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Bluman, Chapter 6 1
Test 6 will be on Thursday Dec 13, 2012
ALL HOMEWORK DUE TUESDAY DEC 11
1Friday, January 25, 13
Bluman, Chapter 6
6.4 The Normal Approximation to the Binomial DistributionA 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.
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Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution
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3Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution
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4Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution In addition, a correction for continuity may be
used in the normal approximation to the binomial.
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4Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution 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.
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Correction for Continuity
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Correction for Continuity
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomial
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NormalUse: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)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:
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NormalUse: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)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:
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NormalUse:
6Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:
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NormalUse:P(a – 0.5 < X < a + 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)
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NormalUse:P(a – 0.5 < X < a + 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)P(X > a + 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)P(X > a + 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)P(X > a + 0.5)P(X < a + 0.5)
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Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)P(X ≤ a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)P(X > a + 0.5)P(X < a + 0.5)
6Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)P(X ≤ a)P(X < a)
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NormalUse:P(a – 0.5 < X < a + 0.5)P(X > a – 0.5)P(X > a + 0.5)P(X < a + 0.5)
6Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)P(X ≤ a)P(X < a)
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NormalUse: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)
6Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial DistributionBinomialWhen finding:P(X = a)P(X ≥ a)P(X > a)P(X ≤ a)P(X < a)
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NormalUse: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)
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Bluman, Chapter 6 7
7Friday, January 25, 13
Bluman, Chapter 6 7
7Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution
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8Friday, January 25, 13
Bluman, Chapter 6
The Normal Approximation to the Binomial Distribution
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Procedure TableStep 1: Check to see whether the normal approximation
can be used.Step 2: Find the mean µ and the standard deviation σ.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.
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Bluman, Chapter 6
9Friday, January 25, 13
Bluman, Chapter 6
Ex. 3: Approximating a Binomial Probability
9Friday, January 25, 13
Bluman, Chapter 6
Ex. 3: Approximating a Binomial ProbabilityThirty-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?
9Friday, January 25, 13
Bluman, Chapter 6
Ex. 3: Approximating a Binomial ProbabilityThirty-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 µ = 5.55 and σ ≈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 µ = 5.55 and σ ≈1.87 and a shaded area to the left of 7.5. The z-score that corresponds to x = 7.5 is
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Continued . . .
10Friday, January 25, 13
Continued . . .
10Friday, January 25, 13
Continued . . .
10Friday, January 25, 13
Continued . . .
Using the Standard Normal Table,
10Friday, January 25, 13
Continued . . .
Using the Standard Normal Table,
P (z<1.04) = 0.8508
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Continued . . .
Using the Standard Normal Table,
P (z<1.04) = 0.8508
So, the probability that fewer than eight people respond yes is 0.8508
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Bluman, Chapter 6
11Friday, January 25, 13
Bluman, Chapter 6
Ex. 4: Approximating a Binomial Probability
11Friday, January 25, 13
Bluman, Chapter 6
Ex. 4: Approximating a Binomial ProbabilityTwenty-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?
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Bluman, Chapter 6
Ex. 4: Approximating a Binomial ProbabilityTwenty-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
11Friday, January 25, 13
Bluman, Chapter 6
Ex. 4: Approximating a Binomial ProbabilityTwenty-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
11Friday, January 25, 13
Bluman, Chapter 6
Ex. 4: Approximating a Binomial ProbabilityTwenty-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
and
11Friday, January 25, 13
Bluman, Chapter 6
Ex. 4: Approximating a Binomial ProbabilityTwenty-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
and
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Bluman, Chapter 6
12Friday, January 25, 13
Bluman, Chapter 6
Ex. 4 Continued
12Friday, January 25, 13
Bluman, Chapter 6
Ex. 4 ContinuedUsing 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 µ = 58 and σ = 6.42, and a shaded area to the right of 49.5.
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Bluman, Chapter 6
Ex. 4 ContinuedUsing 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 µ = 58 and σ = 6.42, and a shaded area to the right of 49.5.
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Bluman, Chapter 6
13Friday, January 25, 13
Bluman, Chapter 6
Ex. 4 Continued
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Bluman, Chapter 6
Ex. 4 ContinuedThe z-score that corresponds to 49.5 is
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Bluman, Chapter 6
Ex. 4 ContinuedThe z-score that corresponds to 49.5 is
13Friday, January 25, 13
Bluman, Chapter 6
Ex. 4 ContinuedThe 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
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Bluman, Chapter 6
Chapter 6Normal Distributions
Section 6-4Example 6-16Page #343
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14Friday, January 25, 13
Bluman, Chapter 6
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.
Example 6-16: Reading While Driving
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15Friday, January 25, 13
Bluman, Chapter 6
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.
Example 6-16: Reading While Driving
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15Friday, January 25, 13
Bluman, Chapter 6
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) = 282Since np ≥ 5 and nq ≥ 5, we can use the normal distribution.
Example 6-16: Reading While Driving
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15Friday, January 25, 13
Bluman, Chapter 6
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) = 282Since 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
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15Friday, January 25, 13
Bluman, Chapter 6
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) = 282Since 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
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15Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
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16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
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P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe area between the two z values is 0.9656 - 0.9429 = 0.0227, or 2.27%.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe area between the two z values is 0.9656 - 0.9429 = 0.0227, or 2.27%.
Example 6-16: Reading While Driving
16
P(X = 25)
16Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(24.5 < X < 25.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
16
P(X = 25)
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Bluman, Chapter 6
Chapter 6Normal Distributions
Section 6-4Example 6-17Page #343
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Bluman, Chapter 6
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.
Example 6-17: Widowed Bowlers
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18Friday, January 25, 13
Bluman, Chapter 6
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.
Example 6-17: Widowed Bowlers
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18Friday, January 25, 13
Bluman, Chapter 6
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) = 180Since np ≥ 5 and nq ≥ 5, we can use the normal distribution.
Example 6-17: Widowed Bowlers
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18Friday, January 25, 13
Bluman, Chapter 6
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) = 180Since 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
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18Friday, January 25, 13
Bluman, Chapter 6
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) = 180Since 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
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Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
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19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
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P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe area to the right of the z value is 1.0000 - 0.0066 = 0.9934, or 99.34%.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe area to the right of the z value is 1.0000 - 0.0066 = 0.9934, or 99.34%.
Example 6-17: Widowed Bowlers
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Step 3: Write in probability notation.Step 4: Rewrite using the continuity correction factor.
P(X > 9.5)Step 5: Find the corresponding z values.
Step 6: Find the solutionThe 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
19
P(X ≥ 10)
19Friday, January 25, 13
Bluman, Chapter 6
Study Tip
20Friday, January 25, 13
Bluman, Chapter 6
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.
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Bluman, Chapter 6
homework
Sec 6.4 page 346 #5-13 odds
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