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Math Help - normal distribution

  1. #1
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    normal distribution

    im having problem with this question:

    The mass of a grade A papaya follows a normal distribution with mean 1.8 kg and standard deviation 200 g. The mass of a grade B papaya follows a normal distribution with mean 1.5 kg and standard deviation 140 g. Determine the probability that
    a)3 grade A papayas chosen at random has a total mass of less than 5 kg,

    b)3 times the mass of a grade A papaya chosen at random is less than 5 kg,

    c)the total mass of 6 grade A papayas is more than that of 7 grade B papayas, all chosen at random.

    please help...
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  2. #2
    Junior Member
    Joined
    Nov 2008
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    This seems more like a prob or stats question than calculus but whatever. Let's call A the mass distribution of grade A papayas and B the mass distribution of grade B papayas. Well then,

    A\sim N(1.8,0.2^2) and B\sim N(1.5, 0.14^2)

    a) We want 3 grade A papayas so the distribution is A+A+A\sim N(1.8+1.8+1.8,0.2^2+0.2^2+0.2^2)=N(5.4,0.12). So, \frac{A+A+A-5.4}{\sqrt{0.12}}\sim N(0,1). To find the probability, use a normal distribution table to find:

    <br />
P(A+A+A<5)=\Phi (\frac{5-5.4}{\sqrt{0.12}})=\Phi (-1.15)=0.1251<br />

    b) Same logic, 3A\sim N(3(1.8),(3\cdot 0.2)^2)=N(5.4,0.36)

    <br />
P(3A<5)=\Phi (\frac{5-5.4}{\sqrt{0.36}})=\Phi (-0.67)=0.2514<br />

    c) Let X= A+A+...+A (six times) -B-B-...-B (seven times)

    then, X\sim N(6(1.8)-7(1.5), 6(0.2^2)+7(0.14^2))=N(0.3, 0.3772)

    So to have 6 A papayas heavier than 7 B papayas means that X>0.

    <br />
P(X>0)=1-P(X<0)=1-\Phi (\frac{0-0.3}{\sqrt{0.3772}})=1-\Phi (-0.49)=1-0.3121=0.6879<br />
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