A study of long distance phone calls made from General Elect

A study of long distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minute. a) What fraction of the calls lasts between 4.2 and 5 minutes? b) What fraction of the calls lasts more than 5 minutes? c) What fraction of the calls lasts between 5 and 6 minutes? d) What fraction of the calls lasts between 4 and 6 minutes? e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

HOW DO I DO THIS IS EXCEL?

Solution

For solving the normal distribution question in excel, we use the function NORM.DIST

Here Mean = 4.2, SD = 0.60,

a) What fraction of the calls lasts between 4.2 and 5 minutes?

Type in a cell

= NORM.DIST(5 , 4.2 , 0.60, TRUE) and hit enter.

This will give us the area to the right of the value X = 5. The last argument TRUE ensures that we get a cumulative area to the left.

Since we need area between X = 5 and X = 4.2, we give the command as:

= NORM.DIST(5, 4.2, 0.60, TRUE) - NORM.DIST(4.2, 4.2, 0.60, TRUE) and hit enter.

Thus the solutions to part

a) What fraction of the calls lasts between 4.2 and 5 minutes?

=  NORM.DIST(5, 4.2, 0.60, TRUE) - NORM.DIST(4.2, 4.2, 0.60, TRUE)

= 0.408789

b) b) What fraction of the calls lasts more than 5 minutes?

= 1 - NORM.DIST(5, 4.2, 0.60, TRUE) [ 1 - area to the left = area to the right)

= 0.091211

c) What fraction of the calls lasts between 5 and 6 minutes?

= NORM.DIST(6, 4.2, 0.60, TRUE) - NORM.DIST(5, 4.2, 0.60, TRUE)

= 0.089861

d) What fraction of the calls lasts between 4 and 6 minutes?

= NORM.DIST(6, 4.2, 0.60, TRUE) - NORM.DIST(4, 4.2, 0.60, TRUE)

= 0.629209

E)

The part e is a tricky one. Here we need longest 4% of the calls or that time duration which gives the probability of 96% of area to it\'s left.

i.e we need to find X where P(x<X) = 0.96

This can be done by inverse function

= NORM.INV(0.96 , 4.2 , 0.60) where the first argument is a probability, then we have mean and SD

We will get the value of NORM.INV(0.96 , 4.2 , 0.60) = 5.25 min

Thus, longest 4% of the calls last 5.25 minutes or longer

Hope this helps. Ask if you have any doubts.