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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 74 students in the high school and found a mean savings of 4500 dollars with a standard deviation of 1300 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 $74$ students in the high school and found a mean savings of $4500$ dollars with a standard deviation of $1300$ 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

74

students in the high school and found a mean savings of

4500

dollars with a standard deviation of

1300

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: Identify the given values from the problem. $\newline$ We are given: $\newline$ - Mean savings ( $\bar{x}$ ) = \(4500\)\(\newline\)- Standard deviation (\(s\)) = $1300$ $\newline$ - Sample size ( $n$ ) = $74$ $\newline$ - Confidence level = $95$ %
Determine Confidence Level: Determine the z-score that corresponds to the $95\%$ confidence level. $\newline$ For a $95\%$ confidence level, the z-score that corresponds to the confidence interval is approximately $1.96$ . This is because $95\%$ of the data falls within $1.96$ standard deviations from the mean in a normal distribution.
Calculate Margin of Error: Calculate the margin of error using the formula for the margin of error at a given confidence level. $\newline$ The margin of error (E) is calculated using the formula: $\newline$ $E = z \times \frac{s}{\sqrt{n}}$ $\newline$ where $z$ is the z-score, $s$ is the standard deviation, and $n$ is the sample size.
Calculate Margin of Error: Plug in the values and calculate the margin of error. $\newline$ $E = 1.96 \times \frac{1300}{\sqrt{74}}$ $\newline$ $E = 1.96 \times \frac{1300}{8.6023}$ (rounded to four decimal places for the square root of $74$ ) $\newline$ $E = 1.96 \times 151.1365$ (rounded to four decimal places for the division) $\newline$ $E = 296.2275$
Round Margin of Error: Round the margin of error to the nearest whole number. $\newline$ The margin of error rounded to the nearest whole number is $296$ .

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