A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 51 residents and found the mean weight to be 176 pounds with a standard deviation of 36 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.
Question
A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of 51 residents and found the mean weight to be 176 pounds with a standard deviation of 36 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.
Solution
To calculate the margin of error for the mean at the 95% confidence level, we will use the formula for the margin of error (E) in a normal distribution, which is:
E = Z * (σ/√n)
Where:
- Z is the Z-score, which corresponds to the desired confidence level. For a 95% confidence level, the Z-score is approximately 1.96 (from Z-table or empirical rule).
- σ is the standard deviation, which is given as 36 pounds.
- n is the sample size, which is given as 51 residents.
Substituting the given values into the formula, we get:
E = 1.96 * (36/√51)
Calculating the above expression will give us the margin of error.
Let's calculate it:
E = 1.96 * (36/√51) = 1.96 * (36/7.141) = 1.96 * 5.04 = 9.8784
Rounding to the nearest tenth, the margin of error for the mean at the 95% confidence level is approximately 9.9 pounds.
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