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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 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 [tex]\mu = 36, \sigma = 2[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.054 = \frac{X - 36}{2}[/tex]
[tex]X - 36 = -2.054*2[/tex]
[tex]X = 31.89[/tex]
Rounding to the closest month, 32.
They should guarantee the lifetime of their batteries for 32 months.
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