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Answer :
Final answer:
a. The percent of the population weighing between 165 and 170 pounds is approximately 15.68%. b. Approximately 12.10% of the population would be expected to weigh more than 182 pounds. c. Out of a group of 9000 people, approximately 1090 would be expected to weigh more than 182 pounds.
Explanation:
a. To find the percent of the population that weighs between 165 and 170 pounds, we need to calculate the z-scores for both weights and use a standard normal distribution table. The formula for calculating z-score is z = (x - μ) / σ, where x is the weight, μ is the mean, and σ is the standard deviation. Once we have the z-scores, we can find the corresponding percentiles using the standard normal distribution table. For 165 pounds, z = (165 - 175) / 6 = -1.67, and for 170 pounds, z = (170 - 175) / 6 = -0.83. Looking up these z-scores in the table, we find that the percentile for -1.67 is 0.0465 and the percentile for -0.83 is 0.2033. To find the percent between these two weights, we subtract the lower percentile from the higher percentile: 0.2033 - 0.0465 = 0.1568, or 15.68%.
b. To find the percent of the population that weighs more than 182 pounds, we calculate the z-score for 182 pounds using the formula: z = (x - μ) / σ = (182 - 175) / 6 = 1.17. Looking up this z-score in the table, we find that the percentile is 0.8790. To find the percent weighing more than 182 pounds, we subtract this percentile from 1: 1 - 0.8790 = 0.1210 or 12.10%.
c. To find the number of people expected to be heavier than 182 pounds out of a group of 9000, we can use the percent from part b. We expect approximately 0.1210 (or 12.10%) of the population to weigh more than 182 pounds. Multiplying this percent by the total population gives us the expected number: 0.1210 * 9000 = 1090. Therefore, we would expect around 1090 people to be heavier than 182 pounds.
d. To find the percent of the population that weighs between 172 and 180 pounds, we calculate the z-scores for both weights using the formula: z = (x - μ) / σ. For 172 pounds, z = (172 - 175) / 6 = -0.50, and for 180 pounds, z = (180 - 175) / 6 = 0.83. Looking up these z-scores in the table, we find the corresponding percentiles: 0.3085 for -0.50 and 0.7967 for 0.83. To find the percent between these two weights, we subtract the lower percentile from the higher percentile: 0.7967 - 0.3085 = 0.4882, or 48.82%.
e. To find the percent of the population that weighs heavier than 163 pounds, we calculate the z-score for 163 pounds using the formula: z = (x - μ) / σ = (163 - 175) / 6 = -2.00. Looking up this z-score in the table, we find that the percentile is 0.0228. To find the percent weighing heavier than 163 pounds, we subtract this percentile from 1: 1 - 0.0228 = 0.9772 or 97.72%.
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