Answer:
A tree with a height of 6.2 ft is 3 standard deviations above the mean
Step-by-step explanation:
⇒
statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)
an X value is found Z standard deviations from the mean mu if:
![\frac{X-\mu}{\sigma} = Z](https://tex.z-dn.net/?f=%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%20%3D%20Z)
In this case we have: ![\mu=5\ ft](https://tex.z-dn.net/?f=%5Cmu%3D5%5C%20ft)
![\sigma=0.4\ ft](https://tex.z-dn.net/?f=%5Csigma%3D0.4%5C%20ft)
We have four different values of X and we must calculate the Z-score for each
For X =5.4\ ft
![Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5C%5CZ%3D%5Cfrac%7B5.4-5%7D%7B0.4%7D%3D1)
Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.
⇒
statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean.
(FALSE)
For X =4.6 ft
![Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5C%5CZ%3D%5Cfrac%7B4.6-5%7D%7B0.4%7D%3D-1)
Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean
.
⇒
statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean
(FALSE)
For X =5.8 ft
![Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5C%5CZ%3D%5Cfrac%7B5.8-5%7D%7B0.4%7D%3D2)
Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.
⇒
statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean.
(TRUE)
For X =6.2\ ft
![Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5C%5CZ%3D%5Cfrac%7B6.2-5%7D%7B0.4%7D%3D3)
Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.