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A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 98 graduating seniors and found the mean score to be 492 with a standard deviation of 117 . 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 by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 9898 graduating seniors and found the mean score to be 492492 with a standard deviation of 117117 . 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 by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 9898 graduating seniors and found the mean score to be 492492 with a standard deviation of 117117 . 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: 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 9595% confidence level. For a 9595% confidence level in a normal distribution, the Z-score is approximately 1.961.96. This value can be found in Z-score tables or using a standard normal distribution calculator.
  3. Plug in Values: Next, we plug in the values we have into the margin of error formula:\newlineσ=117\sigma = 117 (standard deviation)\newlinen=98n = 98 (sample size)\newlineZ=1.96Z = 1.96 (Z-score for 9595% confidence)\newlineME=1.96×(117/98)ME = 1.96 \times (117/\sqrt{98})
  4. Calculate Margin of Error: Now we calculate the margin of error:\newlineME = 1.96×(11798)1.96 \times (\frac{117}{\sqrt{98}})\newlineME = 1.96×(1179.8995)1.96 \times (\frac{117}{9.8995}) (rounded 98\sqrt{98} to four decimal places)\newlineME = 1.96×11.81821.96 \times 11.8182 (rounded 1179.8995\frac{117}{9.8995} to four decimal places)\newlineME = 23.164523.1645 (rounded to four decimal places)
  5. Round Margin of Error: Finally, we round the margin of error to the nearest tenth: ME≈23.2ME \approx 23.2

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