Question

Suppose the lifetime of a cell phone battery is normally distributed with a mean of 36 months and a standard deviation of 2 months. 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

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 36, \sigma = 2$

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.

$Z = \frac{X - \mu}{\sigma}$

$-2.054 = \frac{X - 36}{2}$

$X - 36 = -2.054*2$

$X = 31.89$

Rounding to the closest month, 32.

They should guarantee the lifetime of their batteries for 32 months.