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Math Help - Normal distribution probability question

  1. #1
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    Question Normal distribution probability question

    I want somebody to tell me, do my workings follow the correct method? I'm fairly certain they do but I want another opinion. Thanks.

    Assume the scores on an aptitude are normally distributed with mean=500 and standard deviation=100

    What is the middle 40%?

    My workings

    p(x1≤x≤x2)=p(z1≤z≤z2)

    => p(z1≤z≤z2)=p(z≤z2)-p(z≥z1)=p(z≤z2)-[1-p(z≤z1)]

    p(z≤z2)=0.7 p(z≤z1)=0.3
    from statistical tables
    => z2= -0.525 z1= 0.525

    z1=(x-500)/100=0.525 => x=500+52.5=552.5

    z2=(x-500)/100=-0.525 => x=500-52.5=447.5

    therefore the middle 40% is between 447.5 and 552.5.

    Now my question. Is that the correct method and approach, it is a bonus if I got it right. I only want to check the method really. Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fobster
    I want somebody to tell me, do my workings follow the correct method? I'm fairly certain they do but I want another opinion. Thanks.

    Assume the scores on an aptitude are normally distributed with mean=500 and standard deviation=100

    What is the middle 40%?

    My workings

    p(x1≤x≤x2)=p(z1≤z≤z2)

    => p(z1≤z≤z2)=p(z≤z2)-p(z≥z1)=p(z≤z2)-[1-p(z≤z1)]

    p(z≤z2)=0.7 p(z≤z1)=0.3
    from statistical tables
    => z2= -0.525 z1= 0.525

    z1=(x-500)/100=0.525 => x=500+52.5=552.5

    z2=(x-500)/100=-0.525 => x=500-52.5=447.5

    therefore the middle 40% is between 447.5 and 552.5.

    Now my question. Is that the correct method and approach, it is a bonus if I got it right. I only want to check the method really. Thanks.
    You want to find d such that:

    p(\bar x -d<x<\bar x +d)=p(-d/ \sigma <z< d/ \sigma)=0.4,

    where z\sim N(0,1).

    Now:

    p(-d/ \sigma <z< d/ \sigma)=Q(d/ \sigma)-Q(-d/ \sigma),

    where Q is the cumulative standard normal distribution which
    is what is tabulated in normal distribution tables.

    But by symmetry:

    Q(-d/ \sigma)=1-Q(d/ \sigma),

    so:

    p(-d/ \sigma <z< d/ \sigma)=2Q(d/ \sigma)-1=0.4

    hence:

    Q(d/ \sigma)=0.7,

    Looking this up in a table gives:

    d/ \sigma= 0.7580,

    and as \sigma =100,

    d=75.8.

    So 40% of scores lie in (500-75.8,500+75.8)=(424.2,575.8)

    RonL
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