Question

Consider a normal distribution with mean of 500 and a standard deviation of 50.
a. Below what value can we expect to have the lowest 20%?
b. Between what values can we expect to find the middle 80%?

Answer 1

Let X be a random variable having a normal distribution with mean of 500 and a
standard deviation of 50.

Let also Z = (X-500)/50.
Then Z has a
standard normal distribution, i.e. mean=0 and standard deviation =
1.

a. We should find value X0 such that

P(X<X0)=0.2

Denote
Z0 = (X0-500)/50
Thus
0.2 = P(X<X0)

= P( (X-500)/50 < (X0-500)/50 ) =
= P( Z < (X0-500)/50 )
=
= P( Z < Z0 )
= F(Z0),

where F is the cummulative
distribution function for standard normal distribution.
The values of F can
be taken from tables.
We have that
F(-0.84162) = 0.2,
and so
Z0
= -0.84162,
whence
X0 = 500 + 50 * Z0
= 500 - 50 *
0.84162=
= 457.91894.



b. We should find B>0 such
that

P(500-B < X < 500+B) = 0.8

Then
0.8 = P(500-B
< X < 500+B)
= P( (500-B-500)/50 < (X-500)/50 <
(500+B-500)/50 )
= P( -B/50 < Z < B/50 )
= F(B/50) -
F(-B/50)

Notice that
F(-x) = 1-F(x),
whence
F(x) -
F(-x) = F(x) - 1+ F(x) = 2F(x) - 1.
Thus
0.8 = 2F(B/50)-1,

F(B/50) = (1+0.8)/2 = 0.9

From tables of values of F we get that

B/50 = 1.28155,
whence
B = 50* 1.28155 = 64.07758

And so

500-B = 435.92242
500+B = 564.07758

Thus between the values

435.92242 and 564.07758
we expect to find the middle 80%?