Respuesta :
Answer:
0.0668 = 6.68% probability that the height of a randomly selected tree is as tall as mine or shorter.
0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.
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.
In this question, we have that:
[tex]\mu = 137.1, \sigma = 3.2[/tex]
A tree of this type grows in my backyard, and it stands 132.3 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 = 132.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{132.3 - 137.1}{3.2}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
0.0668 = 6.68% probability that the height of a randomly selected tree is as tall as mine or shorter.
My neighbor also has a tree of this type growing in her backyard, but hers stands 143.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.
This is 1 subtracted by the pvalue of Z when X = 143.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{143.5 - 137.1}{3.2}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
0.0228 = 2.28% probability that the full height of a randomly selected tree is at least as tall as hers.
