### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

# Stats Project

No description
by

## Lauren Ossola

on 13 November 2012

Report abuse

#### Transcript of Stats Project

By Lauren Ossola Poisson Distribution THE END Formula Lambda- average rate of mean
x- number of successes
e- base of the natural logarithm function Use Poisson if...... if you have an average rate of an event happening over a period of time! For Example: Ex. On an average Friday, a waitress gets no tip from 5 customers. Find the probability that she will get no tip from 7 customers this Friday.

The waitress averages 5 customers that leave no tip on Fridays :
lambda = 5.

Random Variable : The number of customers that leave her no tip this Friday.
We are interested in .

So, the probability that 7 customers will leave no tip this Friday is 0.1044 or 10.44% Now you try!

Ex. Once every two days a dog in Denver dies. What is the probability of there being two dogs die in one week? = 1/2 death/days

= 0.5x7= 3.5

We are interested in P(X=2) Your answer: 0.185 or there is an 18.5% chance that two dogs in Denver will die in one week P(x=2)= ((3.5^2)/2!))(e^-3.5) Example 2:
Twenty sheets of aluminum alloy were examined for surface flaws. The frequency of the number of sheets with a given number of flaws per sheet was as follows:

Number of flaws-Frequency
0 - 4
1 - 3
2 - 5
3 - 2
4 - 4
5 - 1
6 - 1
What is the probability of finding a sheet chosen at random which contains 3 or more surface flaws? Solution: The total number of flaws is given by:

(0×4)+(1×3) +(2×5)+(3×2) +(4×4)+(5×1) +(6×1) =46

So the average number of flaws for the 20 sheets is given by:

μ=46/20=2.3
The required probability is:

Probability=P(X≥3)
=1−(P(x0)+P(x1)+P(x2))
=1−(e−2.32.300!+e−2.32.311!+e−2.32.322!)
=0.40396

In other words, the probability that a random sheet will contain three or more surface flaws is 40.4%
Full transcript