To answer your questions, let's break down both parts:
a) Probability that an individual man's weight is greater than 180 lb:
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We need to convert the weight of 180 lb to a z-score using the formula: \[ z = \frac{X - \mu}{\sigma} \] where \(X\) is the weight (180 lb), \(\mu\) is the mean (172 lb), and \(\sigma\) is the standard deviation (29 lb).
\[ z = \frac{180 - 172}{29} \approx \frac{8}{29} \approx 0.276 \]
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Now, we can look up this z-score in a standard normal distribution table or use a calculator to find the probability:
- The probability that a randomly selected man weighs less than 180 lb is given by the cumulative distribution function (CDF) for \(z = 0.276\).
- From a standard normal table or calculator, the CDF value \(P(Z < 0.276) \approx 0.609\).
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To find the probability that a randomly selected man weighs greater than 180 lb, calculate: \[ P(X > 180) = 1 - P(X < 180) = 1 - 0.609 = 0.391 \]
Probability for part (a): Approximately 0.391 or 39.1%.
b) Probability that the mean weight of 20 randomly selected men is greater than 180 lb:
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For the mean of the sample, we'll use the Central Limit Theorem. According to the CLT, the distribution of the sample mean will be approximately normal if the sample size is large enough (n=20 is generally considered large enough).
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The mean of the sample means (\(\mu_{\bar{X}}\)) is the same as the population mean: \(\mu_{\bar{X}} = 172\).
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The standard deviation of the sample means (also called the standard error, \(SE\)) is: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{29}{\sqrt{20}} \approx \frac{29}{4.472} \approx 6.49 \]
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Now, calculate the z-score for a sample mean of 180 lb: \[ z = \frac{\bar{X} - \mu}{SE} = \frac{180 - 172}{6.49} \approx \frac{8}{6.49} \approx 1.23 \]
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Now find the cumulative probability for this z-score:
- From a standard normal table or calculator, \(P(Z < 1.23) \approx 0.8907\).
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To find the probability that the mean weight is greater than 180 lb: \[ P(\bar{X} > 180) = 1 - P(\bar{X} < 180) = 1 - 0.8907 \approx 0.1093 \]
Probability for part (b): Approximately 0.1093 or 10.93%.
Conclusion:
Now, regarding which question requires the Central Limit Theorem:
- Question a concerns individual weights, which follows the normal distribution directly as specified.
- Question b involves the mean weight of a sample of 20 men, which relies on the Central Limit Theorem to justify using the normal distribution for the sample mean.
Therefore, the answer to which question requires the use of the Central Limit Theorem is:
b
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