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Assume that men's weights are normally distributed with a mean of 172 pounds and a standard deviation of 30 pounds.

a) Find the probability that a randomly selected man weighs over 200 pounds.

Answer :

Final answer:

The probability that a randomly selected man weighs over 200 pounds is 0.1762, or 17.62%, found by converting the weight of 200 pounds to a z-score, using that score to find the appropriate area under the curve of the normal distribution, and subtracting that value from 1.

Explanation:

To find the probability that a randomly selected man weighs over 200 pounds, we need to convert this weight to a z-score first. The z-score is a measure of how many standard deviations an observation is above or below the mean. It is calculated using the formula: z = (X - μ)/σ, where X is the observed value, μ is the mean, and σ is the standard deviation.

Here, X is 200 pounds, μ (mean) is 172 pounds, and σ (standard deviation) is 30 pounds. So the z-score calculation would look like this:

z = (200 - 172)/30 = 0.93.

Next, we look this z-score up in a standard normal distribution table, or use a calculator or statistical software capable of finding the area under the curve to the right of this score. This will provide us with the probability that a randomly selected man weighs over 200 pounds.

The z-table tells us that the area to the LEFT of z = 0.93 corresponds to approximately 0.8238, which means there is an 82.38% chance a randomly selected man weighs 200 pounds or less. However, we are looking for people who weigh MORE than 200 pounds, so we need to subtract this from 1. Thus, the final probability that a man weighs more than 200 pounds is

1 - 0.8238 = 0.1762, or 17.62%.

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