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
and standard deviation
, the zscore of a measure X is given by
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
.
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 ![Z = 1.645](https://tex.z-dn.net/?f=Z%20%3D%201.645)
Since we need the mean sample, we need to find the standard deviation of the sample, that is:
![s = \frac{\sigma}{\sqrt{4}} = 0.5](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B4%7D%7D%20%3D%200.5)
So:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![1.645 = \frac{X - 11}{0.5}](https://tex.z-dn.net/?f=1.645%20%3D%20%5Cfrac%7BX%20-%2011%7D%7B0.5%7D)
![X - 11 = 0.5*1.645](https://tex.z-dn.net/?f=X%20-%2011%20%3D%200.5%2A1.645)
![X = 11.8225](https://tex.z-dn.net/?f=X%20%3D%2011.8225)
The lifetime value needed is 11.8225 hours.