Questions: Based on data from a college, scores on a certain test are normally distributed with a mean of 1523 and a standard deviation of 315. Complete parts (a) through (c) below a. Find the percentage of scores greater than 1586. % (Round to two decimal places as needed.)

Solution Steps

To find the Z-score for the value \( X = 1586 \), we use the formula:

\[ z = \frac{X - \mu}{\sigma} = \frac{1586 - 1523}{315} = 0.2 \]

Thus, the Z-score for the value 1586 is:

\[ Z = 0.2 \]

Next, we calculate the probability of scores greater than 1586. This can be expressed as:

\[ P(X > 1586) = P(Z > 0.2) = 1 - P(Z \leq 0.2) \]

Using the cumulative distribution function \( \Phi \):

\[ P = \Phi(\infty) - \Phi(0.2) = 1 - 0.4207 = 0.5793 \]

To express this probability as a percentage, we multiply by 100:

\[ \text{Percentage} = P(X > 1586) \times 100 = 0.5793 \times 100 = 57.93\% \]

Final Answer