Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to know what percent of his sales were more than 84 inches tall. He can use the standard normal distribution to help him.

mean = 68

By "expected distribution," I would assume you mean ± 3SD.

Z = ±3 = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to your Z s cores.

94.5

5.5
1.6
44.5
68
50

44.5

yalll not even putting the right answer!!!!!!!!!!!!!!

According to the information given, the mean length of the trees purchased was 68 inches. So, the mean = 68 inches.

The expected distribution about the mean is as follows:
- 38 inches
- 48 inches
- 58 inches
- 68 inches
- 78 inches
- 88 inches
- 98 inches

Now, to find the percentage of sales below 68 inches, Jim needs to calculate the z-score. The z-score represents the number of standard deviations a particular value is from the mean.

Jim calculates the z-score as follows:
z-score = (84 - 68) / 10 = 1.6

To find the percentage associated with a z-score of 1.6 in the standard normal distribution table, Jim can refer to the table or use a statistical software/tool. The percentage associated with a z-score of 1.6 is approximately 94.4%.

So, Jim now knows that approximately 94.4% of his sales were 84 inches or less.

To find the remaining percentage of sales that were more than 84 inches, Jim can subtract the percentage of sales below 68 inches from 100%:
Remaining percentage = 100% - 94.4% = 5.6%

Therefore, Jim knows that approximately 5.6% of his sales were more than 84 inches.