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Math Help - Am I right on this proportion

  1. #1
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    Am I right on this proportion

    The question is:
    The weights of trout in a certain river follow a normal distribution with mean 3 pounds and standard deviation 0.8 pounds. What proportion of trout weigh between 3.2 pounds and 4.8 pounds?

    Answers are:
    .3891
    .9972
    .9452
    .5865

    I think it's either .9972 or .9452 but can't be sure
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  2. #2
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    Re: Am I right on this proportion

    If a random variable X has a normal distribution with mean \mu and standard deviation \sigma, then the cumulative distribution function (CDF) F_X(z)=\Phi\left(\frac{z-\mu}{\sigma}\right) where \Phi is the CDF of the standard normal distribution (with mean 0 and standard deviation 1) (Wikipedia). You need to find F_X(4.8)-F_X(3.2). The function \Phi(z) can be computed, for example, by Excel function NORMSDIST. There is also a function NORMDIST to compute the CDF of an arbitrary normal distribution.
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  3. #3
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    Re: Am I right on this proportion

    Quote Originally Posted by jay25 View Post
    The question is:
    The weights of trout in a certain river follow a normal distribution with mean 3 pounds and standard deviation 0.8 pounds. What proportion of trout weigh between 3.2 pounds and 4.8 pounds?

    Answers are:
    .3891
    .9972
    .9452
    .5865

    I think it's either .9972 or .9452 but can't be sure
    It's the first answer.
    Intuitively, if the mean is 3 and your range of values is above that,
    the probability of trout being in that weight range must be less than half.

    You need to work it out using the Z-tables of the Normal Distribution,
    first by converting to Z using

    Z=\frac{x-\mu}{\sigma}

    So the solution is

    P\left(Z\le \frac{4.8-3}{0.8}\right)-P\left(Z\le \frac{3.2-3}{0.8}\right)
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