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Properties of a Normal Distribution 5.1 Introduction to Normal Distributions x • 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%

Properties of a Normal Distribution

Means and Standard Deviations Curves with different means, same standard deviation

Inflection point

Means?

Inflection point

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.

10 11

12 13 14

15 16 17 18 19

20

Curves with different means, different standard deviations

• 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.

9 10 11 12 13 14 15 16 17 18 19 20 21 22

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Determining Intervals

Empirical Rule

68%

About 68% of the area lies within 1 standard deviation of the mean 3.3 3.6 3.9 4.2

4.5 4.8 5.1

x

Example: An instruction manual claims that assembly time for a product is normally distributed with a mean of 4.2 hours and a standard deviation of 0.3 hour. Determine the interval in which 95% of the assembly times fall.

About 95% of the area lies within 2 standard deviations

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

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

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.

The Standard Normal Distribution

Chptr 2: The Standard Score

Standard normal distribution: mean = 0, standard deviation = 1

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

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

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

–4 –3 –2 –1 0 1

2 3

4

z

If a normal distribution is standardized using tables, then each value must be standardized to find probabilities.

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Cumulative Areas The total area under the curve is one.

Cumulative Areas Using a standard normal table, find the cumulative area for a z-score of –1.25

Sum left to right

z

–3 –2 –1 0 1 2 3

z

–3 –2 –1 0 1 2 3

Cumulative area is close to 0 for z-scores close to –3.49

Pg. A16: down the z column on the left to z = –1.2 and across to the cell under .05 = 0.1056, the cumulative area. The probability that z is at most –1.25 is 0.1056.

Cumulative area for z = 0 is 0.50 Cumulative area is close to 1 for z-scores close to 3.49

Finding Probabilities

Finding Probabilities

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

To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1. Find P(z > –1.24).

Find P(z < –1.45) P (z < –1.45) = 0.0735

0.1075 0.8925

–3 –2 –1

0 1

2 3

z

Read down the z-column to –1.4 and across to .05 = 0.0735

z –3 –2 –1 0 1 2 3 The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925. P(z > –1.24) = 0.8925

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Finding Probabilities

Summary

The probability that z is between two values: find the cumulative areas for each and subtract the smaller area from the larger.

To find the probability that z is less than a given value, read the corresponding cumulative area.

Find P(–1.25 < z < 1.17)

z

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

–3 –2 –1 0 1 2 1. P(z < 1.17) = 0.8790

3

z

2. P(z < –1.25) = 0.1056

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

Probabilities can’t be negative, so subtract smaller from larger

z

-3 -2 -1 0 1 2 3

To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. -3 -2 -1

*cdf*

0 1 2 3

z

Probabilities and Normal Distributions

Section 5.2 Normal Distributions Finding Probabilities

If a random variable, x, is normally distributed, then the probability that x will fall within an interval is equal to the area under the curve in the interval. Example: IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a person selected at random will have an IQ score less than 115.

100 115

To find the area, first find the standard score equivalent to x = 115

z

115  100 1 15

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Probabilities and Normal Distributions

Monthly utility bills in a city are normally distributed with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected. Find the probability it is between $80 and $115.

Normal Distribution

Normal Distribution:  = 100;  = 12 100 115

P(80 < x < 115)

SAME

SAME

Find P(x < 115). Standard Normal Distribution

Application Example

P(–1.67 < z < 1.25) Find P(z < 1).

0 1 From Standard Normal Table: P(z < 1) = 0.8413, so P(x 30 the sampling distribution of will be normal mean standard deviation

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

z

2.14

There is a 0.0162 or 1.62% probability that a sample of 60 sockeye will have a mean length greater than 70 cm. What is probability that 1 fish will be > 70 cm? P(z>0.28) = 1-0.6103 = 0.3897  39%

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Application Central Limit Theorem A long time ago, the mean price of gasoline in California was $1.164 per gallon. What is the probability that the mean price for a sample of 38 gas stations in California is between $1.169 and $1.179? Assume the standard deviation = $0.049.

Since n > 30 the sampling distribution of

will be normal

Application Central Limit Theorem P( 0.63 < z < 1.90) = 0.9713 – 0.7357 = 0.2356

z

mean standard deviation

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

.63 1.90 The probability is 0.2356 that the mean for the sample is between $1.169 and $1.179. Hint: drawing the distribution, values, and area of interest will help keep calculations clear.

Central Limit Theorem Section 5.5 Creature Cast Central Limit Theorem video http://vimeo.com/75089338

Normal Approximation to Binomial Distributions

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Binomial Distribution Characteristics • There are a fixed number of independent trials, n. • Each trial has 2 outcomes, Success or Failure. • The probability of S on a single trial is p and the probability of F is q. In total:

p+q=1

• We can find the probability of exactly x successes out of n trials. Where x = 0 or 1 or 2 … n. • x is a discrete random variable representing a count of the number of S’s in n trials.

Application 34% of Americans have type A+ blood. If 500 Americans are sampled at random, what is the probability at least 300 have type A+ blood? Using Chapter 4 you could calculate the probability that exactly 300, exactly 301… exactly 500 Americans have A+ blood type and then add the probabilities (but this should drive you crazy). Alternatively…use normal curve probabilities to approximate binomial probabilities. If np  5 and nq  5, then the binomial random variable x is approximately normally distributed with mean

= np and standard deviation

Why np  5 and nq  5?

0

1

2

3

4

5

n=5 p = 0.25, q = .75 np =1.25 nq = 3.75

Binomial Probabilities The binomial distribution is discrete with a probability histogram graph. The probability that a specific value of x will occur is equal to the area of the rectangle with midpoint at x. Example: If n = 50 and p = 0.25 find

n = 20 p = 0.25 np = 5 nq = 15

0.111

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

n = 50 p = 0.25 np = 12.5 nq = 37.5 0

10

20

30

40

Add the areas of the rectangles with midpoints at x = 14, x = 15, and x = 16: 0.111 + 0.089 + 0.065 = 0.265

14

0.089

15

0.065

16

50

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Correction for Continuity

Correction for Continuity

Use the normal approximation to the binomial distribution to find .

14 14

15

16

Values for the binomial random variable x are 14, 15 and 16.

Normal Approximation to the Binomial Use the normal approximation to the binomial to find . Find the mean and standard deviation using binomial distribution formulas:

15

16

The interval of values under the normal curve is

To ensure the boundaries of each rectangle are included in the interval, subtract 0.5 from a left-hand boundary and add 0.5 to a right-hand boundary.

Application A survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation. Since np = 150  5 and nq = 50  5, can use the normal approximation to the binomial distribution.

Adjust the endpoints to correct for continuity P(13.5 ≤ x  16.5). Convert each endpoint to a standard score:

The binomial phrase of “fewer than 140” means up to 139: 0, 1, 2, 3…139. Use the correction for continuity to translate to the continuous variable in the interval . Find P(x< 139.5).

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Application A survey of Internet users found that 75% favored government regulations of “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation. Use the correction for continuity P(x