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Applications of the Normal Distribution Section 6.4 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 Objectives 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: Key Concept original x-value mean z standard deviation xm s Always round z-scores to 2 decimal places Conversion Formula Converting to Standard Normal Distribution z= x-m s P P (a) m x (b) 0 z 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 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 Cautions!!!! 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 Finding Areas given a specified variable, x 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 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? Example 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 Substitute z, mean, and standard deviation in z-score formula and solve for x. Finding variable x given a specific probability 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 Example