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?

Answer:0.1971Step-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 hoursz = (20 – 16.81)/3.74 = 0.852 p (z>0.852) = 0.1971