Q:

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma = 14 inches. What is the probability that a randomly chosen tree is greater than 140 inches? For this problem we want just the answer. Please give up to 4 significant decimal places, and use the proper rules of rounding.

Accepted Solution

A:
Answer:The probability that a randomly chosen tree is greater than 140 inches is 0.0228.Step-by-step explanation:Given : Cherry trees in a certain orchard have heights that are normally distributed with [tex]\mu = 112[/tex] inches and [tex]\sigma = 14[/tex] inches.To find : What is the probability that a randomly chosen tree is greater than 140 inches? Solution : Mean - [tex]\mu = 112[/tex] inchesStandard deviation - [tex]\sigma = 14[/tex] inchesThe z-score formula is given by, [tex]Z=\frac{x-\mu}{\sigma}[/tex]Now, [tex]P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})[/tex][tex]P(X>140)=P(Z>\frac{140-112}{14})[/tex][tex]P(X>140)=P(Z>\frac{28}{14})[/tex][tex]P(X>140)=P(Z>2)[/tex][tex]P(X>140)=1-P(Z<2)[/tex]The Z-score value we get is from the Z-table,[tex]P(X>140)=1-0.9772[/tex][tex]P(X>140)=0.0228[/tex]Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.