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What is the z-score with a confidence level of 95% when finding the margin of error for the mean of a normally distributed population from a sample? A. 0.99 B. 1.65 C. 1.96 D. 2.58
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Step 1
: Recall the formula for calculating the margin of error (ME) for the mean of a normally distributed population.

where $z$ is the z-score, $\sigma$ is the standard deviation of the population, and $n$ is the sample size.

Step 2
: Determine the z-score for a 95% confidence level.

For a 95% confidence level, the z-score is 1.96. This value corresponds to the number of standard errors a given proportion is away from the mean for a normal distribution.

Final Answer

z = 1.96 So, the correct answer is option C: 1.96.

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