Respuesta :
Using the normal distribution, it is found that:
a) 64.8% of years will have an annual rainfall of less than 44 inches.
b) 68.4% of years will have an annual rainfall of more than 39 inches.
c) 31.1% of years will have an annual rainfall of between 37 inches and 42 inches.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
In this problem:
- The mean is of 41.8 inches, hence [tex]\mu = 41.8[/tex].
- The standard deviation is of 5.8 inches, hence [tex]\sigma = 5.8[/tex]
Item a:
The proportion is the p-value of Z when X = 44, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{44 - 41.8}{5.8}[/tex]
[tex]Z = 0.38[/tex]
[tex]Z = 0.38[/tex] has a p-value of 0.648.
0.648 x 100% = 64.8%
64.8% of years will have an annual rainfall of less than 44 inches.
Item b:
The proportion is 1 subtracted by the p-value of Z when X = 39, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{39 - 41.8}{5.8}[/tex]
[tex]Z = -0.48[/tex]
[tex]Z = -0.48[/tex] has a p-value of 0.316.
1 - 0.316 = 0.684
0.684 x 100% = 68.4%
68.4% of years will have an annual rainfall of more than 39 inches.
Item c:
The proportion is the p-value of Z when X = 42 subtracted by the p-value of Z when X = 37, hence:
X = 42:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{42 - 41.8}{5.8}[/tex]
[tex]Z = 0.035[/tex]
[tex]Z = 0.035[/tex] has a p-value of 0.514.
X = 37:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{37 - 41.8}{5.8}[/tex]
[tex]Z = -0.83[/tex]
[tex]Z = -0.83[/tex] has a p-value of 0.203.
0.514 - 0.203 = 0.311
0.311 x 100% = 31.1%
31.1% of years will have an annual rainfall of between 37 inches and 42 inches.
A similar problem is given at https://brainly.com/question/24663213
