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Citrus2011 [14]
3 years ago
7

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

Mathematics
1 answer:
arsen [322]3 years ago
6 0

Answer:

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

Step-by-step 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:

\frac{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=\frac{X-\mu}{\sigma}\\Z=\frac{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=\frac{X-\mu}{\sigma}\\Z=\frac{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=\frac{X-\mu}{\sigma}\\Z=\frac{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=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3

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

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