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6 August, 18:27

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?

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  1. 6 August, 21:09
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    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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