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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 71 students in the high school and found a mean savings of 2600 dollars with a standard deviation of 1500 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 $71$ students in the high school and found a mean savings of $2600$ dollars with a standard deviation of $1500$ 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

71

students in the high school and found a mean savings of

2600

dollars with a standard deviation of

1500

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: Identify the values given in the problem. $\newline$ We are given: $\newline$ - The mean savings of the sample ( $\bar{x}$ ) = \(2600\)\(\newline\)- The standard deviation of the sample (s) = $1500$ $\newline$ - The sample size (n) = $71$ $\newline$ - The confidence level ( $95$ %)
Determine z-score: 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 value is obtained from a standard normal distribution table or z-score table.
Calculate SEM: Calculate the standard error of the mean (SEM). $\newline$ The standard error of the mean is calculated using the formula: $\newline$ SEM = $\frac{s}{\sqrt{n}}$ $\newline$ Where $s$ is the standard deviation and $n$ is the sample size. $\newline$ SEM = $\frac{\$1500}{\sqrt{71}}$ $\newline$ SEM $\approx \frac{\$1500}{8.426}$ $\newline$ SEM $\approx \$177.83$
Calculate ME: Calculate the margin of error (ME) using the z-score and the standard error of the mean. $\newline$ $ME = z \times SEM$ $\newline$ $ME = 1.96 \times \$177.83$ $\newline$ $ME \approx \$348.53$
Round margin: Round the margin of error to the nearest whole number. $ME \approx \$(349)$

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