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Math Help - Normal distribution problem

  1. #1
    Junior Member Fnus's Avatar
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    Normal distribution problem

    Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!

    The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
    Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Fnus View Post
    Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!

    The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
    Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.
    \Pr(25 < X < x) = 0.3

    \Rightarrow \Pr(X < x) - \Pr(X < 25) = 0.3

    \Rightarrow \Pr(X < x) - 0.5 = 0.3

    \Rightarrow \Pr(X < x) = 0.8

    which is now just a routine inverse normal problem.
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  3. #3
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    First thing you do it simply picture a normal distribution curve. Right in the middle is the average, which is 25 kg in this case. You are looking for an area under the curve in between x and the mean, and since x > 25 it will be the right side of the mean.

    Since both on the left an the right side of the mean you have exactly 50% you know that left of the mean you have 50%, then you have the area mean to x which is 30% and then you have a remaining 20%.

    I assume you have a TI calculator, if not, what I'll say is useless. You press [2ND] [Vars] [2] [ENTER] and you'll have the program 'normalcdf'. If you enter a left boundy, a right boundry, the mean, the standard deviation and it will give you the area contained withing the boundaries. (for instance: normalcdf(25,10^90,25,3) will give exactly .5 because that's half of the whole area.)

    You are looking for not halve, but 30% = .3 of the area. Go to [tex] solver and enter eqn: 0= normalcdf(25,x,25,3) - 0,3 and then press down [APLHA] [ENTER] and it will give you the answer.

    I hope you understand. And also, you mentioned plotting, which is what most schools teach, but using the solver is a LOT easier and faster.
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