Respuesta :
Answer:
8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours.
Step-by-step explanation:
Problems of normally distributed samples 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.
In this problem, we have that:
[tex]\mu = 7500, \sigma = 220[/tex]
What percentage of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours?
This is the probability that X is lower or equal than 7200 hours. So this is the pvalue of Z when X = 7200.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{7200 - 7500}{220}[/tex]
[tex]Z = -1.36[/tex]
[tex]Z = -1.36[/tex] has a pvalue of 0.0869
So 8.69% of their bulbs would they have to replace under warranty if they offered a warranty of 7200 hours.
