Questions: Question 3 of 18, Step 1 of 1 2/18 Correct The Arc Electronic Company had an income of 20 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 15 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. Normal Table → to -z Normal Table → to z Submit Answer © 2024 Hawkes Learning

Solution Steps

To find the Z-score for Arc's income, we use the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \(X = 20\) (Arc's income),
• \(\mu = 15\) (mean income of the industry),
• \(\sigma = 7\) (standard deviation of the industry).

Substituting the values, we have:

\[ z = \frac{20 - 15}{7} = \frac{5}{7} \approx 0.7143 \]

Thus, the Z-score for Arc's income is:

\[ \text{Z-score for Arc's income: } 0.7143 \]

Next, we need to find the probability that a randomly selected firm will earn more than Arc did last year. This is given by:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) \]

In this case, \(Z_{start} = 0.7143\) and \(Z_{end} = \infty\). Therefore, we can express the probability as:

\[ P = \Phi(\infty) - \Phi(0.7143) \]

Since \(\Phi(\infty) = 1\), we have:

\[ P = 1 - \Phi(0.7143) \approx 1 - 0.7625 = 0.2375 \]

Thus, the probability that a randomly selected firm will earn more than Arc did last year is:

\[ \text{Probability: } 0.2375 \]

Final Answer