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5.3 Applications of Normal Distributions

5.3 Applications of Normal Distributions. Statistics Mrs. Spitz Fall 2008. Objectives/Assignment. How to compare data from two normal distributions How to find probabilities for normally distributed variables using a table and using technology

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5.3 Applications of Normal Distributions

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  1. 5.3 Applications of Normal Distributions Statistics Mrs. Spitz Fall 2008

  2. Objectives/Assignment • How to compare data from two normal distributions • How to find probabilities for normally distributed variables using a table and using technology • How to find a specific data entry of a normal distribution given the probability. Assignment: pp. 217-219 #1-16

  3. Comparing Normal Distributions • In 5.2, you learned how to transform a normally distributed variable, x to a z-score using the equation

  4. continued • You also learned to find areas under the standard normal curve. In this section, you will learn how to apply those skills. For instance, transforming x-values to z-scores allows you to compare variables in two normal distributions as shown in the following example.

  5. Ex. 1: Comparing scores from two distributions • The Graduate Record Exam (GRE) and the Miller Analogy Test (MAT) are tests that graduate schools use to evaluate applicants. GRE scores are normally distributed with = 1500 and =300, while MAT scores are normally distributed with =50 and =5. You decide to take both tests. You score 1875 on the GRE and a 57 on the MAT. On which test did you score better? Explain

  6. Solution: You can transform each score to a standard z-score to determine which is better. The area to the left of the z-score is the percentile. The area to the left of z=1.25 is .8944 and the area to the left of z=1.4 is .9222. The MAT z-score is in the 92nd percentile which is greater than the GRE z-score (89th percentile). So you scored better on the MAT.

  7. Probability and Normal Distributions • To find the area under any normal curve, first convert each data value to a z-score. Then use the standard normal distribution. For instance, consider a normal curve with = 500 and = 100, as shown at the upper left. The value of x one standard deviation above the mean is  +  = 500 + 100 = 600. • If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the ara under the normal curve for the given interval.

  8. Probability and Normal Distributions • Now consider the standard normal curve shown at the lower left. The v alue of z one standard deviation above the mean is  + =0 + 1= 1. Because a z-score of 1 corresponds to an x-value of 600 and areas are not changed with a transformation to a standard normal curve, the shaded areas in the graph are equal

  9. Ex. 2: Finding probabilities for Normal Distributions • A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is .5 year. If a computer owner is selected at random, find the probability that he or she will use it for less than two years before upgrading. Assume that the variable x is normally distributed SOLUTION: The graph shows a normal curve with = 2.5 and  = .5 and a shaded area for x less than 2. The z-score that corresponds to two years is:

  10. Ex. 2: Finding probabilities for Normal Distributions • Using the Standard Normal Table, P(z x -0.8) = 0.2119. The probability that the computer will be upgraded in less than 2 years is 0.2119. So, 21.19% of new owners will upgrade in less than two years. SOLUTION: The graph shows a normal curve with = 2.5 and  = .5 and a shaded area for x less than 2. The z-score that corresponds to two years is:

  11. Ex. 3: Finding probabilities for Normal Distributions • 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. 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 the lengths of time 1) between 24 and 54 minutes and 2) more than 39 minutes. SOLUTION: The graph shows a normal curve with = 45 minutes and  = 12 minutes. The area for x between 24 and 54 minutes is shaded. The z-score that corresponds to 24 minutes and to 54 minutes are:

  12. Ex. 3: Finding probabilities for Normal Distributions (continued) • So the probability that a shopper will be in the store between 24 and 54 minutes is: Another way of interpreting this probability is to say that 77.33% of the shoppers will be in the store between 24 and 54 minutes

  13. Ex. 3: Finding probabilities for Normal Distributions • The graph at the right shows a normal curve with  = 45 minutes and  = 12 minutes. The area for x greater than 39 minutes is shaded. The z-score that corresponds to 39 minutes is: So, the probability that a shopper will be in the store more than 39 minutes is:

  14. Ex. 4: Using technology to find Normal Distributions Another way to find normal distributions is using a calculator or computer. You can find normal probabilities with a TI-83 or TI-84 or using Excel. • Cholesterol levels of American men are normal distributed with a mean of 215 and a standard deviation of 25. If you randomly select an American man, what is the probability that his cholesterol level is less than 175? Use a calculator or computer to find the probability. SOLUTION: The TI-83 has a feature that allows you to find normal probabilities without first converting to a standard z-score. You must specify the mean and standard deviation of the population as well as the x-value(s) that determine the interval.

  15. Ex. 4: Using technology to find Normal Distributions 2nd VARS - #2 Type in 0 followed by 175 (what you are looking for) 215 (the mean) and 25 (Standard Deviation. SOLUTION: From the display, you can see that the probability that his cholesterol is less than 175 is about 0.055, or 5.5%.

  16. Ex. 5: Finding a Specific Data Value You can also use the normal distribution to find a specific data value (x-value) for a given probability shown in ex. 5. Scores for a civil service exam are normally distributed with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment?

  17. Ex. 5: Finding a Specific Data Value - continued Exam scores in the top 5% correspond to the shaded region shown. An exam score in the top 5% is any score above the 95th percentile. To find the score that represents the 95th percentile, you must first find the z-scores that correspond to a cumulative area of 0.95. From the Standard Normal Table, you can find that the areas closest to 0.95 are 0.9495 (z=1.64) and .9508 (z=1.65). Because 0.95 is halway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is z=1.645.

  18. Ex. 5: Finding a Specific Data Value - continued Using the equation, x =  + z x = 75 + 1.645(6.5) x ≈85.69 So, the lowest score you can earn and still be eligible for employment is 86.

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