Average sales for an online textbook distributor were $\$ 74.27$ per customer per purchase. Assume the sales are normally distributed. If the standard deviation of the amount spent on textbooks is $\$ 8.61$, find the percentage of customers of the online textbook distributor that spent:
(a) more than $\$ 82.88$ per purchase.
(b) less than $\$ 100.10$ per purchase.
Round your answer to the nearest tenth of a percent.
Refer to the table of values ( Area Under the Standard Normal Distribution) as needed.
Part: $0 / 2$
Part 1 of 2
(a) The percentage of customers that spent more than $\$ 82.88$ per purchase is $\%$

Step 1 ：First, we need to find the z-scores for each value. The z-score is calculated as \(z = \frac{x - \mu}{\sigma}\), where \(x\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step 2 ：For \(x = 82.88\), we have \(z = \frac{82.88 - 74.27}{8.61} = \frac{8.61}{8.61} = 1\).

Step 3 ：Now, we need to find the area to the right of \(z = 1\) in the standard normal distribution table. The table value for \(z = 1\) is 0.3413, which represents the area to the left of \(z = 1\). To find the area to the right, we subtract this value from 1: \(1 - 0.3413 = 0.6587\).

Step 4 ：To find the percentage of customers that spent more than \$82.88 per purchase, we multiply the area by 100: \(0.6587 \times 100 = 65.87\%\).

Step 5 ：\(\boxed{65.9\%}\) of customers spent more than \$82.88 per purchase.

Step 6 ：For \(x = 100.10\), we have \(z = \frac{100.10 - 74.27}{8.61} = \frac{25.83}{8.61} \approx 3\).

Step 7 ：Now, we need to find the area to the left of \(z = 3\) in the standard normal distribution table. The table value for \(z = 3\) is 0.9987.

Step 8 ：To find the percentage of customers that spent less than \$100.10 per purchase, we multiply the area by 100: \(0.9987 \times 100 = 99.87\%\).

Step 9 ：\(\boxed{99.9\%}\) of customers spent less than \$100.10 per purchase.