Answer:
They should guarantee the lifetime of their batteries for 32 months.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the z-score 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)
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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 36 months and a standard deviation of 2 months.
This means that ![\mu = 36, \sigma = 2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2036%2C%20%5Csigma%20%3D%202)
If the company wants to replace no more than 2% of all batteries, for how many months should they guarantee the lifetime of their batteries?
The guarantee should be the 2th percentile of lengths, which is X when Z has a pvalue of 0.02. So X when Z = -2.054.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-2.054 = \frac{X - 36}{2}](https://tex.z-dn.net/?f=-2.054%20%3D%20%5Cfrac%7BX%20-%2036%7D%7B2%7D)
![X - 36 = -2.054*2](https://tex.z-dn.net/?f=X%20-%2036%20%3D%20-2.054%2A2)
![X = 31.89](https://tex.z-dn.net/?f=X%20%3D%2031.89)
Rounding to the closest month, 32.
They should guarantee the lifetime of their batteries for 32 months.