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    • Thread Starter

    I have no idea how to answer these kind of questions. I've put two of them below. Can anyone help me?

    A sample of 25 bottles is taken from the production line at a local bottling plant. Assume that the fill amounts follow a normal distribution. Which of the following statements is the most accurate estimate of the probability that the sample standard deviation is more than 40% of the population standard deviation?
    A) The probability is smaller than 0.95
    B) The probability is smaller than 0.975
    C) The probability is smaller than 0.995
    D) The probability is greater than 0.995

    The time needed to process a claim at an insurance company is known to follow the normal distribution. For a random sample of 22 insurance claims, what is the probability that the sample variance is more than 60% greater than the population variance?
    A) <0.01
    B) <0.025
    C) <0.05
    D) <0.005

    Hmmm. Does it help knowing that since the fill amounts are normally distributed then

    (n-1)*s^2/(sigma)^2 is chi-squared with n-1 degrees of freedom ?

    Have you done Cochran's theorem?
    • Thread Starter

    It's just come under the topic of normal distribution. We haven't done chi-squared yet. Other questions around it are the usual P(X<50) = (Pz<...)

    "more than 40% of the population standard deviation"...find the probability corresponding to a z-value of 0.4.
    Subtract from 1 (since it's more than). So option A.
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Updated: March 20, 2011
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