bluman, chapter 6 1 - navimath · the normal approximation to the binomial distribution bluman,...

Bluman, Chapter 6 1

Review the following from

Chapter 5

A surgical procedure has an 85% chance of

success and a doctor performs the

procedure on 10 patients, find the following:

a) The probability that the procedure was

successful on exactly two patients?

b) The mean, variance and standard

deviation of number of successful surgeries.

Bluman, Chapter 6 2

Approximating a Binomial Distribution

But what if the doctor performs the surgical procedure on 150 patients and you want to find the probability of fewer than 100 successful surgeries?

To do this using the techniques described in chapter 5, you would have to use the binomial formula 100 times and find the sum of the resulting probabilities. This is not practical and a better approach is to use a normal distribution to approximate the binomial distribution.

6.4 The Normal Approximation to

the Binomial Distribution

A normal distribution is often used to solve

problems that involve the binomial distribution

since when n is large (say, 100), the calculations

are too difficult to do by hand using the binomial

distribution.

Bluman, Chapter 6 4

The Normal Approximation to the

Binomial Distribution

The normal approximation to the binomial

is appropriate when np > 5 and nq > 5 .

I. 𝝁 = 𝒏𝒑

II. 𝝈 = 𝒏𝒑𝒒 𝒘𝒉𝒆𝒓𝒆 𝒒 = 𝟏 − 𝒑

See page 274 for above formulas.

Bluman, Chapter 6 5



In addition, a correction for continuity may be

used in the normal approximation to the

binomial.

The continuity correction means that for any

specific value of X, say 8, the boundaries of X

in the binomial distribution (in this case, 7.5 to

8.5) must be used.

Bluman, Chapter 6 6

Correction for Continuity

Websites

The following websites simulate the

concept.

http://opl.apa.org/contributions/Rice/rvls_sim

/stat_sim/normal_approx/index.html

http://www.rossmanchance.com/applets/Bin

omDist/BinomDist.html

Bluman, Chapter 6 8

http://opl.apa.org/contributions/Rice/rvls_sim/stat_sim/normal_approx/index.html

http://opl.apa.org/contributions/Rice/rvls_sim/stat_sim/normal_approx/index.html

http://www.rossmanchance.com/applets/BinomDist/BinomDist.html




Binomial Distribution; see page 342

Binomial

When finding:

P(X = a)

P(X a)

P(X > a)

P(X a)

P(X < a)

Bluman, Chapter 6 9

Normal

Use:

P(a – 0.5 < X < a + 0.5)

P(X > a – 0.5)

P(X > a + 0.5)

P(X < a + 0.5)

P(X < a – 0.5)

For all cases, , , 5, 5 np npq np nq

Bluman, Chapter 6 10




Procedure Table Step 1: Check to see whether the normal approximation

can be used.

Step 2: Find the mean µ and the standard deviation .

Step 3: Write the problem in probability notation, using X.

Step 4: Rewrite the problem by using the continuity

correction factor, and show the corresponding area

under the normal distribution.

Step 5: Find the corresponding z values.

Step 6: Find the solution.

Ex. 3: Approximating a Binomial Probability

Thirty-seven percent of Americans say they always fly an American flag on the Fourth of July. You randomly select 15 Americans and ask each if he or she flies an American flag on the Fourth of July. What is the probability that fewer than eight of them reply yes?

SOLUTION: From Example 1, you know that you can

use a normal distribution with = 5.55 and ≈1.87 to

approximate the binomial distribution. By applying the

continuity correction, you can rewrite the discrete

probability P(x < 8) as P (x < 7.5). The graph on the

next slide shows a normal curve with = 5.55 and

≈1.87 and a shaded area to the left of 7.5. The z-score

that corresponds to x = 7.5 is

Continued . . .

04.187.1

55.55.7

xz

Using the Standard Normal

Table,

P (z<1.04) = 0.8508

So, the probability that fewer than eight people respond yes is

0.8508

Ex. 4: Approximating a Binomial Probability

Twenty-nine percent of Americans say they are confident that passenger trips to the moon will occur during their lifetime. You randomly select 200 Americans and ask if he or she thinks passenger trips to the moon will occur in his or her lifetime. What is the probability that at least 50 will say yes?

SOLUTION: Because np = 200 ● 0.29 = 58 and nq =

200 ● 0.71 = 142, the binomial variable x is

approximately normally distributed with

58 np

42.671.029.0200 npq

and

Ex. 4 Continued

Using the correction for continuity, you can rewrite

the discrete probability P (x ≥ 50) as the

continuous probability P ( x ≥ 49.5). The graph

shows a normal curve with = 58 and = 6.42,

and a shaded area to the right of 49.5.

Ex. 4 Continued

The z-score that corresponds to 49.5 is

So, the probability that at least 50 will say yes is:

P(x ≥ 49.5) = 1 – P(z -1.32)

= 1 – 0.0934

= 0.9066

32.142.6

585.49

xz

Chapter 6

Normal Distributions

Section 6-4

Example 6-16

Page #343


A magazine reported that 6% of American drivers read the

newspaper while driving. If 300 drivers are selected at

random, find the probability that exactly 25 say they read

the newspaper while driving.

Here, p = 0.06, q = 0.94, and n = 300.

Step 1: Check to see whether a normal approximation can

be used.

np = (300)(0.06) = 18 and nq = (300)(0.94) = 282

Since np 5 and nq 5, we can use the normal distribution.

Step 2: Find the mean and standard deviation.

µ = np = (300)(0.06) = 18

Example 6-16: Reading While Driving


300 0.06 0.94 4.11 npq

Step 3: Write in probability notation.

Step 4: Rewrite using the continuity correction factor.

P(24.5 < X < 25.5)


Step 6: Find the solution

The area between the two z values is

0.9656 - 0.9429 = 0.0227, or 2.27%.

Hence, the probability that exactly 25 people read the

newspaper while driving is 2.27%.

Example 6-16: Reading While Driving


24.5 18 25.5 181.58, 1.82

4.11 4.11

z z

P(X = 25)

Chapter 6

Normal Distributions

Section 6-4

Example 6-17

Page #343


Of the members of a bowling league, 10% are widowed. If

200 bowling league members are selected at random, find

the probability that 10 or more will be widowed.

Here, p = 0.10, q = 0.90, and n = 200.

Step 1: Check to see whether a normal approximation can

be used.

np = (200)(0.10) = 20 and nq = (200)(0.90) = 180

Since np 5 and nq 5, we can use the normal distribution.

Step 2: Find the mean and standard deviation.

µ = np = (200)(0.06) = 20

Example 6-17: Widowed Bowlers


200 0.10 0.90 4.24 npq

Step 3: Write in probability notation.

Step 4: Rewrite using the continuity correction factor.

P(X > 9.5)


Step 6: Find the solution

The area to the right of the z value is

1.0000 - 0.0066 = 0.9934, or 99.34%.

The probability of 10 or more widowed people in a random

sample of 200 bowling league members is 99.34%.

Example 6-17: Widowed Bowlers


9.5 202.48

4.24

z

P(X 10)

Study Tip

In a discrete distribution, there is a

difference between P (x ≥ c) and P( x > c).

This is true because the probability that x

is exactly c is not zero. In a continuous

distribution, however, there is no

difference between P (x ≥ c) and P (x >c)

because the probability that x is exactly c

is zero.

homework

Sec 6.4 page 346

#5-13 odds


bluman, chapter 6 1 - navimath · the normal approximation to the binomial distribution bluman,...

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