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Applications of the Normal Distribution

Applications of the Normal Distribution. Section 6.4 . Objectives. Find the probabilities for a normally distributed variable by transforming it into a standard normal variable Find specific data values given the percentages, using the standard normal distribution. Key Concept.

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Applications of the Normal Distribution

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  1. Applications of the Normal Distribution Section 6.4

  2. Objectives • Find the probabilities for a normally distributed variable by transforming it into a standard normal variable • Find specific data values given the percentages, using the standard normal distribution

  3. Key Concept • This section presents methods for working with normal distributions that are not standard (NON-STANDARD). That is the mean, m, is not 0 or the standard deviation, s is not 1 or both. • The key concept is that we transform the original variable, x, to a standard normal distribution by using the following formula:

  4. Conversion Formula

  5. x -  z =  Converting to Standard Normal Distribution P P  x z 0 (a) (b)

  6. Cautions!!!! • Don’t confuse z-scores and areas • Z-scores are distances on the horizontal scale. Table E lists z-scores in the left column and across the top row • Areas (probabilities or percentages) are regions UNDER the normal curve. Table E lists areas in the body of the table • Choose the correct (left/right) of the graph • Negative z-score implies it is located to the left of the mean • Positive z-score implies it is located to the right of the mean • Area less than 50% is to the left, while area more than 50% is to the right • Areas (or probabilities) are positive or zero values, but they are never negative

  7. Finding Areas given a specified variable, x • Draw a normal distribution curve labeling the mean and x • Shade the area desired • Convert x to standard normal distribution using z-score formula • Use procedures on page 287-288

  8. Example • According to the American College Test (ACT), results from the 2004 ACT testing found that students had a mean reading score of 21.3 with a standard deviation of 6.0. Assuming that the scores are normally distributed: • Find the probability that a randomly selected student has a reading ACT score less than 20 • Find the probability that a randomly selected student has a reading ACT score between 18 and 24 • Find the probability that a randomly selected student has a reading ACT score greater than 30

  9. Example • Women’s heights are normally distributed with a mean 63.6 inches and standard deviation 2.5 inches. The US Army requires women’s heights to be between 58 inches and 80 inches. Find the percentage of women meeting that height requirement. Are many women being denied the opportunity to join the Army because they are too short or too tall?

  10. Finding variable x given a specific probability • Draw a normal distribution labeling the given probability (percentage) under the curve • Using Table E, find the closest area (probability) to the given and then identify the corresponding z-score (WATCH SIGN!!!!) • Substitute z, mean, and standard deviation in z-score formula and solve for x.

  11. Example • According to the American College Test (ACT), results from the 2004 ACT testing found that students had a mean reading score of 21.3 with a standard deviation of 6.0. Assuming that the scores are normally distributed: • Find the 75th percentile for the ACT reading scores

  12. Example • The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. • One classical use of the normal distribution is inspired by a letter to “Dear Abby” in which a wife claimed to have given birth 308 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting 308 days or longer. What does this result suggest? • If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 4%, find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care

  13. Example • Men’s heights are normally distributed with a mean of 69.0 inches and standard deviation of 2.8 inches. • The standard casket has an inside length of 78 inches • What percentage of men are too tall to fit in a standard casket? • A manufacturer of caskets wants to reduce production costs by making smaller caskets. What inside length would fit all men except the tallest 1%?

  14. Assignment • Page 312 #1-27 odd

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