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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 45 residents and found the mean weight to be 191 pounds with a standard deviation of 28 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 $45$ residents and found the mean weight to be $191$ pounds with a standard deviation of $28$ 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

45

residents and found the mean weight to be

191

pounds with a standard deviation of

28

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 given information: Identify the given information and the formula to calculate the margin of error at the $95\%$ confidence level using the normal distribution. $\newline$ Given: $\newline$ - Sample mean ( $\bar{x}$ ) = $191$ pounds $\newline$ - Standard deviation ( $\sigma$ ) = $28$ pounds $\newline$ - Sample size ( $n$ ) = $45$ 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 $\bar{x}$ $0$ is the Z-score corresponding to the desired confidence level. For a $95\%$ confidence level, the Z-score is approximately $\bar{x}$ $2$ .
Calculate margin of error: Calculate the margin of error using the formula. $\newline$ $E = 1.96 \times \left(\frac{28}{\sqrt{45}}\right)$ $\newline$ First, calculate the standard error $\left(\frac{\sigma}{\sqrt{n}}\right)$ : $\newline$ Standard error = $\frac{28}{\sqrt{45}}$ $\newline$ Standard error $\approx \frac{28}{6.7082}$ $\newline$ Standard error $\approx 4.1725$ $\newline$ Now, calculate the margin of error: $\newline$ $E = 1.96 \times 4.1725$ $\newline$ $E \approx 1.96 \times 4.2$ (rounded to one decimal place for intermediate calculation) $\newline$ $E \approx 8.2$ $\newline$ Round the margin of error to the nearest tenth: $\newline$ $E \approx 8.2$

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