Question

Find the value C for each of the following questions:

(a). X is normal with mean 5 and variance 4. P{ X< 4}=C.

(b). X is normal with mean 5 and variance 4, Y is normal with mean 6 and variance 5. P{ Y-X < 4}=C.

(c ). X is t random variable with 4 degrees of freedom . P{ X < C= .99}.

Answer 1

(a)

Mean, m = 5

Standard Deviation = 4^0.5 = 2

At X = 4, the corresponding z-score is:

z = (X-m)/S = (4-5)/2 = -0.5

So, from the cumulative z-table we have:

P(X<4) = P(z<-0.5) = 0.308

(b)

For the variable Y-X, we have:

Mean, m = 6-5 = 1

Standard Deviation = (4+5)^0.5 = 3

At Y-X = 4, the corresponding z-score is:

z = ((Y-X)-m)/S = (4-1)/3 = 1

So, from the cumulative z-table we have:

P((Y-X)<4) = P(z<1) = 0.842

(c)

Degrees of freedom, df = 4

From the cumulative t-table, we have:

P(X<3.747) = 0.99

So,

C = 3.747

Hope this helps !