Respuesta :
Using the normal distribution, it is found that:
a) 0.8599 = 85.99% probability that x is more than 60.
b) 0.1788 = 17.88% probability that x is less than 110.
c) 0.6811 = 68.11% probability that x is between 60 and 110.
d) 0.0643 = 6.43% probability that x is greater than 125.
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 87, thus [tex]\mu = 87[/tex].
- The standard deviation is of 25, thus [tex]\sigma = 25[/tex].
Item a:
This probability is 1 subtracted by the p-value of Z when X = 60, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{60 - 87}{25}[/tex]
[tex]Z = -1.08[/tex]
[tex]Z = -1.08[/tex] has a p-value of 0.1401.
1 - 0.1401 = 0.8599
0.8599 = 85.99% probability that x is more than 60.
Item b:
This probability is the p-value of Z when X = 110, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{110 - 87}{25}[/tex]
[tex]Z = 0.92[/tex]
[tex]Z = 0.92[/tex] has a p-value of 0.8212.
1 - 0.8212 = 0.1788.
0.1788 = 17.88% probability that x is less than 110.
Item c:
This probability is the p-value of Z when X = 110 subtracted by the p-value of Z when X = 60.
From the previous two items, 0.8212 - 0.1401 = 0.6811.
0.6811 = 68.11% probability that x is between 60 and 110.
Item d:
This probability is 1 subtracted by the p-value of Z when X = 125, thus:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{125 - 87}{25}[/tex]
[tex]Z = 1.52[/tex]
[tex]Z = 1.52[/tex] has a p-value of 0.9357.
1 - 0.9357 = 0.0643.
0.0643 = 6.43% probability that x is greater than 125.
A similar problem is given at https://brainly.com/question/24863330
