Questions: The approximate Z-score that corresponds to a right tail area of 0.20 is (Round to two decimal places as needed.)

Solution Steps

We need to find the approximate Z-score that corresponds to a right tail area of \(0.20\). This means we are looking for a Z-score \(Z\) such that the cumulative probability to the left of \(Z\) is \(1 - 0.20 = 0.80\).

The cumulative probability \(P\) for various Z-scores is calculated as follows:

• For \(Z = 0.0\): \[ P = \Phi(0.0) - \Phi(-\infty) = 0.5 \]

• For \(Z = 2.5\): \[ P = \Phi(2.5) - \Phi(-\infty) = 0.9938 \]

• For \(Z = 1.25\): \[ P = \Phi(1.25) - \Phi(-\infty) = 0.8944 \]

• For \(Z = 0.625\): \[ P = \Phi(0.625) - \Phi(-\infty) = 0.734 \]

• For \(Z = 0.9375\): \[ P = \Phi(0.9375) - \Phi(-\infty) = 0.8257 \]

• For \(Z = 0.7812\): \[ P = \Phi(0.7812) - \Phi(-\infty) = 0.7827 \]

• For \(Z = 0.8594\): \[ P = \Phi(0.8594) - \Phi(-\infty) = 0.8049 \]

• For \(Z = 0.8203\): \[ P = \Phi(0.8203) - \Phi(-\infty) = 0.794 \]

• For \(Z = 0.8398\): \[ P = \Phi(0.8398) - \Phi(-\infty) = 0.7995 \]

• For \(Z = 0.8496\): \[ P = \Phi(0.8496) - \Phi(-\infty) = 0.8022 \]

• For \(Z = 0.8447\): \[ P = \Phi(0.8447) - \Phi(-\infty) = 0.8009 \]

• For \(Z = 0.8423\): \[ P = \Phi(0.8423) - \Phi(-\infty) = 0.8002 \]

• For \(Z = 0.8411\): \[ P = \Phi(0.8411) - \Phi(-\infty) = 0.7998 \]

• For \(Z = 0.8417\): \[ P = \Phi(0.8417) - \Phi(-\infty) = 0.8 \]

• For \(Z = 0.8414\): \[ P = \Phi(0.8414) - \Phi(-\infty) = 0.7999 \]

• For \(Z = 0.8415\): \[ P = \Phi(0.8415) - \Phi(-\infty) = 0.8 \]

• For \(Z = 0.8414\): \[ P = \Phi(0.8414) - \Phi(-\infty) = 0.8 \]

From the calculations, we observe that the cumulative probability \(P\) approaches \(0.80\) as \(Z\) approaches \(0.84\). Thus, the approximate Z-score for a right tail area of \(0.20\) is:

\[ Z \approx 0.84 \]

Final Answer