5.1 Introduction to Normal Distributions

Properties of a Normal Distribution

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 Inflection point

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. • 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.

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

Means?

10 11

12 13 14

15 16 17 18 19

20

Curves with different means, different standard deviations

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

Empirical Rule

68%

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

About 95% of the area lies within 2 standard deviations About 99.7% of the area lies within 3 standard deviations of the mean

Determining Intervals

x 3.3 3.6 3.9 4.2

4.5 4.8 5.1

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. 95% of the data will fall within 2 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 Standard normal distribution: mean = 0, standard deviation = 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

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

Chptr 2: The Standard Score The standard- or z-score, represents the number of standard deviations a random variable x falls from the mean:

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

Cumulative Areas The total area under the curve is one.

–3 –2 –1 0 1 2 3

z

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

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

–3 –2 –1 0 1 2 3

z

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.

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.

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

–3 –2 –1

0 1

2 3

z

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

Finding Probabilities 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). 0.1075 0.8925

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

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

Find P(–1.25 < z < 1.17)

–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

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

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 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. -3 -2 -1 0 1 2 3

*cdf*

z

Section 5.2

Normal Distributions Finding Probabilities

Probabilities and Normal Distributions 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

115 − 100 z= =1 15

Probabilities and Normal Distributions Normal Distribution

Standard Normal Distribution

100 115

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:

Interpreting the Central Limit Theorem

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%

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

mean standard deviation

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

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

Section 5.5

Normal Approximation to Binomial Distributions

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

µ= 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 n = 20 p = 0.25 np = 5 nq = 15

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

50

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

0.089 0.065

14

15

16

Correction for Continuity Use the normal approximation to the binomial distribution to find .

14

15

16

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

Correction for Continuity

14

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.

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

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

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.

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).

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