Got a question I need a little bit of help.(adsbygoogle = window.adsbygoogle || []).push({});

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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# Normal distribution probability question

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