Question

Suppose the life of a particular brand of laptop battery is normally distributed with a mean of 8 hours and a standard deviation of 0.6 hours. What is the probability that the battery will last more than 9 hours before running out of power g

Answer 1

Answer:

4.75% probability that the battery will last more than 9 hours before running out of power

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 8, \sigma = 0.6$

What is the probability that the battery will last more than 9 hours before running out of power g

This is 1 subtracted by the pvalue of Z when X = 9. So

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

$Z = \frac{9 - 8}{0.6}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525

1 - 0.9525 = 0.0475

4.75% probability that the battery will last more than 9 hours before running out of power