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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 162 pounds with a standard deviation of 35 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 $162$ pounds with a standard deviation of $35$ 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

162

pounds with a standard deviation of

35

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:

Calculate Z-score: To calculate the margin of error at the $95$ % confidence level using the normal distribution, we need to use the formula for the margin of error (ME) which is: $\newline$ ME = $Z \times (\sigma/\sqrt{n})$ $\newline$ where $Z$ is the Z-score corresponding to the desired confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size.
Find Margin of Error Formula: First, we need to find the Z-score that corresponds to the $95\%$ confidence level. For a normal distribution, the Z-score for a $95\%$ confidence level is approximately $1.96$ . This value is obtained from a Z-table or standard normal distribution table.
Plug in Values: Next, we plug in the values into the margin of error formula: $\newline$ $ME = 1.96 \times \left(\frac{35}{\sqrt{33}}\right)$
Calculate Square Root: Now, we calculate the square root of the sample size, which is $\sqrt{33}$ .
Divide Standard Deviation: The square root of $33$ is approximately $5.7446$ .
Multiply by Z-score: We then divide the standard deviation by the square root of the sample size: $35 / 5.7446 \approx 6.0922$
Final Margin of Error: Finally, we multiply this result by the Z-score to find the margin of error: $ME = 1.96 \times 6.0922 \approx 11.9407$
Round to Nearest Tenth: Rounding to the nearest tenth, the margin of error is approximately $11.9$ pounds.

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