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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 65 students in the high school and found a mean savings of 2700 dollars with a standard deviation of 700 dollars. At the
95% confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
+- ).
Answer:

A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of $65$ students in the high school and found a mean savings of $2700$ dollars with a standard deviation of $700$ dollars. At the $95 \%$ confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write $\pm$ ). $\newline$ Answer:

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Use normal distributions to approximate binomial distributions

Full solution

Q. A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of

65

students in the high school and found a mean savings of

2700

dollars with a standard deviation of

700

dollars. At the

95 \%

confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write

\pm

).

\newline

Answer:

Identify values and formula: Identify the values given in the problem and the formula to use for the margin of error at the $95$ % confidence level. $\newline$ We know: $\newline$ - The mean savings ( $\mu$ ) is $\$2700$ . $\newline$ - The standard deviation ( $\sigma$ ) is $\$700$ . $\newline$ - The sample size ( $n$ ) is $65$ students. $\newline$ - The z-score for a $95$ % confidence level is approximately $1.96$ for a normal distribution. $\newline$ The margin of error ( $E$ ) can be calculated using the formula: $\newline$ $E = z \times (\sigma/\sqrt{n})$
Calculate standard error: Calculate the standard error of the mean $\sigma/\sqrt{n}$ by dividing the standard deviation by the square root of the sample size. $\newline$ Standard error (SE) = $\sigma/\sqrt{n}$ $\newline$ SE = $700 / \sqrt{65}$ $\newline$ SE $\approx 700 / 8.0623$ $\newline$ SE $\approx 86.77$
Calculate margin of error: Calculate the margin of error (E) by multiplying the z-score by the standard error. $\newline$ $E = z \times SE$ $\newline$ $E = 1.96 \times 86.77$ $\newline$ $E \approx 170.07$
Round margin of error: Round the margin of error to the nearest whole number. $\newline$ $E \approx 170.07$ $\newline$ Rounded $E = 170$

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