Central Limit Theorem

x

If a sample of any size is taken from a population with a normal distribution with mean = and standard

deviation =

Application

Distribution of means of sample size 60, will be normal.

The mean height of American men (ages 20-29) is inches. Random samples of 60 such men are selected. Find the mean and standard deviation (standard error) of the sampling distribution.

mean

Standard deviation

69.2

Interpreting the Central Limit Theorem

The mean height of American men (ages 20-29) is = 69.2”. If a random sample of 60 men in this age group is selected, what is the probability the mean height for the sample is greater than 70”? Assume the standard deviation is 2.9”.

Find the z-score for a sample mean of 70:

standard deviation

mean

Since n > 30 the sampling distribution of will be normal

2.14z

There is a 0.0162 probability that a sample of 60 men will have a mean height greater than 70”.

Interpreting the Central Limit Theorem

Application Central Limit Theorem

During a certain week the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for the sample of 38 gas stations in California is between $1.169 and $1.179? Assume the standard deviation = $0.049.

standard deviation

mean

Calculate the standard z-score for sample values of $1.169 and $1.179.

Since n > 30 the sampling distribution of will be normal

.63 1.90

z

Application Central Limit Theorem

P( 0.63 < z < 1.90)

= 0.9713 – 0.7357

= 0.2356

The probability is 0.2356 that the mean for the

sample is between $1.169 and $1.179.

Elementary Statistics Larson Farber 5 Normal Probability Distributions.

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