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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 33 residents and found the mean weight to be 162 pounds with a standard deviation of 35 pounds. At the 95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write +- ). Answer:

A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 3333 residents and found the mean weight to be 162162 pounds with a standard deviation of 3535 pounds. At the 95% 95 \% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write ± \pm ).\newlineAnswer:

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Q. A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 3333 residents and found the mean weight to be 162162 pounds with a standard deviation of 3535 pounds. At the 95% 95 \% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write ± \pm ).\newlineAnswer:
  1. Calculate Z-score: To calculate the margin of error at the 9595% confidence level using the normal distribution, we need to use the formula for the margin of error (ME) which is:\newlineME = Z×(σ/n)Z \times (\sigma/\sqrt{n})\newlinewhere ZZ is the Z-score corresponding to the desired confidence level, σ\sigma is the standard deviation, and nn is the sample size.
  2. Find Margin of Error Formula: First, we need to find the Z-score that corresponds to the 95%95\% confidence level. For a normal distribution, the Z-score for a 95%95\% confidence level is approximately 1.961.96. This value is obtained from a Z-table or standard normal distribution table.
  3. Plug in Values: Next, we plug in the values into the margin of error formula:\newlineME=1.96×(3533)ME = 1.96 \times \left(\frac{35}{\sqrt{33}}\right)
  4. Calculate Square Root: Now, we calculate the square root of the sample size, which is 33\sqrt{33}.
  5. Divide Standard Deviation: The square root of 3333 is approximately 5.74465.7446.
  6. Multiply by Z-score: We then divide the standard deviation by the square root of the sample size: 35/5.7446≈6.092235 / 5.7446 \approx 6.0922
  7. Final Margin of Error: Finally, we multiply this result by the Z-score to find the margin of error: ME=1.96×6.0922≈11.9407ME = 1.96 \times 6.0922 \approx 11.9407
  8. Round to Nearest Tenth: Rounding to the nearest tenth, the margin of error is approximately 11.911.9 pounds.

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