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**15.4 The Normal Distribution Objectives:**

Understand the basic properties of the normal curve. Relate the area under a normal curve to z-scores. Make conversions between raw scores and Use the normal distribution to solve applied problems.

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**The Normal Distribution**

The normal distribution describes many real-life data sets. The histogram shown gives an idea of the shape of a normal distribution.

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**Normal Curve properties**

We represent the mean by μ and the standard deviation by σ.

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**Normal Curve properties cont. (2)**

The “68 – 95 – 99.7 rule” for a normal distribution.

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**The Normal Distribution Applications**

Example: Suppose that the distribution of scores for 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, a) how many scores do we expect to fall between 425 and 475? b) how many scores do we expect to fall above 500?

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**The Normal Distribution Applications cont. (2)**

a) Solution: 425 and 475 are each 1 standard deviation from the mean. Approximately 68% of the scores lie within 1 standard deviation of the mean. So we expect about 0.68 × 1,000 = 680 scores are in the range from 425 to 475.

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**The Normal Distribution Applications cont. (3)**

Solution: We know 5% of the scores lie more than 2 standard deviations above or below the mean, so (by symmetry) we expect to have 0.05 ÷ 2 = of the scores to be more than 2 standard deviations above the mean. For this distribution, μ + 2σ = (25) = 500. 2.5% of 1000 is 25 so we can expect * 1,000 = 25 of the scores to be above 500.

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**z-Scores Every raw score in a normal distribution has its own Z-score.**

The Z-score for a particular raw score is the number of standard deviations that the raw score 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.

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**Standard Normal Distribution**

The standard normal distribution has a mean of μ = 0 and a standard deviation of σ = 1. There are tables that give the area (which represents the percentage of data) under this curve between the mean and a particular z-score.

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Using z-Score tables 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: 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 Solution (a): The area under the curve between z = 0 and z = 1.3 is shown. Using a table we find this area for the z-score We find that A is So we expect 40.3%, of the data to fall between 0 and 1.3 standard deviations above the mean.

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**Using z-Score tables cont. (2)**

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 = Using a table we get A = when z = 2.1, and A = when z = The area is – = or 4.9%

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**Using z-Score tables cont. (3)**

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 = when z = Therefore, 46.6% of the data values lie between 0 and –1.83.

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**Using z-Scores on a calculator**

The equation for the standard normal distribution is Enter the function in Y= Set the WINDOW as Xmin = -5 Xmax = 5 Ymin = -0.2 Ymax = 0.5 To calculate the area under the normal distribution curve find the associated percentages, go to 2nd TRACE choose 7: f(x)dx and then enter the z-values.

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**Using z-Scores on a calculator cont. (2)**

Example: Use a calculator to verify the percentage of data (area under the curve) that lie in the following regions for a standard normal distribution: 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 40.3% 4.9% 46.6%

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**Converting Raw Scores to z-Scores**

See how far the raw score is from the mean. i.e. Take the difference x - μ See how many standard deviations are in that distance. i.e. Divide that difference by σ

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**Converting Raw Scores to z-Scores cont. (2)**

Example: Suppose the mean of a normal distribution is 20 and its standard deviation is 3. a) Find the z-score corresponding to the raw score 25. b) Find the z-score corresponding to the raw score 16.

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Applications 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 μ = 65 and a standard deviation of σ = 4. What percentage of those who took this test had a score below yours? Example: Consider the following information: 1911: Ty Cobb hit .420. Mean average was with standard deviation 1941: Ted Williams hit .406. Mean average was .267 with standard deviation 1980: George Brett hit .390. Mean average was .261 with standard deviation Assuming normal distributions, use z-scores to determine which of the three batters was ranked the highest in relationship to his contemporaries.

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Applications cont. (2) 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? 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 the z-score such that 95% of the area is below this score.

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Applications cont. (3) 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 % of the area underneath the standard normal curve falls below z = By symmetry, 95% of the values lie above –1.64. Since , we obtain Then solving for x gives 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.

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4.3 NORMAL PROBABILITY DISTRIBUTIONS

4.3 NORMAL PROBABILITY DISTRIBUTIONS

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