Poisson Distribution Part II
Strickberger pgs. 146-148 gives an example of the use
of the Poisson distribution. We do not know the probability
of success, only the mean number of successful events.
(Pwa son) Example
Let the mean number of successful events = 0.35. Find
the probability of 0, 1, 2, 3, 4 successes.
m = 0.35, then e-m = e-0.35
e
= 1 
e
= 1
e e
e0.35 = 1.42
e-0.35 = 1/1.42 = 0.705
number
of
successes (X) |
Probability
(X successes) |
| 0 |
e-0.35
(1) = 0.705 (1) = 0.705 |
| 1 |
e-0.35
(m) = 0.705(0.35) = 0.247 |
| 2 |
e-0.35
(m2/2!) = 0.705(0.35)2/2 = 0.043 |
| 3 |
0.705(0.35)3/6 = 0.005 |
| 4 |
0.705(0.35)4/24 = 0.004 |
m = mean number of successes
Prob (success) = m
number of trials
m = Prob (success) x number of trials
Example
We must estimate m from an experiment and then calculate
the probability of x successful events.
150 petri dishes were each plated with one million
bacteria on streptomycin agar.
Mean number of events = m =0.35
e(1,
m, m
,
m,
m
...)
2!
3! 4!
e
= 0.705; e = 2.7183
Example
| Mutant
Colonies |
Number
of Petri Dishes |
Number
of Muntant Colonies |
| 0 |
98 |
0 |
| 1 |
40 |
40 |
| 2 |
8 |
16 |
| 3 |
3 |
9 |
| 4
or more |
1 |
4 |
| |
150 |
69 |
We observed 98 out of 150 petri dishes with zero mutant
colonies. We observed 40 of 150 petri dishes with one
mutant colony per dish.
Out estimate of the mean number of mutant colonies
is:
m = 69 = 0.46
per dish
150
e-0.46 = 0.631. We will calculate the expected
probability of 0, 1, 2, 3, and 4 mutant colonies per
dish.
Mutant
Colonies
per dish |
Expected
Probability |
Observed
Proportion |
| 0 |
e-0.46
(1) = 0.631 |
98/150
= 0.653 |
| 1 |
e-0.46
(0.46) = 0.291 |
40/150
= 0.267 |
| 2 |
e-0.46(.46)2/2=0.0668 |
0.053 |
| 3 |
0.631(46)3/6=0.0102
|
0.020 |
| 4 |
0.631(.46)4/24=0.001
|
0.007 |
| Total |
1.00 |
|
