Respuesta :
Using the normal distribution, it is found that the the travel time such that 26.11% of the 60 days have a travel time that is at least is of 42.19 minutes.
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.
In this problem:
- The mean is of [tex]\mu = 35.6[/tex].
- The standard deviation is of [tex]\sigma = 10.3[/tex].
The travel time such that 26.11% of the 60 days have a travel time that is at least X when Z has a p-value of 1 - 0.2611 = 0.7389, hence X when Z = 0.64, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.64 = \frac{X - 35.6}{10.3}[/tex]
[tex]X - 35.6 = 0.64(10.3)[/tex]
[tex]X = 42.19[/tex]
The travel time such that 26.11% of the 60 days have a travel time that is at least is of 42.19 minutes.
More can be learned about the normal distribution at https://brainly.com/question/24663213
