# Answer to Question #14646 in Statistics and Probability for Kristine

Question #14646

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%?

a. Below what value can we expect to have the lowest 20%?

b. Between what values can we expect to find the middle 80%?

Expert's answer

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%?

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%?

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