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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 76 residents and found the mean weight to be 198 pounds with a standard deviation of 22 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 $76$ residents and found the mean weight to be $198$ pounds with a standard deviation of $22$ 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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Use normal distributions to approximate binomial distributions

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

76

residents and found the mean weight to be

198

pounds with a standard deviation of

22

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. $\newline$ Given values: $\newline$ - Sample mean ( $\bar{x}$ ) = $198$ pounds $\newline$ - Standard deviation ( $\sigma$ ) = $22$ pounds $\newline$ - Sample size ( $n$ ) = $76$ $\newline$ - Confidence level = $95\%$ $\newline$ The formula to calculate the margin of error ( $E$ ) at the $95\%$ confidence level using the normal distribution is: $\newline$ $E = Z \times (\sigma/\sqrt{n})$ $\newline$ Where $198$ $0$ is the Z-score corresponding to the desired confidence level. For a $95\%$ confidence level, the Z-score is typically $198$ $2$ .
Calculate margin of error: Calculate the margin of error using the formula. $\newline$ First, calculate the standard error $\sigma/\sqrt{n}$ : $\newline$ Standard error = $\sigma/\sqrt{n} = 22/\sqrt{76}$ $\newline$ Now, calculate the margin of error (E): $\newline$ E = Z * $\sigma/\sqrt{n}$ = $1$ . $96$ * $22/\sqrt{76}$
Perform calculations: Perform the calculations. $\newline$ Standard error = $\frac{22}{\sqrt{76}} \approx \frac{22}{8.7178} \approx 2.5226$ $\newline$ Margin of error $(E) = 1.96 \times 2.5226 \approx 4.9447$ $\newline$ Round the margin of error to the nearest tenth: $\newline$ $E \approx 4.9$ pounds

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