Find the \( z \)-score corresponding to the given area. Round to two decimals, if necessary, The area to the left of \( z \) is \( 60 \% \). \( z=\square \) The area to the right of \( z \) is \( 10 \% \). The area between \( -z \) and \( z \) is \( 97 \% \). \( z=\square \)

Alright, I need to find the \( z \)-scores corresponding to the given areas. Let's tackle each part one by one. **First Problem:** - **Given:** The area to the left of \( z \) is \( 60\% \). - **To Find:** The \( z \)-score \( z \). Since the area to the left of \( z \) is \( 60\% \), this means that \( z \) is the 60th percentile in the standard normal distribution. To find the exact \( z \)-score, I'll use the standard normal distribution table or a calculator that provides the inverse cumulative distribution function. Looking up \( 60\% \) in the standard normal distribution table: - The table typically provides the cumulative area from the left up to a certain \( z \)-value. - For \( 60\% \), the corresponding \( z \)-score is approximately \( 0.25 \). So, \( z \approx 0.25 \). **Second Problem:** - **Given:** The area to the right of \( z \) is \( 10\% \). - **To Find:** The \( z \)-score \( z \). The area to the right of \( z \) being \( 10\% \) implies that the area to the left of \( z \) is \( 90\% \) (since the total area under the curve is \( 100\% \)). Using the standard normal distribution table: - \( 90\% \) corresponds to a \( z \)-score of approximately \( 1.28 \). Therefore, \( z \approx 1.28 \). **Third Problem:** - **Given:** The area between \( -z \) and \( z \) is \( 97\% \). - **To Find:** The \( z \)-score \( z \). The area between \( -z \) and \( z \) represents the central \( 97\% \) of the distribution. This means that \( 97\% \) of the data lies within \( z \) standard deviations from the mean. To find \( z \), I'll use the property that the area between \( -z \) and \( z \) is \( 2 \times \text{Area from } 0 \text{ to } z \). So, \( 2 \times \text{Area from } 0 \text{ to } z = 97\% \). This simplifies to: \[ \text{Area from } 0 \text{ to } z = 48.5\% \] Looking up \( 48.5\% \) in the standard normal distribution table: - The corresponding \( z \)-score is approximately \( 2.05 \). Therefore, \( z \approx 2.05 \). **Summary of Findings:** 1. For the area to the left of \( z \) being \( 60\% \), \( z \approx 0.25 \). 2. For the area to the right of \( z \) being \( 10\% \), \( z \approx 1.28 \). 3. For the area between \( -z \) and \( z \) being \( 97\% \), \( z \approx 2.05 \).