Suppose that the distribution of the lifetime of a car battery producedby a certain car company is well approximated by a normal distribution with a meanof 1.2*103hours and variance 104. What is the approximate probability that abatch of 100 car batteries will contain at least 20 whose lifetimes are less than 1100hours?

0.1971

Step-by-step explanation:

First, let's find the probability that a single battery last less than 1100 hours

Let's find z

z = (x - mean) / standard deviation

z = (1100 - 1200) / 104

z = -.96

p (z<-0.96) = 0.1684

Let's check if we can use the Normal Approximation to the Binomial to solve this problem

Given

n = sample size = 100

p = probability = 0.1684

q = 1 - p = 0.8316

n * p and n * q has to be greater that 5

n * p = 0.1684 * 100 = 16.81

q * p = 0.8316 * 100 = 83.16

we can use the Normal Approximation to the Binomial

mean = n * p = 16.81

standard deviation = √ (n * p * q) = √ (100 * 0.1684 * 0.8316) = 3.74

now we can find z and the probability that at least 20 batteries has a lifetine less than 1100 hours

z = (20 - 16.81) / 3.74 = 0.852

p (z>0.852) = 0.1971