What is the confidence level of each of the following confidence intervals for $\mu$ ? Complete parts a through $\mathbf{e}$ . $\newline$ ] Click the icon to view the table of normal curve areas. $\newline$ a. $\bar{x} \pm 1.96\left(\frac{\sigma}{\sqrt{n}}\right)$ $\newline$ $\%$ (Round to two decimal places as needed.)

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Q. What is the confidence level of each of the following confidence intervals for

\mu

? Complete parts a through

\mathbf{e}

\newline

] Click the icon to view the table of normal curve areas.

\newline

\bar{x} \pm 1.96\left(\frac{\sigma}{\sqrt{n}}\right)

\newline

\%

(Round to two decimal places as needed.)

Understanding z-scores and confidence levels: To determine the confidence level of the given confidence interval, we need to understand that the value $1.96$ corresponds to the z-score for a certain confidence level in a normal distribution. This z-score typically corresponds to a $95\%$ confidence level.
Calculating total area for confidence level: The z-score of $1.96$ is associated with the area to the left of the z-score in the standard normal distribution. To find the total area (which corresponds to the confidence level), we need to consider the area to the left of $z = 1.96$ and to the right of $z = -1.96$ . Since the normal distribution is symmetric, the area to the left of $z = 1.96$ is equal to the area to the right of $z = -1.96$ .
Determining correct confidence level calculation: The area to the left of $z = 1.96$ in the standard normal distribution is approximately $0.975$ . This is because $1.96$ z-score cuts off the top $2.5\%$ of the distribution, leaving $97.5\%$ to the left.
Determining correct confidence level calculation: The area to the left of $z = 1.96$ in the standard normal distribution is approximately $0.975$ . This is because $1.96$ z-score cuts off the top $2.5\%$ of the distribution, leaving $97.5\%$ to the left.To find the total area between $z = -1.96$ and $z = 1.96$ , we double the area to the left of $z = 1.96$ because the normal distribution is symmetric. So, we calculate $0.975 \times 2$ to get the confidence level.
Determining correct confidence level calculation: The area to the left of $z = 1.96$ in the standard normal distribution is approximately $0.975$ . This is because $1.96$ z-score cuts off the top $2.5\%$ of the distribution, leaving $97.5\%$ to the left.To find the total area between $z = -1.96$ and $z = 1.96$ , we double the area to the left of $z = 1.96$ because the normal distribution is symmetric. So, we calculate $0.975 \times 2$ to get the confidence level.However, since we are doubling the area to the left of $z = 1.96$ , we must subtract the overlapping area once, which is the area beyond $z = 1.96$ , which is $0.975$ $1$ . So the calculation is $0.975$ $2$ .
Determining correct confidence level calculation: The area to the left of $z = 1.96$ in the standard normal distribution is approximately $0.975$ . This is because $1.96$ z-score cuts off the top $2.5\%$ of the distribution, leaving $97.5\%$ to the left.To find the total area between $z = -1.96$ and $z = 1.96$ , we double the area to the left of $z = 1.96$ because the normal distribution is symmetric. So, we calculate $0.975 \times 2$ to get the confidence level.However, since we are doubling the area to the left of $z = 1.96$ , we must subtract the overlapping area once, which is the area beyond $z = 1.96$ , which is $0.975$ $1$ . So the calculation is $0.975$ $2$ .Performing the calculation gives us $0.975$ $3$ or $0.975$ $4$ . This is incorrect because the confidence level cannot exceed $0.975$ $5$ . We need to correct this by understanding that the area to the left of $z = 1.96$ already accounts for half the confidence interval, and we only need to double the one-tail area of $0.975$ $1$ to find the area outside the confidence interval.