Normal Distributions

ChapterNormal Probability Distributions

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6

© 2012 Pearson Education, Inc.All rights reserved.

Chapter Outline

• 6.1 Graphs of Normal Probability Distributions• 6.2 Standard Units and Areas Under the Standard

Normal Distribution• 6.3 Areas Under Any Normal Curve• 6.4 Normal Approximations to Binomial

Distributions

© 2012 Pearson Education, Inc. All rights reserved. 2 of 105

Section 6.1

Graphs of Normal Probability Distributions


Section 6.1 Objectives

• Graph a normal curve and summarize its important properties

• Apply the empirical rule to solve real-world problems


Continuous Probability Distribution

Continuous random variable • Has an infinite number of possible values that can be

represented by an interval on the number line.

Continuous probability distribution• The probability distribution of a continuous random

variable.

Hours spent studying in a day

0 63 9 1512 18 2421

The time spent studying can be any number between 0 and 24.


Properties of Normal Distributions

Normal distribution • A continuous probability distribution for a random

variable, x. • The most important continuous probability

distribution in statistics.• The graph of a normal distribution is called the

normal curve.

x


Properties of Normal Distributions

1. The mean, median, and mode are equal.2. The normal curve is bell-shaped and is symmetric

about the mean.3. The total area under the normal curve is equal to 1.4. The normal curve approaches, but never touches, the

x-axis as it extends farther and farther away from the mean.

x

Total area = 1

μ© 2012 Pearson Education, Inc. All rights reserved. 7 of 105

Properties of Normal Distributions5. 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, or transition points.


μ – 3σ μ + σμ – σ μ μ + 2σ μ + 3σμ – 2σ

2

2( )

21( )2

x

f x e

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


Exercise 1: Understanding Mean and Standard Deviation

1. Which normal curve has the greater mean?

Solution: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.)


Exercise 2: Understanding Mean and Standard Deviation

2. Which curve has the greater standard deviation?

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


Example: Interpreting Graphs

The scaled test scores for the New York State Grade 8 Mathematics Test are normally distributed. The normal curve shown below represents this distribution. What is the mean test score? Estimate the standard deviation.Solution:


Because a normal curve is symmetric about the mean, you can estimate that μ ≈ 675.

Because the inflection points are one standard deviation from the mean, you can estimate that σ ≈ 35.

Example: Two Normal Curves

Both curves have the samemean, µ = 6.

Curve A has a standarddeviation of σ = 1.

Curve B has a standarddeviation of σ = 3.

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The Empirical Rule

The Empirical Rule

Exercise 3: Sketch Distribution


The playing life of a Sunshine radio is normally distributed with a mean of 600 hours and a standard deviation of 100 hours. Sketch a normal curve showing the distribution of the playing life of the Sunshine radio. Scale and label the axis; include the transition points.

Exercise 4: Use Empirical Rule


The playing life of a Sunshine radio is normally distributed with a mean of 600 hours and a standard deviation of 100 hours. Use the empirical rule to compute the probabilities that a randomly selected radio will last as specified:Life of Randomly Selected Radio

Probability Expression

Probability Calculation

Between 600 and 700 hours


Greater than 700 hours



The playing life of a Sunshine radio is normally distributed with a mean of 600 hours and a standard deviation of 100 hours. Use the empirical rule to compute the probabilities that a randomly selected radio will last as specified:Life of Randomly Selected Radio

Probability Expression Probability Calculation


½ of 68% = 34%


13.5%

Greater than 700 hours

13.5% + 2.35% = 15.85%

The annual wheat yield per acre on a farm is normally distributed with a mean of 35 bushels and a standard deviation of 8 bushels. Sketch a normal curve and shade in the area that represents the probability that an acre will yield between 19 and 35 bushels.



= Exercise 5: Use Empirical Rule


Section 6.1 Summary

• Graph a normal curve and summarize its important properties

• Apply the empirical rule to solve real-world problems


Section 6.2

Areas Under the Standard Normal Distribution



• Convert raw data to z scores• Convert z scores to raw data• Graph the standard normal distribution• Find areas under the standard normal curve


The Standard Normal Distribution

Standard normal distribution • A normal distribution with a mean of 0 and a standard

deviation of 1.

–3 1–2 –1 0 2 3

z

Area = 1

z Value Mean

Standard deviation x

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


z scores

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• z scores also have a normal distribution µ = 0 σ = 1


• 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

σ

0

σ 1

z

Standard Normal Distribution

• Use the Standard Normal Table to find the cumulative area under the standard normal curve.


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.

z = –3.49

Area is close to 0

z

–3 1–2 –1 0 2 3


z = 3.49

Area is close to 1

Properties of the Standard Normal Distribution

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.

Area is 0.5000z = 0

z

–3 1–2 –1 0 2 3


Standard Normal Table – Appendix II, Table 5 – Page A22

Table gives the cumulative area for a given z value.• When calculating a z Score, round to 2 decimal

places.• If a z-score is halfway between two values in the

table, round to 3 decimal places• For a z-score less than -3.49, use 0.0000 to

approximate the area.• For a z-score greater than 3.49, use 1.0000 to

approximate the area.

Area to the Left of a Given z-score

Area to the Right of a Given z-score

Area Between Two z-scores

Exercise 1a: Using The Standard Normal Table

Find the cumulative area that corresponds to a z-score of 1.15.

The area to the left of z = 1.15 is 0.8749.Move across the row to the column under 0.05

Solution:Find 1.1 in the left hand column.


Exercise 1b: Using The Standard Normal 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.© 2012 Pearson Education, Inc. All rights reserved. 35 of 105

Move across the row to the column under 0.04

Exercise 1c: Finding Area Under the Standard Normal Curve

Find the area under the standard normal curve to the left of z = –0.99.

From the Standard Normal Table, the area is equal to 0.1611.

–0.99 0z

0.1611

Solution:


Exercise 1d: Finding Area Under the Standard Normal Curve

Find the area under the standard normal curve to the right of z = 1.06.


1 – 0.8554 = 0.1446

1.060z

Solution:

0.8554


Find the area under the standard normal curve between z = –1.5 and z = 1.25.

Exercise 1e: Finding Area Under the Standard Normal Curve


1.250z

–1.50

0.89440.0668

Solution:0.8944 – 0.0668 = 0.8276


Using Technology to find Normal Probabilities

Must specify the mean and standard deviation of the population, and the x-value(s) that determine the interval.




Re-compute Exercise 1 (parts a-e) using the normalcdf function on the TI-84 calculator.

Exercise 2: Finding Area Under the Standard Normal Curve


Description Cumulative Area

a) Find the cumulative area that corresponds to a z-score of 1.15.

0.8749

b) Find the cumulative area that corresponds to a z-score of -0.24.

0.4052

c) Find the cumulative area that corresponds to a z-score of -0.99.

0.1611

d) Find the cumulative area to the right of z=1.06

1 – 0.8554 = 0.1446

e) Find the area under the standard normal curve between z = –1.5 and z = 1.25.

0.8944 – 0.0668 = 0.8276

Suppose Tina and Jack are in two different sections of the same course and they recently took midterms. Tina’s class average was 64 (with S.D.=3) and she got a 74. Jack’s class average was 72 (with S.D.=5) and he got an 82. Assuming that all scores are normally distributed, who did better relative to the class? • Interpretive Statement: Although each score was 10

points above the average, Tina’s score was far better with respect to the other students in her section.

Exercise 3: Standardize Data


A pizza parlor chain claims a large pizza has 8 oz. of cheese with a standard deviation of 0.5 oz. An inspector ordered a pizza and found it only had 6.9 oz. of cheese. Franchisee’s can lose their store if they make pizzas with 3 standard deviations (or more) of cheese below the mean. Assume the distribution of weights is normally distributed.

Exercise 4: Standardize Data


Exercise 4 (contd.): Standardize Data


a) Find the z-score for x = 6.9 oz. of cheese.b) Is the franchisee in danger of losing its store? Why?c) Find the minimum amount of cheese a franchisee can put on a

large pizza so it is not in danger of losing its store. b) No.

𝜇=8 ,𝜎=0.5

Practice #1: Area Under Standard Normal Curve


Practice #1:Find the area under the standard normal curve

Lower Upper Area Type

Bound BoundLeft Tail

Right Tail

Between Bounds

1) -3 3 0.99732) 1 0.1587 3) 0 2.53 0.49434) 2.53 0.0057 5) -2.34 0.0096 6) -2 2 0.9545

Finding Areas Under The Normal Curve


The probability that z equals a certain number is always 0.P(z = a) = 0

Therefore, < and ≤ can be used interchangeably. Similarly, > and ≥ can be used interchangeably. P(z < b) = P(z ≤ b)P(z > c) = P(z ≥ c)

Section 6.2 Summary

• Converted raw data to z scores• Converted z scores to raw data• Graphed the standard normal distribution• Found areas under the standard normal curve


Section 6.3

Areas Under Any Normal Curve



• Find probabilities for “standardized events”• Find a z score from a given normal probability

(inverse normal)• Use the inverse normal function to find guarantee

problems


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 μ = 500σ = 100

600μ = 500x


Probability and Normal Distributions

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

Normal Distribution

600μ =500

P(x < 600)

μ = 500 σ = 100

x


600 500 1100

xz

1μ = 0

μ = 0 σ = 1

z

P(z < 1)

Same Area


Exercise 1: Finding Probabilities for Normal Distributions

A survey indicates that people use their cellular phones an average of 1.5 years before buying a new one. The standard deviation is 0.25 year. A cellular phone user is selected at random. Find the probability that the user will use their current phone for less than 1 year before buying a new one. Assume that the variable x is normally distributed. (Source: Fonebak)


Solution: Finding Probabilities for Normal Distributions

P(x < 1) = P(z < -2.00) = 0.0228

Normal Distribution

1 1.5

P(x < 1)

μ = 1.5 σ = 0.25

x

1 1.5 20.25

xz



–2 0

μ = 0 σ = 1

z

P(z < –2)

0.0228


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

z1

x

24 4512

1.75

24 45

P(24 < x < 54)

x

Normal Distribution μ = 45 σ = 12

0.0401

54

z2

x

54 4512

0.75

–1.75z

Standard Normal Distribution μ = 0 σ = 1

0

P(–1.75 < z < 0.75)

0.75

0.7734



If 200 shoppers enter the store, how many shoppers would you expect to be in the store between 24 and 54 minutes?

Solution:Recall P(24 < x < 54) = 0.7333

200(0.7333) =146.66 (or about 147) shoppers



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 = 0.6915

z x

39 45

12 0.50

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



If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes?

Solution:Recall P(x > 39) = 0.6915

200(0.6915) =138.3 (or about 138) shoppers



Must specify the mean and standard deviation of the population, and the x-value(s) that determine the interval.


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Example: Using Technology to find a Proportion in Normal Distribution

Larson/Farber 5th ed

A certain set of normally distributed exam scores as a mean of 74 with a standard deviation of 5. What is the proportion of scores less than 82?

Standard Normal tables tell you the proportion of data that is below a given z-score in a normally distributed population.


If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? Use the TI-84 calculator.

Solution:P(x > 39) = normalcdf(-0.50,1E+99) = 0.6915 200(0.6915) =138.3 (or about 138) shoppers


Section 6.3

Inverse Normal Lookups


Find Values, Given a Probability

• In the previous section, we were given a normally distributed random variable x and we were asked to find a probability.

• In this section, we will be given a probability and we will be asked to find the value of the random variable x.

x z probability


Find Values, Given a Probability


Exercise 1: Finding a z-Score Given an Area

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

z 0z

0.3632

Solution:


Solution: Finding a z-Score Given an Area

• Locate 0.3632 in the body of the Standard Normal Table.

• The values at the beginning of the corresponding row and at the top of the column give the z-score.

The z-score is –0.35.


Inverse Normal Lookup viaTI-84


TI-84 Solution: Finding a z-Score

z = invNorm(0.3632) = -0.35.


z 0z

0.3632

Exercise 2: Finding a z-Score Given a Right Tail Area

Find the z-score that has 10.75% of the distribution’s area to its right.

z0z

0.1075

Solution:

1 – 0.1075 = 0.8925

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


Exercise 2: Finding a z-Score Given a Right Tail Area

• Locate 0.8925 in the body of the Standard Normal Table.

• The values at the beginning of the corresponding row and at the top of the column give the z-score.

The z-score is 1.24.


Exercise 3: Finding a z-Score Given a Left Tail Area via TI-84

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

z 0z

0.3632

Solution: z = invNorm(0.3632) = -0.35.


Exercise 4: Finding a z-Score Given a Percentile

Find the z-score that corresponds to P5.Solution:The z-score that corresponds to 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.

z 0z

0.05


Exercise 5: Find z Scores, Given Center Area

Find the z value such that 90% of the area under the standard normal curve lies between –z and z.


Solution: Find z Scores given Center Area


Exercise 5: Find z Scores given Center Area

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Looking in Table 5, we see that 0.0500 lies exactly between areas 0.0495 and 0.0505. The halfway value between z=1.65 and z=1.64 is z=1.645. Therefore, we conclude that 90% of the area under the standard normal curve lies between the z values -1.645 and 1.645.

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


Exercise 6: Finding an x-ValueA veterinarian records the weights of cats treated at a clinic. The weights are normally distributed, with a mean of 9 pounds and a standard deviation of 2 pounds. Find the weights x corresponding to z-scores of 1.96, –0.44, and 0.Solution: Use the formula x = μ + zσ

z = 1.96: x = 9 + 1.96(2) = 12.92 pounds

z = –0.44: x = 9 + (–0.44)(2) = 8.12 pounds

z = 0: x = 9 + (0)(2) = 9 poundsNotice 12.92 pounds is above the mean, 8.12 pounds is below the mean, and 9 pounds is equal to the mean.


Exercise 7: Finding a Specific Data ValueScores for the California Peace Officer Standards and Training test are normally distributed, with a mean of 50 and a standard deviation of 10. An agency will only hire applicants with scores in the top 10%. What is the lowest score you can earn and still be eligible to be hired by the agency?Solution:

An exam score in the top 10% is any score above the 90th percentile. Find the z-score that corresponds to a cumulative area of 0.90.


Solution: Finding a Specific Data ValueFrom the Standard Normal Table, the area closest to 0.9 is 0.8997. So the z-score that corresponds to an area of 0.9 is z = 1.28.


Solution: Finding a Specific Data Value

Using the equation x = μ + zσ

x = 50 + 1.28(10) = 62.8

The lowest score you can earn and still be eligible to be hired by the agency is about 63.


Solution via TI-84

x = invNorm(0.90,50,10) = 62.8

The lowest score you can earn and still be eligible to be hired by the agency is about 63.


Exercise 8: Finding a Specific Data ValueMagic Video Games Inc. sells expensive computer games and wants to advertise an impressive, full-refund warranty period. It has found that the mean life for its’ computer games is 30 months with a standard deviation of 4 months. If the life spans of the computer games are normally distributed, how long of a warranty period (to the nearest month) can be offered so that the company will not have to refund the price of more than 7% of the computer games?

:


Exercise 8: Finding a Specific Data ValueMagic Video Games Inc. sells expensive computer games and wants to advertise an impressive, full-refund warranty period. It has found that the mean life for its’ computer games is 30 months with a standard deviation of 4 months. If the life spans of the computer games are normally distributed, how long of a warranty period (to the nearest month) can be offered so that the company will not have to refund the price of more than 7% of the computer games?

:


Exercise 8: Finding a Specific Data Value


Interpretive Statement: The warranty period should be 24 months if the company does not want to refund the price of more than 7% of the games.

Exercise 8: Finding a Specific Data Value


Interpretive Statement: The warranty period should be 24 months if the company does not want to refund the price of more than 7% of the games.

Practice #2: Inverse Normal Lookups


.

Area Left TailRight Tail Center Area

µ σ Components Area Area A (1-A)/2 var value valuea) 0.0 1.0 0.50 0.32 0.82 a = 0.92 b) 0.0 1.0 0.94 0.03a = -1.88 c) 90.0 7.0 0.82 0.09z = -1.34 1.34

90.0 7.0 bL 80.61

90.0 7.0 bR 99.39d) 45.0 5.0 0.88 0.06z = -1.55 1.55

45.0 5.0 xL 37.23

45.0 5.0 xR 52.77 45.0 5.0 b = 7.77

Section 6.3 Summary

• Found probabilities for “standardized events”• Found a z score from a given normal probability

(inverse normal)• Used the inverse normal function to solve guarantee

problems


Section 6.4

Normal Approximations to Binomial Distributions



• Determine when the normal distribution can approximate the binomial distribution

• Find the continuity correction• Use the normal distribution to approximate binomial

probabilities


Normal Approximation to the Binomial

Normal Approximation to a Binomial• Binomial distribution: p = 0.25

• As n increases the histogram approaches a normal curve.© 2012 Pearson Education, Inc. All rights reserved. 92 of 105

1. Sixty-two percent of adults in the U.S. have an HDTV in their home. You randomly select 45 adults in the U.S. and ask them if they have an HDTV in their home.

Example: Approximating the Binomial

Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation.


Solution: Approximating the Binomial

• You can use the normal approximationn = 45, p = 0.62, q = 0.38

np = (45)(0.62) = 27.9nq = (45)(0.38) = 17.1

• Mean: μ = np = 27.9• Standard Deviation: 45 0.62 0.38 3.26 σ npq


2. Twelve percent of adults in the U.S. who do not have an HDTV in their home are planning to purchase one in the next two years. You randomly select 30 adults in the U.S. who do not have an HDTV and ask them if they are planning to purchase one in the next two years.

Example: Approximating the Binomial

Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation.


Solution: Approximating the Binomial

• You cannot use the normal approximationn = 30, p = 0.12, q = 0.88

np = (30)(0.12) = 3.6nq = (30)(0.88) = 26.4

• Because np < 5, you cannot use the normal distribution to approximate the distribution of x.


Correction for Continuity

• The binomial distribution is discrete and can be represented by a probability histogram.

• To calculate exact binomial probabilities, the binomial formula is used for each value of x and the results are added.

• Geometrically this corresponds to adding the areas of bars in the probability histogram.


Correction for Continuity• When you use a continuous normal distribution to

approximate a binomial probability, you need to move 0.5 unit to the left and right of the midpoint to include all possible x-values in the interval (continuity correction).


Exact binomial probability

P(r = c) P(c – 0.5 < x < c + 0.5)

c c – 0.5 c c + 0.5

Normal approximation

Correction for Continuity

Exercise 1: Using a Correction for Continuity

Use a continuity correction to convert the binomial interval to a normal distribution interval.

1. The probability of getting between 270 and 310 successes, inclusive.

Solution:• The discrete midpoint values are 270, 271, … 310• The corresponding interval for the continuous normal

distribution is269.5 < x < 310.5




2. The probability of getting at least 158 successes.

Solution:• The discrete midpoint values are 158, 159, 160 …. • The corresponding interval for the continuous normal

distribution is x 157.5




3. The probability of getting fewer than 63 successes.

Solution:• The discrete midpoint values are …, 60, 61, 62.• The corresponding interval for the continuous normal

distribution is x < 62.5

© 2012 Pearson Education, Inc. All rights reserved.102 of 105

Using the Normal Distribution to Approximate Binomial Probabilities

1. Verify that the binomial distribution applies.

2. Determine if you can use the normal distribution to approximate x, the binomial variable.

3. Find the mean µ and standard deviation σ for the distribution.

npq np

Is np > 5?Is nq > 5?

Specify n, p, and q. In Words In Symbols


Using the Normal Distribution to Approximate Binomial Probabilities

4. Apply the appropriate continuity correction. Shade the corresponding area under the normal curve.

5. Find the correspondingz-score(s).

6. Find the probability.

xz

Add or subtract 0.5 from endpoints.

Use the Standard Normal Table.

In Words In Symbols


Exercise 4: Approximating a Binomial Probability

Sixty-two percent of adults in the U.S. have an HDTV in their home. You randomly select 45 adults in the U.S. and ask them if they have an HDTV in their home. What is the probability that fewer than 20 of them respond yes? (Source: Opinion Research Corporation)

Solution:• Can use the normal approximation (see slide 91) μ = 45 (0.62) = 27.9 450.620.38 3.26


Solution: Approximating a Binomial Probability

z x

19.5 27.9

3.26 2.58


Normal Distribution μ = 27.9 σ ≈ 3.26

0.0049–2.58 μ = 0

P(z < –2.58)

Standard Normalμ = 0 σ = 1

z


A survey reports that 62% of Internet users use Google Chrome as their browser. You randomly select 150 Internet users and ask them whether they use Chrome as their browser. What is the probability that exactly 96 will say yes?

Solution:• Can use the normal approximation np = 150∙0.62 = 93 > 5 nq = 150∙0.38 = 57 > 5

150 0.62 0.38 5.94 σ μ = 150∙0.62 = 93


Solution: Approximating a Binomial Probability

z1

x

95.5 935.94

0.42

0.6628 z2

x

96.5 935.94

0.59

0.59μ = 0

P(0.42 < z < 0.59)

Standard Normalμ = 0 σ = 1

z

0.42

0.7224

P(0.42 < z < 0.59) = 0.7224 – 0.6628 = 0.0596© 2012 Pearson Education, Inc. All rights reserved. 108 of 105

Normal Distribution μ = 27.9 σ = 3.26

The owner of a new apartment building needs to have 25 new water heaters installed. Assume the probability that a water heater will last 10 years is 0.25. a) What is the probability that 8 or more will last at

least 10 years? Use the binomial distributionb) Can the binomial probability distribution be

approximated by a normal distribution? Explain.c) If so, use a normal distribution to approximate the

binomial distributiond) Find the error between the two calculations.





𝑃 (𝑟 ≥ 8 )=𝑃 (8 ≤𝑟 )=𝑃 (7.5 ≤𝑥 )

• What is the probability that 8 or more will last at least 10 years? Use the binomial distribution.

• Can the binomial probability distribution be approximated by a normal distribution? Explain.

• Yes.



c) If so, use a normal distribution to approximate the binomial distribution

0.58 z ) = normalcdf(0.58, 1E+99) d) Find the error between the two calculations.0.2810 – 0.2735 = 0.0075. The normal approximation to the binomial is quite good.



Section 6.4 Summary

• Determined when the normal distribution can approximate the binomial distribution

• Found the continuity correction• Used the normal distribution to approximate binomial

probabilities


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