Respuesta :
Answer:
0.7931 = 79.31% probability that a truck drives between 43 and 141 miles in a day.
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 problem, we have that:
[tex]\mu = 100, \sigma = 38[/tex]
Find the probability that a truck drives between 43 and 141 miles in a day.
This is the pvalue of Z when X = 141 subtracted by the pvalue of Z when X = 43. So
X = 141
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{141 - 100}{38}[/tex]
[tex]Z = 1.08[/tex]
[tex]Z = 1.08[/tex] has a pvalue of 0.8599
X = 43
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{43 - 100}{38}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
0.8599 - 0.0668 = 0.7931
0.7931 = 79.31% probability that a truck drives between 43 and 141 miles in a day.
