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------------------------------------------------ If the average (mean) mid-twenties male weighs 169 pounds and a weight of 151 is in the 31st percentile, what is the standard deviation of the weights in this age group? Round your answer to two decimal places.

To find the standard deviation of the weights in this age group, use the formula σ = (x - μ) / z, where x is the weight, μ is the mean, and z is the z-score. The standard deviation is approximately 45.23 pounds.

Explanation:

To find the standard deviation of the weights in this age group, we need to use the information provided. We know that the average weight is 169 pounds and a weight of 151 is in the 31st percentile.

First, we need to find the z-score for a weight of 151 using the formula z = (x - μ) / σ, where x is the weight, μ is the mean, and σ is the standard deviation. Rearranging the formula, we have σ = (x - μ) / z.

Using the percentile-to-z-score conversion table, we can find that the z-score corresponding to the 31st percentile is approximately -0.398. Plugging in the values, we have σ = (151 - 169) / -0.398 ≈ 45.23 pounds. Therefore, the standard deviation of the weights in this age group is approximately 45.23 pounds.

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