Question

The lifetime of a certain brand of battery is normally distributed with a mean value of 9 hr and a standard deviation of 0.8 hr when it is used in a particular DVD player. Suppose that two new batteries are independently selected and put into the player. The player stops functioning as soon as one of the batteries fails. (Round your answers to four decimal places.)

Required:
a. What is the probability that the player functions for at least 8 hr?
b. What is the probability that the cassette player works for at most 11 hr?
c. Find a number z* such that only 10% of all cassette players will function without battery replacement for more than z* hr.

Answer 1

Answer:

a. 0.8944 = 89.44% probability that the player functions for at least 8 hr

b. 0.9938 = 99.38% probability that the cassette player works for at most 11 hr

c. z* = 10.024 hr

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 value of 9 hr and a standard deviation of 0.8 hr

This means that $\mu = 9, \sigma = 0.8$

a. What is the probability that the player functions for at least 8 hr?

This is 1 subtracted by the pvalue of Z when X = 8. So

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

$Z = \frac{8 - 9}{0.8}$

$Z = -1.25$

$Z = -1.25$ has a pvalue of 0.1056

1 - 0.1056 = 0.8944

0.8944 = 89.44% probability that the player functions for at least 8 hr.

b. What is the probability that the cassette player works for at most 11 hr?

This is the pvalue of Z when X = 11. So

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

$Z = \frac{11 - 9}{0.8}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

0.9938 = 99.38% probability that the cassette player works for at most 11 hr.

c. Find a number z* such that only 10% of all cassette players will function without battery replacement for more than z* hr.

This is the 90th percentile, which is X when Z has a pvalue of 0.9, that is, X when Z = 1.28.

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

$1.28 = \frac{X - 9}{0.8}$

$X - 9 = 1.28*0.8$

$X = 10.024$

So

z* = 10.024 hr