Question

The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard

deviation o = 0.4 it. Which statement must be true?​

Answer 1

D. A tree with a height of 6.2 ft is 3 standard deviations above the mean

Explanation:

Answer 2

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Explanation:

⇒
`1^s^t` 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:

`(X-\mu)/(\sigma) = Z`

In this case we have:
`\mu=5\ ft`
`\sigma=0.4\ ft`

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

`Z=(X-\mu)/(\sigma)\\Z=(5.4-5)/(0.4)=1`

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒
`2^n^d` statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft

`Z=(X-\mu)/(\sigma)\\Z=(4.6-5)/(0.4)=-1`

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒
`3^r^d` 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=(X-\mu)/(\sigma)\\Z=(5.8-5)/(0.4)=2`

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒
`4^t^h` statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

`Z=(X-\mu)/(\sigma)\\Z=(6.2-5)/(0.4)=3`

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.