The following is the probability distribution for the number of daily requests a health secretary receives for a health clearance.

Question

Find the expected daily number of health clearance requests and the standard deviation of the daily number health clearance requests.

NUMBER OF REQUESTS ----- PROBABILITY

0 0.02

1 0.25

2 0.33

3 0.17

4 0.14

5 0.09

Answer 1

The formula is s^2=1/(n-1)∑(x1-x)^2.

Mean=0 0.02

1 0.25

2 0.33

3 0.17

4 0.14

+5 0.09=.90/6=.15

s^2=1/5((.02-.15)^2+(.25-.15)^2+(.33-.15)^2+(.17-.15)^2+(.14-.15)^2+(.09-.15)^2)=standard deviation.

s^2=1/(5(-.13)^2+(.1)^2+(.18)^2+(.02)^2+(-.01)^2+(-.06)^2)

=1/(5(.0169+ .01+ .0324+ .0004+ .0001+ .0036))

=1/(5(.0634))

=1/.317

=3.1545....

Answer 2

Let X be the number of health clearance requests. The expected value of X is

E[X]=∑x*P[X=x], where x ranges from 0 to 5. Hence, E[X]=0*0.02+1*0.25+2*0.33+3*0.17+4*0.14+5*0.09=2.43.

Also, E[X²]=0*0.02+1*0.25+4*0.33+9*0.17+16*0.14+25*0.09=7.59

let S be the standard deviation, then S=squareroot(E[X²]- (E[X])²)=1.298