Answer:
68.3% of the trees are between 20 and 30 years old.
Step-by-step explanation:
We are given that the ages of trees in a forest are normally distributed with a mean of 25 years and a standard deviation of 5 years.
Let X = ages of trees in a forest
So, X ~ N([tex]\mu = 25 , \sigma^{2}=5^{2}[/tex])
The standard normal z score distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So, percent of the trees between 20 and 30 years old is given by;
P(20 < X < 30) = P(X < 30) - P(X <= 20)
P(X < 30) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{30-25}{5}[/tex] ) = P(Z < 1) = 0.84134 {using z table}
P(X <= 20) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{20-25}{5}[/tex] ) = P(Z <= -1) = 1 - P(Z < 1)= 1 - 0.84134 = 0.15866
So, P(20 < X < 30) = 0.84134- 0.15866 = 0.68268 or 68.3%
Therefore, 68.3% of the trees are between 20 and 30 years old.
