Respuesta :
Using the normal distribution, it is found that there is a 0.7452 = 74.52% probability of randomly selecting a vehicle with an engine between than 3.1 L and 4.2 L.
Normal Probability Distribution
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.
Researching the problem on the internet, it is found that:
- The mean is [tex]\mu = 3.59804[/tex].
- The standard deviation is [tex]\sigma = 0.47986[/tex].
The probability is the p-value of Z when X = 4.2 subtracted by the p-value of Z when X = 3.1, hence:
X = 4.2
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4.2 - 3.59804}{0.47986}[/tex]
[tex]Z = 1.25[/tex]
[tex]Z = 1.25[/tex] has a p-value of 0.8944.
X = 3.1
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3.1 - 3.59804}{0.47986}[/tex]
[tex]Z = -1.04[/tex]
[tex]Z = -1.04[/tex] has a p-value of 0.1492.
0.8944 - 0.1492 = 0.7452.
0.7452 = 74.52% probability of randomly selecting a vehicle with an engine between than 3.1 L and 4.2 L.
More can be learned about the normal distribution at https://brainly.com/question/24663213
