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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 199 graduating seniors and found the mean score to be 537 with a standard deviation of 82 . 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 $199$ graduating seniors and found the mean score to be $537$ with a standard deviation of $82$ . 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

199

graduating seniors and found the mean score to be

537

with a standard deviation of

82

. 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. $\newline$ Given: $\newline$ Mean $M = 537$ $\newline$ Standard Deviation $SD = 82$ $\newline$ Sample Size $n = 199$ $\newline$ Confidence Level = $95\%$ $\newline$ The margin of error $E$ at a certain confidence level can be calculated using the formula: $\newline$ $E = Z \times (SD / \sqrt{n})$ $\newline$ Where $Z$ is the Z-score corresponding to the desired confidence level. For a $95\%$ confidence level, the Z-score is typically $1.96$ .
Calculate Margin of Error: Calculate the margin of error using the formula. $\newline$ $E = 1.96 \times (\frac{82}{\sqrt{199}})$ $\newline$ First, calculate the denominator: $\newline$ $\sqrt{199} \approx 14.1067$ $\newline$ Then, divide the standard deviation by the square root of the sample size: $\newline$ $\frac{82}{14.1067} \approx 5.8114$ $\newline$ Finally, multiply this result by the Z-score: $\newline$ $E = 1.96 \times 5.8114 \approx 11.3911$ $\newline$ Round to the nearest tenth: $\newline$ $E \approx 11.4$

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