# Normal distribution probability question

• Jan 17th 2006, 10:55 AM
fobster
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.
• Jan 19th 2006, 07:56 PM
CaptainBlack
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 $\displaystyle d$ such that:

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

where $\displaystyle z\sim N(0,1)$.

Now:

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

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

But by symmetry:

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

so:

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

hence:

$\displaystyle Q(d/ \sigma)=0.7$,

Looking this up in a table gives:

$\displaystyle d/ \sigma= 0.7580$,

and as $\displaystyle \sigma =100$,

$\displaystyle d=75.8$.

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

RonL