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Suppose that people's weights are normally distributed, with mean 175 pounds and a standard deviation of 6 pounds.

a. What percent of the population would weigh between 165 and 170 pounds?
b. What percent of the population would you expect to weigh more than 182 pounds?
c. Out of a group of 9000 people, how many would you expect to be heavier than 182 pounds?
d. What percent of the population would you expect to weigh between 172 and 180 pounds?
e. What percent of the population would you expect to be heavier than 163 pounds?

User Dan Bliss
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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.

Step-by-step 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%.

User Sherif Ahmed
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