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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?

Respuesta :

Answer:

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

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