Question

Jim Tree sells Christmas 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.

Complete the following statements to find what percent of Jim's sales were more than 84 inches tall.
First, find the mean. The mean is inches.Then find the distribution about the mean: % of the sales will be below 68 inches.You will need find the percentage between 68 and 84 to add to that. To do so, calculates a-score:. Then find the percentage associated with that value in the table.To the nearest tenth, the percentage associated with that value is %. Now Jim knows that % of his sales were 84 inches or less. Therefore, the remaining % were more than 84 inches.

Answer 1

Using the mean and standard deviation of tree lengths, a z-score calculation shows that approximately 5.5% of Jim's Christmas tree sales were for trees more than 84 inches tall.

Step-by-step explanation:

Jim Tree wants to know what percent of his Christmas tree sales were more than 84 inches tall. The mean length of the trees is 68 inches with a standard deviation of 10 inches. To find out the percentage of trees taller than 84 inches, we first acknowledge that the mean length is 68 inches. Since the tree lengths follow a normal distribution, 50% of the sales will be below the mean of 68 inches.

Next, we calculate the z-score for a tree of 84 inches using the formula z = (X - μ) / σ.

This gives us:

z = (84 - 68) / 10

= 16 / 10

= 1.6

We then look up the z-score of 1.6 in the standard normal distribution table, which gives us a percentage of about 94.5%. This percentage represents trees that are 68 inches or shorter and those between 68 and 84 inches.

To find the percentage of trees taller than 84 inches, we subtract this value from 100%, resulting in approximately 5.5%. Therefore, Jim knows that 5.5% of his sales were of trees more than 84 inches tall.