Questions: The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be greater than 200 pounds. A. 0.8815 B. 0.7613 C. 0.0985 D. 0.9015

The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be greater than 200 pounds. A. 0.8815 B. 0.7613 C. 0.0985 D. 0.9015
Transcript text: The average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (Source: American Dietetic Association). If a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be greater than 200 pounds. A. 0.8815 B. 0.7613 C. 0.0985 D. 0.9015
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Solution

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Solution Steps

Step 1: Given Information

We are given the following parameters:

  • Mean of the population (\( \mu \)): 196 pounds
  • Standard deviation of the population (\( \sigma \)): 22 pounds
  • Sample size (\( n \)): 50 individuals
  • Threshold for the sample mean: 200 pounds
Step 2: Calculate Z-scores

To find the probability that the sample mean is greater than 200 pounds, we first calculate the Z-score for the sample mean using the formula:

\[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]

Where:

  • \( \bar{X} \) is the sample mean (200 pounds)
  • \( \mu \) is the population mean (196 pounds)
  • \( \sigma \) is the population standard deviation (22 pounds)
  • \( n \) is the sample size (50)

Calculating the Z-score for the sample mean of 200 pounds:

\[ Z_{start} = \frac{200 - 196}{22 / \sqrt{50}} \approx 1.2856 \]

The Z-score for the upper limit is \( Z_{end} = \infty \).

Step 3: Calculate Probability

Using the Z-scores, we can find the probability that the sample mean is greater than 200 pounds:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(\infty) - \Phi(1.2856) \]

From the output, we have:

\[ P \approx 0.0993 \]

Thus, the probability that the sample mean is greater than 200 pounds is:

\[ P(X > 200) = 1 - P(X \leq 200) = 1 - 0.0993 \approx 0.9007 \]

Final Answer

The probability that the mean of the sample will be greater than 200 pounds is approximately \( \boxed{0.9007} \).

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