Question

The heights of the pine trees in a certain forest are normally distributed, with a mean of 86 feet and a standard deviation of 8 feet.
Approximately what percentage of the pine trees in this forest are taller than 100 feet?

Answer 1

Answer:

4%

Step-by-step explanation:

We solve this question using z score formula

z = (x-μ)/σ,

where

x is the raw score = 100 feet

μ is the population mean = 86 feet

σ is the population standard deviation = 8 feet

Hence,

x > 100 feet

z = (100 - 86)/8

z = 1.75

Probability value from Z-Table:

P(x<100) = 0.95994

P(x>100) = 1 - P(x<100) = 0.040059

Converting to Percentage

= 0.040059 × 100

= 4.0059%

Approximately = 4%

Therefore, the percentage of the pine trees in this forest are taller than 100 feet is 4%