Question

For questions 2 and 3. It is known that the mean diameter of pine trees in a national forest is 9.6 (inches) with a standard deviation 2.4 (inches). Assuming that the tree diameters are distributed normally. If a pine tree is to be selected randomly from this forest, what is the probability that its diameter will fall between 7.2 inches and 14.4 inches?

Answer 1

Answer: 0.8185

Step-by-step explanation:

Let X represents the tree diameters that are distributed normally.

Given: Mean $\mu=9.6\text{ inches}$, Standard deviation  $\sigma=2.4\text{ inches}$

The probability that its diameter will fall between 7.2 inches and 14.4 inches will be:

$P(7.2<X<14.4)=P(\dfrac{7.2-9.6}{2.4}<\dfrac{X-\mu}{\sigma}<\dfrac{14.4-9.6}{2.4})\\\=P(-1<Z<2)\ \ \ \ [Z=\dfrac{X-\mu}{\sigma}]\\\\=P(Z<2)-P(Z<-1)\\\\=P(Z<2)-(1-P(Z<1))\\\\=0.9772-(1-0.8413)=0.8185\ [\text{By p-value table}]$

Hence, required probability = 0.8185