If X is χ2(23), find the following:

(a) P(14.85 < X < 32.01).

(b) Constants a and b such that P(a < X < b) = 0.95 and P(X < a) = 0.025.

(c) The mean and variance of X.

Step 1 of 5:

Given that X has a distribution with 23 degrees of freedom.That is n=23.

Therefore mean of X is 23 and variance of X is 2*23=46. That is =23 and =46.

Standard deviation of X becomes =6.7823.That is =6.7823.

Step 2 of 5:

(a)

Here we have to find the value of P(14.85<X<32.01).

To find this value we make use of standard normal approximation.

It is gives by,

P(14.85<X<32.01)=P(<<)

=P(<<)

=P(<Z<)

=P( -1.2017<Z<1.3285)

=P(Z<1.3285)-P(Z<-1.2017)

We have to use the table of standard normal distribution table representing the area to the left of the Z score statistic to get the probability values.

=0.90658-0.11314

=0.79344

Thus,P(14.85<X<32.01)=0.79344.

Step 3 of 5:

(b)

Here we have to find the values of the constants ‘a’ and ‘b’ using the fact that P(a<X<b)=0.95 and P(X<a)=0.025).

We know that P(-1.96<Z<1.95)=0.95. That is 95% of the area under the normal curve lies between -1.96 and +1.96. Using this we can find the values of a and b.

Using standard normal approximation, we get

P(a<X<b)=P( <<)

=P(<Z<)

It is given that P(<Z<)=0.95.

=-1.96 and =1.96

Consider

=-1.96

a-23=(-1.96)*6.7823

=-13.293308

a=-13.293308+23

=9.70662.

Now consider

=1.96

b-23=1.96*6.7823

=13.293308

Threfore

b=13.293308+23

=36.293308

Thus the value of constant a is 9.70662 and value of constant b is 36.293308.