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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 103 residents and found the mean weight to be 159 pounds with a standard deviation of 33 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 103103 residents and found the mean weight to be 159159 pounds with a standard deviation of 3333 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 103103 residents and found the mean weight to be 159159 pounds with a standard deviation of 3333 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. Identify Z-score for 9595% Confidence: 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. Calculate 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. Substitute Values into Formula: Next, we plug in the values into the margin of error formula:\newlineME=1.96×(33103)ME = 1.96 \times (\frac{33}{\sqrt{103}})
  4. Calculate Square Root of Sample Size: Now, we calculate the square root of the sample size, which is 103\sqrt{103}. Using a calculator, we find that 103\sqrt{103} is approximately 10.1510.15.
  5. Divide Standard Deviation by Square Root: We then divide the standard deviation by the square root of the sample size: 3310.15≈3.25\frac{33}{10.15} \approx 3.25
  6. Multiply Result by Z-score: Finally, we multiply this result by the Z-score to find the margin of error:\newlineME=1.96×3.25≈6.37ME = 1.96 \times 3.25 \approx 6.37
  7. Round to Nearest Tenth: Since we need to round to the nearest tenth, the margin of error is approximately 6.46.4 pounds.

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