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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 178 graduating seniors and found the mean score to be 487 with a standard deviation of 119 . 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 $178$ graduating seniors and found the mean score to be $487$ with a standard deviation of $119$ . 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 $\pm$ ). $\newline$ Answer:

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Use normal distributions to approximate binomial distributions

Full solution

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

178

graduating seniors and found the mean score to be

487

with a standard deviation of

119

. 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

\pm

).

\newline

Answer:

Identify Given Information: Identify the given information and the formula to calculate the margin of error. $\newline$ We have the mean $\mu$ = $487$ , the standard deviation $\sigma$ = $119$ , and the sample size $n$ = $178$ . For a $95$ % confidence level, the z-score associated with it is approximately $1$ . $96$ (from the z-table or standard normal distribution table). $\newline$ The formula for the margin of error $E$ when using the z-score is: $\newline$ $E = z \times (\sigma/\sqrt{n})$
Calculate Margin of Error: Calculate the margin of error using the formula. $\newline$ $E = 1.96 \times (\frac{119}{\sqrt{178}})$ $\newline$ First, calculate the denominator: $\newline$ $\sqrt{178} \approx 13.3417$ $\newline$ Now, divide the standard deviation by the square root of the sample size: $\newline$ $\frac{119}{13.3417} \approx 8.9175$ $\newline$ Finally, multiply this result by the z-score: $\newline$ $E = 1.96 \times 8.9175 \approx 17.4799$
Round Margin of Error: Round the margin of error to the nearest tenth. $E \approx 17.5$

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