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# Stats Project

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by

Tweet## Lauren Ossola

on 13 November 2012#### 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 transcriptx- 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%