Normal distribution probability question

[fobster]

Got a question I need a little bit of help.

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.

 

[maverick6664]

I think it's correct.

Normal distribution is given $$\frac 1 {\sigma \sqrt{\pi}} exp(-(x-\mu)^2/\sigma^2)$$, where $$\mu$$ is the mean and $$\sigma$$ is the standard deviation.

I looked at table and center 20% probability is (standard deviation)*(between 0.52 and 0.53)

So your calculation looks fine.

 

[Architeuthis Dux]

Your approach and method are correct. You correctly used the standard normal distribution and statistical tables to find the z-scores corresponding to the middle 40% of the distribution. Then, you used the formula for converting z-scores to raw scores to find the corresponding scores on the aptitude test. This is a valid and commonly used method for finding the middle percentage of a normal distribution. Good job!