Question

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.

Answer 1

Answer:

Follows are the solution to this question:

Step-by-step explanation:

Given value:

$\to X = 84\\\\\to \mu = 68\\\\\to \sigma= 10$

Using formula:

$\bold{Z= \frac{X-\mu}{\sigma}}$

    $= \frac{84-68}{10}\\\\ = \frac{16}{10}\\\\=1.6$

Calculating the percent of the sales which is less than or equal to 84 inches:

$\to P(Z \leq 1.6)=0.9452$

                      $= 0.9452 \times 100\\\\ =94.52\% \approx 95.5\%$

Calculating the remaining value $100-94.5=5.5\%$ were more than 84 inches.