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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 60 students in the high school and found a mean savings of 2500 dollars with a standard deviation of 1400 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 $60$ students in the high school and found a mean savings of $2500$ dollars with a standard deviation of $1400$ 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

60

students in the high school and found a mean savings of

2500

dollars with a standard deviation of

1400

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 Given Values and Formula: Identify the given values and the formula to use for the margin of error at the $95$ % confidence level. $\newline$ Given values: $\newline$ - Mean ( $\mu$ ) = $\$2500$ $\newline$ - Standard deviation ( $\sigma$ ) = $\$1400$ $\newline$ - Sample size ( $n$ ) = $60$ $\newline$ - Confidence level = $95\%$ $\newline$ For a normal distribution, the empirical rule states that approximately $95\%$ of the data falls within $2$ standard deviations of the mean. Therefore, we can use the z-score associated with a $95\%$ confidence level, which is approximately $\$2500$ $0$ . $\newline$ The formula for the margin of error ( $\$2500$ $1$ ) is: $\newline$ $\$2500$ $2$
Calculate Margin of Error: Calculate the margin of error using the formula. $\newline$ $E = 1.96 \times (\frac{1400}{\sqrt{60}})$ $\newline$ First, calculate the denominator: $\newline$ $\sqrt{60} \approx 7.746$ $\newline$ Then, divide the standard deviation by the square root of the sample size: $\newline$ $\frac{1400}{7.746} \approx 180.72$ $\newline$ Finally, multiply by the z-score: $\newline$ $E = 1.96 \times 180.72 \approx 354.21$
Round Margin of Error: Round the margin of error to the nearest whole number. $\newline$ Rounded margin of error = $354$ (to the nearest whole number)

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