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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 33 residents and found the mean weight to be 166 pounds with a standard deviation of 26 pounds. 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 was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of $33$ residents and found the mean weight to be $166$ pounds with a standard deviation of $26$ pounds. 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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Find values of normal variables

Full solution

Q. A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of

33

residents and found the mean weight to be

166

pounds with a standard deviation of

26

pounds. 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 values and formula: Identify the values given in the problem and the formula to calculate the margin of error. $\newline$ We are given: $\newline$ - Sample mean ( $\bar{x}$ ) = $166$ pounds $\newline$ - Standard deviation ( $\sigma$ ) = $26$ pounds $\newline$ - Sample size ( $n$ ) = $33$ residents $\newline$ - Confidence level = $95\%$ $\newline$ The formula for the margin of error ( $E$ ) when using the normal distribution is: $\newline$ $E = Z \times (\sigma/\sqrt{n})$ $\newline$ Where $Z$ is the Z-score corresponding to the desired confidence level. For a $95\%$ confidence level, the Z-score is approximately $166$ $1$ .
Calculate margin of error: Calculate the margin of error using the formula. $\newline$ First, calculate the standard error $\sigma/\sqrt{n}$ : $\newline$ Standard error = $\sigma/\sqrt{n} = 26/\sqrt{33} \approx 26/5.7446 \approx 4.526$ $\newline$ Now, calculate the margin of error (E): $\newline$ E = Z * $\sigma/\sqrt{n}$ = $1$ . $96$ * $4$ . $526$ \approx $8$ . $87096$ $\newline$ Round the margin of error to the nearest tenth: $\newline$ E \approx $8$ . $9$ pounds

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