Respuesta :
Given:
[tex]\begin{gathered} \mu=10\text{ }inches \\ \sigma=2.2\text{ inches} \end{gathered}[/tex]To find- P(X>15)
Explanation-
We know that a z-score is given by-
[tex]z=\frac{x-\mu}{\sigma}[/tex]where x is the raw score, mu is the mean and sigma is the standard deviation.
Hence, the proportion of trees having a diameter greater than 15 inches will be-
[tex]\begin{gathered} P(x>15)=P(\frac{x-\mu}{\sigma}>\frac{15-\mu}{\sigma}) \\ P(x>15)=P(Z>\frac{15-10}{2.2}) \end{gathered}[/tex]On further solving, we get
[tex]\begin{gathered} P(x\gt15)=P(Z\gt\frac{5}{2.2}) \\ P(x\gt15)=P(Z\gt2.2727) \end{gathered}[/tex]With the help of an online tool, the probability will be
[tex]P(x>15)=0.0115[/tex]Since the significance level is not mentioned, we assumed it is 0.05.
Thus, the proportion of trees having a diameter greater than 15 inches is 0.0115.
The answer is 0.0115.
