Respuesta :
Answer:
The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
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 = 4000, \sigma = 200[/tex]
What lifetime should the company promote for these bulbs, whereby only 2% burnout before the claimed lifetime?
This is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.055 = \frac{X - 4000}{200}[/tex]
[tex]X - 4000 = -2.055*200[/tex]
[tex]X = 3589[/tex]
The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
