Answer :
Answer:
0.0082 = 0.82% probability that the height of a randomly selected tree is as tall as mine or shorter.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Assume that X has a normal distribution with a mean of 191.8 ft and a standard deviation of 4 ft.
This means that [tex]\mu = 191.8, \sigma = 4[/tex]
A tree of this type grows in my backyard, and it stands 182.2 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.
This is the pvalue of Z when X = 182.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{182.2 - 191.8}{4}[/tex]
[tex]Z = -2.4[/tex]
[tex]Z = -2.4[/tex] has a pvalue of 0.0082
0.0082 = 0.82% probability that the height of a randomly selected tree is as tall as mine or shorter.