Respuesta :
Answer:
a) N(33,15).
b) 37.33% probability that a randomly selected LA worker has a commute that is longer than 38 minutes
c) 45.6 minutes.
Step-by-step explanation:
Problems of normally distributed samples are solved using 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 = 33, \sigma = 15[/tex]
a. What is the distribution of X?
Normal with mean 33 and standard deviaton 15. So
N(33,15).
b. Find the probability that a randomly selected LA worker has a commute that is longer than 38 minutes
This is 1 subtracted by the pvalue of Z when X = 38. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{38 - 33}{15}[/tex]
[tex]Z = 0.333[/tex]
[tex]Z = 0.333[/tex] has a pvalue of 0.6267.
1 - 0.6267 = 0.3733
37.33% probability that a randomly selected LA worker has a commute that is longer than 38 minutes
c. Find the 80th percentile for the commute time of LA workers.
This is X when Z has a pvalue of 0.8. So it is X when Z = 0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 33}{15}[/tex]
[tex]X - 33 = 0.84*15[/tex]
[tex]X = 45.6[/tex]
45.6 minutes.
