Post on 14-Jan-2016
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Dr. Fowler AFM Unit 8-4The Normal Distribution
• Understand the basic properties of the normal curve.• Relate the area under a normal curve to z-scores.
• Make conversions between raw scores and z-scores.• Use the normal distribution to solve applied problems.
The Normal Distribution
The normal distribution describes many real-life data sets. The histogram shown gives an idea ofthe shape of a normal distribution.
Write all Notes – All slides today – notes are short
The Normal Distribution
The Normal Distribution
We represent the mean by μ and the standard deviation by σ.
Section 15.4, Slide 6
• Example: Suppose that the distribution of scores of 1,000 students who take a standardized intelligence test is a normal distribution. If the distribution’s mean is 450 and its standard deviation is 25,
The Normal Distribution
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a) how many scores do we expect to fall between 425 and 475?
b) how many scores do we expect to fall above 500?
• Solution (a): 425 and 475 are each 1 standard deviation from the mean. Approximately 68% of the scores lie within 1 standard deviation of the mean. We expect about 0.68 × 1,000 = 680 scores are in the range 425 to 475.
The Normal Distribution
Solution (b): We know 5% of the scores lie more than 2 standard deviations above orbelow the mean, so we expect to have 0.05 ÷ 2 = 0.025 of the scores to be above 500. Multiplying by 1,000, we can expect that 0.025 * 1,000 = 25 scores to be above 500.
The Normal Distribution
z-Scores
The standard normal distribution has a mean of 0 and a standard deviation of 1.
There are tables (see next slide) that give the area under this curve between the mean and a number called a z-score. A z-score represents the number of standard deviations a data value is from the mean.
For example, for a normal distribution with mean 450 and standard deviation 25, the value 500 is 2 standard deviations above the mean; that is, the value 500 corresponds to a z-score of 2.
Section 15.4, Slide 11
Converting Raw Scores to z-Scores
• Example: Suppose you take a standardized test. Assume that the distribution of scores is normal and you received a score of 72 on the test, which had a mean of 65 and a standard deviation of 4. What percentage of those who took this test had a score below yours?
• Solution: We first find the z-score that corresponds to 72.
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Applications
z-ScoresTableOf NormalDistribution
Using the z-score table, we have that A = 0.4591 when z = 1.75. The normal curve is symmetric, so another 50% of the scores fall below the mean. So, there are 50% + 46% = 96% of the scores below 72.
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Applications
• Example: Consider the following information:1911: Ty Cobb hit .420. Mean average was .266
with standard deviation .0371.1941: Ted Williams hit .406. Mean average was .267 with standard deviation .0326.1980: George Brett hit .390. Mean average was .261 with standard deviation .0317.
Assuming normal distributions, use z-scores to determine which of the three batters was ranked the highest in relationship to his contemporaries.(continued on next slide)
Applications
• Solution:Ty Cobb’s average of .420 corresponded to a z-score of
Ted Williams’s average of .406 corresponded to a z-score of
George Brett’s average of .390 corresponded to a z-score of
Compared with his contemporaries, Ted Williams ranks as the best hitter.
Applications
Excellent Job !!!Well Done
Stop Notes for Today.Do Worksheet
• Example: Suppose the mean of a normal distribution is 20 and its standard deviation is 3.
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Converting Raw Scores to z-Scores
a) Find the z-score corresponding to theraw score 25.
b) Find the z-score corresponding to theraw score 16.
• Solution (a): We have
We compute
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Converting Raw Scores to z-Scores
Section 15.4, Slide 21
• Solution (b): We have
We compute
Converting Raw Scores to z-Scores
© 2010 Pearson Education, Inc. All rights reserved. Section 15.4, Slide 22
z-Scores
Command SummaryCalculates the inverse of the cumulative normal distribution function.
Command SyntaxinvNorm(probability[,μ, σ])
Menu LocationPress:2ND DISTR to access the distribution menu3 to select invNorm(, or use arrows.
© 2010 Pearson Education, Inc. All rights reserved. Section 15.4, Slide 23
z-Scores
• Example: Use a table to find the percentage of the data (area under the curve) that lie in the following regions for a standard normal distribution:
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a) between z = 0 and z = 1.3
b) between z = 1.5 and z = 2.1
c) between z = 0 and z = –1.83
z-Scores
• Solution (a): The area under the curve betweenz = 0 and z = 1.3 is shown. Using a table we find this area for the z-score 1.30.We find that A is 0.403
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when z = 1.30. We expect 40.3%, of the data to fall between 0 and 1.3 standard deviations above the mean.
Section 15.4, Slide 25
z-Scores
• Solution (b): The area under the curve between z = 1.5 and z = 2.1 is shown. We first find the area from z = 0 to z = 2.1 and then subtract the area from z = 0 to z = 1.5
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Using a table we get A = 0.482 when z = 2.1, and A = 0.433 when z = 1.5. The area is 0.482 – 0.433 = 0.049 or 4.9%
© 2010 Pearson Education, Inc. All rights reserved. Section 15.4, Slide 26
z-Scores
• Solution (c): Due to the symmetry of the normal distribution, the area between z = 0 and z = –1.83 is the same as the area between z = 0 and z = 1.83. Using a table, we see that A = 0.466 when z = 1.83. Therefore, 46.6% of the data values lie between 0 and –1.83.
© 2010 Pearson Education, Inc. All rights reserved. Section 15.4, Slide 27
• Example: A manufacturer plans to offer a warranty on an electronic device. Quality control engineers found that the device has a mean time to failure of 3,000 hours with a standard deviation of 500 hours. Assume that the typical purchaser will use the device for 4 hours per day. If the manufacturer does not want more than 5% to be returned as defective within the warranty period, how long should the warranty period be to guarantee this?
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Applications
• Solution: We need to find a z-score such that at least 95% of the area is beyond this point. This score is to the left of the mean and is negative.By symmetry we find thez-score such that 95% of the area is below this score.
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Applications
50% of the entire area lies below the mean, so our problem reduces to finding a z-score greater than 0 such that 45% of the area lies between the mean and that z-score. If A = 0.450, the corresponding z-score is 1.64. 95% of the area underneath the standard normal curve falls below z = 1.64. By symmetry, 95% of the values lie above –1.64.
Since , we obtain
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Applications
Solving the equation for x, we get
Owners use the device about 4 hours per day, so we divide 2,180 by 4 to get 545 days. This is approximately 18 months if we use 31 days per month. The warranty should be for roughly 18 months.
Applications