Question

When used in a particular DVD player, the lifetime of a certain brand of battery is normally distributed with a mean value of 12 hours and a standard deviation of 0.8 hours. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. (Use a table or technology.)

(a) What is the probability that the DVD player functions for at least 10 hours? (Round your answer to four decimal places.)

(b) What is the probability that the DVD player works for at most 13 hours? (Round your answer to four decimal places.)

(c) Find a number x* such that only 10% of all DVD players will function without battery replacement for more than x* hours. (Round your answer to two decimal places.) x* = hr

Answer 1

Solution:

a) Probability that the DVD player functions for at least 10 hours

b) Probability that the DVD player works for at most 13 hours

c) number x* such that only 10% of all DVD players will function without battery replacement for more than x* hours.

Given that

µ = 12

σ = 0.8

for given 0.10 value z = 1.645

? = ?? + ?

= 0.8 * 1.645 + 12
= 13.316