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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 133 residents and found the mean weight to be 165 pounds with a standard deviation of 25 pounds. Use the normal distribution/empirical rule to estimate a
95% confidence interval for the mean, rounding all values to the nearest tenth.

A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of $133$ residents and found the mean weight to be $165$ pounds with a standard deviation of $25$ pounds. Use the normal distribution/empirical rule to estimate a $95 \%$ confidence interval for the mean, rounding all values to the nearest tenth.

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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

133

residents and found the mean weight to be

165

pounds with a standard deviation of

25

pounds. Use the normal distribution/empirical rule to estimate a

95 \%

confidence interval for the mean, rounding all values to the nearest tenth.

Identify Data Parameters: Identify the sample mean, standard deviation, and sample size. The sample mean ( $\bar{x}$ ) is $165$ pounds, the standard deviation ( $s$ ) is $25$ pounds, and the sample size ( $n$ ) is $133$ residents.
Calculate Standard Error: Determine the standard error of the mean (SEM). The standard error of the mean is calculated by dividing the standard deviation by the square root of the sample size. $SEM = \frac{s}{\sqrt{n}}$ $SEM = \frac{25}{\sqrt{133}}$ $SEM \approx \frac{25}{11.5326}$ $SEM \approx 2.1689$ Round to the nearest tenth: $SEM \approx 2.2$ pounds
Find Z-Score: Find the z-score that corresponds to a $95\%$ confidence level. $\newline$ For a $95\%$ confidence interval, the z-score is typically $1.96$ (this value comes from standard normal distribution tables).
Calculate Margin of Error: Calculate the margin of error (ME) using the z-score and the standard error. $\newline$ $ME = z \times SEM$ $\newline$ $ME = 1.96 \times 2.2$ $\newline$ $ME \approx 4.312$ $\newline$ Round to the nearest tenth: $ME \approx 4.3$ pounds
Determine Confidence Interval Bounds: Calculate the lower and upper bounds of the $95$ % confidence interval. $\newline$ Lower bound = $\bar{x} - ME$ $\newline$ Lower bound = $165 - 4.3$ $\newline$ Lower bound $\approx 160.7$ pounds $\newline$ Upper bound = $\bar{x} + ME$ $\newline$ Upper bound = $165 + 4.3$ $\newline$ Upper bound $\approx 169.3$ pounds
Final Confidence Interval: Round the lower and upper bounds to the nearest tenth and state the final $95\%$ confidence interval. $\newline$ The $95\%$ confidence interval for the mean weight of the residents is approximately $(160.7, 169.3)$ pounds.

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