Statistic Word Problem using Excel functions NORM. DIST and NORM.INV. Help!!!

Question

Suppose that a random variable X follows a normal distribution with mean 81.0 and standard deviation 7.3. Find the following using NORM. DIST and NORM.INV:

A) p(x<85.6)

B) p(x greater than or equal to 92.3)

C) p(70.0 < x <92.3)

D) Give the value C for which P(x > c)= .685

Answer 1

Kiki,

I don't have the EXCEL functions, but I'll explain the process mathematically, and give you the approximate correct answers.

 

The Normal distribution (also called the Gaussian distribution) is probably a "standardized" one, whose mean =0 and standard deviation s=1.  But your random variable X follows a non-standardized Normal distribution because its center or mean is 81.0, not zero, and its standard deviation is given as 7.3, not 1.  So you'll have to take any X-distribution values and convert them to standardized ones before you can use the EXCELL functions or a table lookup in a book.

 

Let Z = equivalent standardized variable corresponding to an X value

Then, Z = (X - mean)/(std. deviation) = (X - 81.0)/7.3

 

Now to answer the four lettered questions:

 

(A): p(X < 85.6)

The corresponding Z-value for X=85.6  is (85.6 - 81.0)/7.3 = 4.6/7.3 = 0.63

The question then becomes p(Z < 0.63) = 0.7357           I looked it up in a table of the standardized Normal and it represents the area under the Normal from Z = minus infinity to Z = 0.63 

 

(B): p(X >= 92.3)

Z = (92.3 - 81.0)/7.3 = 11.3/7.3 = 1.548  so its asking for the area under the right tail of the Normal.  Remember that the area on either side of mean = 0 is 0.5 since the Normal distribution is symmetrical.

p(Z >= 1.548) = 0.50 - p(0 <= Z < 1.548) Also note that the area is the same whether we use <= or just =.

p( ) = 0.50 - 0.4392 = 0.0608

 

(C): p(70.0 < X < 92.3)

Z₁ = (70.0 - 81.0)/ 7.3 = -11.0/7.3 = - 1.507  and as in (B) X = 92.3 corresponds to Z₂ = 1.548

So we get p(Z1 < Z < Z2) = p(-1.507 < Z < +1.548), so just throw away the probabilities of the two tails

p( ) = 1.0 - p(Z < -1.507) - p(Z > 1.548)  and we already have the value for the right tail from (B)

p( ) = 1.0 - 0.0659 - 0.0608 = 0.8733

 

(D): p(X > C) = 0.685

First we find what value of negative Z gives a 0.685 probability to the right of it or 0.315 to the left of it.

Z = - 0.481

Then find X from Z = -0.481 = (C - 81.0)/7.3 

So  C = 81.0 - (-0.481)(7.3) = 81.0 - 3.51 = 77.49