In: Statistics and Probability
Suppose that our company performs DNA analysis for a law enforcement agency. We currently have one machine that is essential to performing the analysis. When an analysis is performed, the machine is in use for a third of the day. Thus, each machine can perform at most THREE DNA analyses per day. Based on past experience, the distribution of analyses needing to be performed on any given day are as follows: (Fill in the table for any hope of credit.)
Jobs |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 or more |
f(x) |
.05 |
.08 |
.11 |
.17 |
.23 |
.18 |
.11 |
.07 |
Answer:(Yearly) |
|
We are considering purchasing a second machine. For each analysis that the machine performs, we profit $1200. What is the yearly expected value of this new machine? (Assume 365 days per year – no weekends or holidays.)
Answer:
Given that
jobs | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
f(x) | 0.05 | 0.08 | 0.11 | 0.17 | 0.23 | 0.18 | 0.11 | 0.07 |
1st row = the number of jobs we need to perform per day
2nd row = f(x) that is probability of we needing to to perform the given jobs...
Now to find the yearly expected value(μ) of this new machine? (Assume 365 days per year – no weekends or holidays.) :
We know that
The formula of expected value( μ ) is given by
Now,
E(X) = 0(0.05)+1(0.08)+2(0.11)+3(0.17)+4(0.23)+5(0.18)+6(0.11)+7(0.07)
E(X) = 0+0.08+0.22+0.51+0.92+0.9+0.66+0.49
E(X) = 3.78
Therefore the value of E(X) is 3.78
Here we already have 1 machine can perfom at most 3 jobs per day
Therefore, new machine will perform =3.78- 3
= 0.78 jobs per day
Now yearly expected value of new machine = 1200*0.78*365
yearly expected value of new machine = 341640
Therefore the yearly expected value of this new machine is 341640