Respuesta :
Answer:
3.59% of all tires produced by this company will have a lifetime of 3270 or more miles
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
Normally distributed random variable with a mean of 3000 miles and a standard deviation of 150 miles.
This means that [tex]\mu = 3000, \sigma = 150[/tex]
What percentage of all tires produced by this company will have a lifetime of 3270 or more miles?
The proportion is 1 subtracted by the pvalue of Z when X = 3270. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{3270 - 3000}{150}[/tex]
[tex]Z = 1.8[/tex]
[tex]Z = 1.8[/tex] has a pvalue of 0.9641
1 - 0.9641 = 0.0359
0.0359*100% = 3.59%
3.59% of all tires produced by this company will have a lifetime of 3270 or more miles
