Respuesta :
Answer:
The lifetime value needed is 11.8225 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]
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 This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
The lifetime of a certain type of battery is normally distributed with mean value 11 hours and standard deviation 1 hour. This means that [tex]\mu = 11, \sigma = 1[/tex].
What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?
This is the value of THE MEAN SAMPLE X when Z has a pvalue of 0.95. That is between Z = 1.64 and Z = 1.65. So we use [tex]Z = 1.645[/tex]
Since we need the mean sample, we need to find the standard deviation of the sample, that is:
[tex]s = \frac{\sigma}{\sqrt{4}} = 0.5[/tex]
So:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.645 = \frac{X - 11}{0.5}[/tex]
[tex]X - 11 = 0.5*1.645[/tex]
[tex]X = 11.8225[/tex]
The lifetime value needed is 11.8225 hours.
