5.1 Normal Probability Distributions

Normal distribution

• A continuous probability distribution for a continuous random variable, x.

• The most important continuous probability distribution in statistics.

• The graph of a normal distribution is called the normal curve.

x

1Larson/Farber

Properties of Normal Distributions1. The mean, median, and mode are equal.

2. The normal curve is bell-shaped and symmetric about the mean.

3. The total area under the curve is equal to one.

4. The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean.

Total area = 1

2

5. Between μ – σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ – σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points.

μ 3σ μ + σμ 2σ μ σ μ μ + 2σ μ + 3σx

Inflection points

Means and Standard Deviations

• A normal distribution can have any mean and any positive standard deviation.

• The mean gives the location of the line of symmetry.

• The standard deviation describes the spread of the data.

μ = 3.5σ = 1.5

μ = 3.5σ = 0.7

μ = 1.5σ = 0.7

3Larson/Farber

Example: Understanding Mean and Standard Deviation

Which curve has the greater mean and Standard Deviation?Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.)

4Larson/Farber

Curve B has the greater standard deviation (Curve B is more spread out than curve A.)

Interpreting Graphs

The heights of fully grown white oak trees are normally distributed. The curve represents the distribution.

μ = 90 (A normal curve is symmetric about the mean)

σ = 3.5 (The inflection points are one standard deviation away from the mean)

5Larson/Farber

0 1-1 2-2 33Z-Scores:

The Standard Normal Distribution

Standard normal distribution

• A normal distribution with a mean of 0 and a standard deviation of 1.

3 12 1 0 2 3

z

Area = 1

Value - Mean -Standard deviation

xz

• Any x-value can be transformed into a z-score by using the formula

6Larson/Farber

The Standard Normal Distribution

• If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution.

Normal Distribution

x

z

Standard Normal Distribution

-xz

• Use Standard Normal Table or Calculator function: NormalCdf() to find cumulative area under the standard normal curve.

7Larson/Farber

Properties of the Standard Normal Distribution

1. The cumulative area is close to 0 for z-scores close to z = 3.49.

2. The cumulative area increases as the z-scores increase.

3. The cumulative area for z = 0 is 0.5000.

4. The cumulative area is close to 1 for z-scores close to z = 3.49.

z = 3.49

Area is close to 0

z

3 12 1 0 2 3

8Larson/Farber

z = 3.49

Area is close to 1

Z = 0Area = 0.500

** Note: Cumulative Area = Area under the curve to the ‘LEFT’ of the z-score.

Using The Standard Normal TableFind the cumulative area that corresponds to a z-score of 1.15 and –0.24

The area to the left of z = 1.15 is 0.8749

9Larson/Farber

The area to the left of z = 0.24 is 0.4052

TI-83/84<2nd><DISTR>2-NormalCdfNormalcdf(-10000, 1.15)

Why use –10000 in normal cdf()?

Finding Areas Under the Standard Normal Curve

#1 To find the area to the left of z, find the area that corresponds to z in the standard normal table or Ti83/84 normalcdf(-10000, zscore)

The area to the left of z = 1.23 is 0.8907

10Larson/Farber

Area to the leftof z = 1.23 is 0.8907.

#2 To find area to the right of z, find area corresponding to z in table & subtract from 1 (total area) or

Ti83/84 normalcdf (zscore, 10000)

Area to the right is1 – 0.8907 = 0.1093

#3 To find the area between two z-scores, find both z-scores in table & subtract smaller from larger area, or Ti83/84 normalcdf(low-zscore, hi-zscore)

The area to the left of z = 0.75 is 0.2266.

Subtract to find the area of the region between the two zscores: 0.8907 0.2266 = 0.6641.

Practice

#1: Find the area under the standard normal curve to the left of z = -0.99

0.99 0 z

Larson/Farber

Answer: 0.1611

#2: Find the area under the standard normal curve to the right of z = 1.06

Answer: 0.1446

1.060 z

#3: Find the area under the standard normal curve between z=-1.5 and 1.25

Answer: 0.8276

1.250z1.50

μ = 500σ = 100

600μ =500x

5.2 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.

P(x < 600) = Area

12Larson/Farber

600μ =500

P(x < 600)

Normal Distributionμ = 500 σ = 100

x

1μ = 0

Standard Normal Distribution μ = 0 σ = 1

z

P(z < 1)

600 5001

100

xz

Same AreaP(x < 600) = P(z < 1)

Example:

Example1: Finding Probabilities for Normal Distributions

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.

13Larson/Farber

2 2.4

P(x < 2)

Normal Distribution μ = 2.4 σ = 0.5

x-0.80 0

Standard Normal Distribution μ = 0 σ = 1

z

P(z < -0.80)2 2.40.80

0.5

xz

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

Find the probability using your Ti-84/83 Calculator___________________

0.2119

Example2: Finding Probabilities for Normal Distributions

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.

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

Find the probability using your Ti-84/83 Calculator_________________

124 45 1 75

12xz - - - .

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

Example3: Finding Probabilities for Normal Distributions

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

P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915Find the probability using your Ti-84/83 Calculator__________________

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

15

If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? 200(.6915) = 138.3 (approx. 138 shoppers)

You Try this one!Assume that cholesterol levels of men in the United States are normally distributed, with a mean of 215 milligrams per deciliter and a standard deviation of 25 milligrams per deciliter. You randomly select a man from the United States. Draw pictures, use Z-scores and use your calculator to answer the questions below:

1. A lower risk of heart attack is associated with a cholesterol level below 200. What is the probability that the man’s cholesterol level is less than 200?

2. A moderate risk of heart attack is associated with a cholesterol level between 200 and 239. What is the probability that the man’s cholesterol level is between 200 and 239?

3. A higher risk of heart attack is associated with cholesterol levels above 239. What is the probability that the man’s cholesterol level is above 239?

(How would you do this using a complement?)

16Answers: 1) .2743 2) .5572 3) .1685

5.3 Finding a z-Score Given an Area

Example1: Find the z-score that corresponds to a cumulative area of 0.3632.

z 0z

0.3632

17Larson/Farber 4th ed

The z-score is -0.35.

TI 83/84<2nd><DISTR>3:InvNorm(.3632)

Finding a z-Score Given an AreaExample2: Find the z-score that has 10.75% of the distribution’s area to its right.

z0z

0.10751 – 0.1075 = 0.8925

Because the area to the right is 0.1075, the cumulative area is 0.8925.

Locate .8925 in the Standard Normal Table. Or InvNorm (.8925)

Answer: Z-score = 1.24

18Larson/Farber 4th ed

Finding a z-Score Given a Percentile

Find the z-score that corresponds to P5 (or the 5th Percentile)

(Recall: 5th percentile refers to a value that is higher than 5% of the data)

The z-score for P5 is the same z-score that corresponds to an area of 0.05.

The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = -1.64). Because 0.05 is halfway between the two areas in the table, use the z-score that is halfway between -1.64 and -1.65. The z-score is -1.645.

OR

InvNorm (.05) = -1.645

z 0z

0.05

19Larson/Farber 4th ed

Transforming a z-Score to an x-Score

To transform a standard z-score to a data value x in a given population, use the formula : x = μ + zσ

20Larson/Farber 4th ed

The speeds of vehicles along a stretch of highway are normally distributed, with a mean of 67 miles per hour and a standard deviation of 4 miles per hour. Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0.

• z = 1.96:x = 67 + 1.96(4) = 74.84 miles per hour (Above the mean)

• z = -2.33: x = 67 + (-2.33)(4) = 57.68 miles per hour (Below the mean)

• z = 0: x = 67 + 0(4) = 67 miles per hour (Equal to the mean)

Finding a Specific Data ValueScores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment?

? 0z

5%

?75x

1 – 0.05 = 0.95

An exam score in the top 5% is any score above the 95th percentile. Find the z-score that corresponds to a cumulative area of 0.95.

From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645 OR

Using the equation x = μ + zσ : x = 75 + 1.645(6.5) ≈ 85.69

The lowest score you can earn and still be eligible for employment is 86.

invNorm (.95)