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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 142 residents and found the mean weight to be 164 pounds with a standard deviation of 26 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 142142 residents and found the mean weight to be 164164 pounds with a standard deviation of 2626 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 142142 residents and found the mean weight to be 164164 pounds with a standard deviation of 2626 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. Understand the problem: Understand the problem and what is being asked.\newlineWe need to calculate the margin of error for the mean weight of the residents in a town at the 95%95\% confidence level using the normal distribution/empirical rule.
  2. Identify the formula: Identify the formula to calculate the margin of error using the standard deviation and the z-score for the 95%95\% confidence level.\newlineThe margin of error (EE) can be calculated using the formula E=z×(σ/n)E = z \times (\sigma/\sqrt{n}), where zz is the z-score corresponding to the confidence level, σ\sigma is the standard deviation, and nn is the sample size.
  3. Find the z-score: Find the z-score corresponding to the 95%95\% confidence level.\newlineFor a 95%95\% confidence level, the z-score is typically 1.961.96. This value is obtained from a z-table or standard normal distribution table.
  4. Calculate the margin of error: Calculate the margin of error using the given values.\newlineWe have the standard deviation σ\sigma = 2626 pounds, the sample size nn = 142142, and the z-score for 9595% confidence zz = 1.961.96.\newlineNow, plug these values into the formula:\newlineE=z×(σ/n)E = z \times (\sigma/\sqrt{n})\newlineE=1.96×(26/142)E = 1.96 \times (26/\sqrt{142})
  5. Perform the calculations: Perform the calculations.\newlineFirst, calculate the denominator:\newline142≈11.916\sqrt{142} \approx 11.916\newlineNow, divide the standard deviation by the square root of the sample size:\newline2611.916≈2.181\frac{26}{11.916} \approx 2.181\newlineFinally, multiply this result by the z-score:\newlineE=1.96×2.181E = 1.96 \times 2.181\newlineE≈4.275E \approx 4.275
  6. Round the margin of error: Round the margin of error to the nearest tenth.\newlineE≈4.275E \approx 4.275 rounds to 4.34.3 when rounded to the nearest tenth.

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