Respuesta :
Answer:
84.13% probability a particular tire of this brand will last longer than 57,100 miles
Step-by-step explanation:
When the distribution is normal, we use 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 = 60000, \sigma = 2900[/tex]
What is the probability a particular tire of this brand will last longer than 57,100 miles
This is 1 subtracted by the pvalue of Z when X = 57100. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57100 - 60000}{2900}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587
1 - 0.1587 = 0.8413
84.13% probability a particular tire of this brand will last longer than 57,100 miles
