Question

A particular tree falls over in a bonsai forest and is considered to be an average tree (in the middle 20%). If the mean height is 10cm and the standard deviation is 2 between what heights must the tree be?

Answer 1

Using the normal distribution, it is found that the tree must have heights between 9.49 cm and 10.51 cm.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 10, \sigma = 2$

Considering the symmetry of the normal distribution, the middle 20% is between the 40th percentile(Z = -0.253) and the 60th percentile(Z = 0.253), hence:

$Z = \frac{X - \mu}{\sigma}$

$-0.253 = \frac{X - 10}{2}$

X - 10 = -0.253 x 2

X = 9.49.

$Z = \frac{X - \mu}{\sigma}$

$0.253 = \frac{X - 10}{2}$

X - 10 = 0.253 x 2

X = 10.51.

More can be learned about the normal distribution at https://brainly.com/question/24663213
#SPJ1