The Arc Electronic Company had an income of $\mathbf{4 5}$ million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 40 million dollars with a standard deviation of 7 million dollars. If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year? Round your answer to four decimal places. Answer If you would like to look up the value in a table, select the table you want to view, then either click the cell at the intersection of the row and column or use the arrow keys to find the appropriate cell in the table and select it using the Space key.

Step 1 ：Given that the mean income of firms in the same industry as Arc for a year is \(40\) million dollars and the standard deviation is \(7\) million dollars. Arc's income is \(45\) million dollars.

Step 2 ：We need to calculate the z-score for Arc's income. The z-score is calculated as \((X - \mu) / \sigma\), where \(X\) is the value we're interested in, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 3 ：Substituting the given values, we get \((45 - 40) / 7 = 0.7142857142857143\). So, the z-score for Arc's income is \(0.7142857142857143\).

Step 4 ：The z-score tells us how many standard deviations an element is from the mean. In this case, Arc's income is approximately \(0.7142857142857143\) standard deviations above the mean income.

Step 5 ：We then need to find the probability that a randomly selected firm will have a higher income. This is equivalent to finding the area under the normal distribution curve to the right of the z-score.

Step 6 ：Looking up the z-score of \(0.7142857142857143\) in the z-table, we find that the area to the left of the z-score is approximately \(0.7625\). Since the total area under the curve is \(1\), the area to the right of the z-score is \(1 - 0.7625 = 0.2375\).

Step 7 ：So, the probability that a randomly selected firm will earn more than Arc did last year is \(0.2375\).

Step 8 ：Final Answer: \(\boxed{0.2375}\)