Quiz Active 1 2 3 4 5 6 7 8 9 10 TIME REMAINING 57:14 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? 0.99 1.65 1.96 2.58

The \( z \)-score for a \( 95\% \) confidence level is \( 1.96 \). This value is derived from the standard normal distribution, where \( 95\% \) of the data falls within \( 1.96 \) standard deviations from the mean. This means that if you're constructing confidence intervals, you'll use this \( z \)-score to determine how far out from the sample mean you'll go to capture the true population mean. When computing the margin of error using the \( z \)-score, the formula is: \[ \text{Margin of Error} = z \times \left( \frac{\sigma}{\sqrt{n}} \right) \] Here, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. Using \( 1.96 \) ensures your interval is wide enough to be \( 95\% \) confident that it contains the population mean!