What is the probability that one randomly selected adult male will weigh less than 185 pounds, assuming the weight of adult males is normally distributed with a mean of 172 pounds and a standard deviation of 29 pounds?

The probability that a randomly selected adult male will weigh less than 185 pounds, if the weight of adult males is normally distributed with a mean weight of 172 pounds and a standard deviation of 29 pounds, is approximately 67.57%. This is calculated using the Z-score method and using statistical tables or a calculator for determining the cumulative probability associated with that Z-score.

Explanation:

This question involves a concept in probability and statistics known as the normal distribution. In a normally distributed variable, knowing the mean and standard deviation allows us to predict the likelihood of certain outcomes.

In this case, we need to calculate the probability that a randomly selected adult male weighs less than 185 pounds, given a mean weight of 172 pounds and a standard deviation of 29 pounds. First, we have to standardize the weight of 185 pounds into what is known as a Z-score. The Z-score is calculated as follows: (X - μ) / σ, where X is the random variable (185 pounds in this case), μ is the mean (172 pounds), and σ is the standard deviation (29 pounds).

By substituting our values into the formula, we get Z = (185-172)/29 = 0.45. This Z-score tells us that the weight of 185 pounds is 0.45 standard deviations above the mean.

Using standard statistical tables or a calculator with statistical functions, you can determine that a Z-score of 0.45 corresponds to a cumulative probability (the probability that a randomly selected value is less than or equal to X) of approximately 0.6757. Therefore, there's about a 67.57% chance that a randomly selected adult male will weigh less than 185 pounds.

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