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2. The Poisson Probability Distribution

The Poisson distribution was developed by the French mathematician Simeon

Denis Poisson in 1837.

The Poisson random variablesatisfies the following conditions:

1. The number of successes in two disjoint time intervals is independent.2. The probability of a success during a small time interval is proportional to

the entire length of the time interval.

Apart from disjoint time intervals, the Poisson random variable also applies todisjoint regions of space.

Applications

the number of deaths by horse kicking in the Prussian army (firstapplication)

birth defects and genetic mutations rare diseases (like Leukemia, but not AIDS because it is infectious and so not

independent) - especially in legal cases

car accidents traffic flow and ideal gap distance number of typing errors on a page hairs found in McDonald's hamburgers spread of an endangered animal in Africa failure of a machine in one month

The probability distribution of a Poisson random variableX representing the

number of successes occurring in a given time interval or a specified region of

space is given by the formula:

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where

x= 0, 1, 2, 3...

e= 2.71828 (but use your calculator's ebutton)

= mean number of successes in the given time interval or region of space

Mean and Variance of Poisson Distribution

If is the average number of successes occurring in a given time interval or region

in the Poisson distribution, then the mean and the variance of the Poisson

distribution are both equal to .

E(X) =

and

V(X) = 2=

Note: In a Poisson distribution, only oneparameter, is needed to determine the

probability of an event.

EXAMPLE 1

A life insurance salesman sells on the average 3 life insurance policies per week.

Use Poisson's law to calculate the probability that in a given week he will sell

(a) some policies

(b) 2 or more policies but less than 5 policies.

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(c) Assuming that there are 5 working days per week, what is the probability that

in a given day he will sell one policy?

Answer

Here, = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1

minus the "zero policies" probability:

P(X> 0) = 1 P(x0)

Now So o

b)

c) Average number of policies sold per day:

So on a given day,

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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?

Answer

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:

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The required probability is:

Histogram of Probabilities

We can see the predicted probabilities for each of "No flaws", "1 flaw", "2 flaws",etc on this histogram.


[The histogram was obtained by graphing the following function for integer values

ofxonly.

Then the horizontal axis was modified appropriately.]

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EXAMPLE 3

If electricity power failures occur according to a Poisson distribution with an

average of 3 failures every twenty weeks, calculate the probability that there will

not be more than one failure during a particular week.

Answer

The average number of failures per week is:

"Not more than one failure" means we need to include the probabilities for "0

failures" plus "1 failure".

EXAMPLE 4

Vehicles pass through a junction on a busy road at an average rate of 300 per hour.

1. Find the probability that none passes in a given minute.2. What is the expected number passing in two minutes?3. Find the probability that this expected number actually pass through in a

given two-minute period.

Answer

The average number of cars per minute is:

(a)
http://www.intmath.com/counting-probability/ans-13.php?a=3http://www.intmath.com/counting-probability/ans-13.php?a=3http://www.intmath.com/counting-probability/ans-13.php?a=3

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(b) E(X) = 5 2 = 10

(c) Now, with = 10, we have:


Based on the function

we can plot a histogram of the probabilities for the number of cars per minute:

EXAMPLE 5

A company makes electric motors. The probability an electric motor is defective is

0.01. What is the probability that a sample of 300 electric motors will contain

exactly 5 defective motors?

Answer

The average number of defectives in 300 motors is = 0.01 300 = 3

The probability of getting 5 defectives is:
http://www.intmath.com/counting-probability/ans-13.php?a=4http://www.intmath.com/counting-probability/ans-13.php?a=4http://www.intmath.com/counting-probability/ans-13.php?a=4

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NOTE:This problem looks similar to abinomial distributionproblem, that we met

in the last section.

If we do it using binomial, with n= 300,x= 5,p= 0.01 andq= 0.99, we get:

P(X= 5) = C(300,5)(0.01)5(0.99)

295= 0.10099

We see that the result is very similar. We can use binomial distribution to

approximate Poisson distribution (and vice-versa) under certain circumstances.


3.The Normal Distribution

A random variableXwhose distribution has the shape of a normal curveis called a

normal random variable.
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Normal Curve

This random variableXis said to be normally distributed with mean and standard

deviation if its probability distribution is given by

Properties of a Normal Distribution

1. The normal curve is symmetrical about the mean ;2. The mean is at the middle and divides the area into halves;3. The total area under the curve is equal to 1;4. It is completely determined by its mean and standard deviation (or

variance 2)

Note:

In a normal distribution, only 2parameters are needed, namely and 2.

Area Under the Normal Curve using Integration

The probability of a continuous normal variableXfound in a particular interval [a,

b] is the area under the curve bounded byx = aandx = band is given by

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and the area depends upon the values of and .

The Standard Normal Distribution

It makes life a lot easier for us if we standardizeour normal curve, with a mean of

zero and a standard deviation of 1 unit.

If we have the standardized situationof = 0and = 1, then we have:

Standard Normal Curve = 0, = 1

We can transform all the observations of any normal random variableXwith mean

and variance to a new set of observations of another normal random variableZ

with mean 0 and variance 1 using the following transformation:

We can see this in the following example.

Example

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Say = 2 and = 1/3in a normal distribution.

The graph of the normal distribution is as follows:

= 2, = 1/3

The following graph represents the same information, but it has been

standardizedso that = 0 and = 1:

= 0, = 1

The two graphs have different and , but have the same shape (if we tweak the

axes).

The new distribution of the normal random variableZwith mean 0 and variance 1

(or standard deviation 1) is called a standard normal distribution. Standardizing

the distribution like this makes it much easier to calculate probabilities.

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If we have mean and standard deviation , then

Since all the values of X falling between x1 and x2 have corresponding Z values

between z1and z2, it means:

The area under theXcurve betweenX = x1andX = x2equals:

The area under theZcurve betweenZ = z1andZ = z2.

Hence, we have the following equivalent probabilities:

P(x1

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The area above is exactly the same as the area

z1= 0.5 to z2= 2

in the standard normal curve:

= 0, = 1

Percentages of the Area Under the Standard Normal Curve

A graph of this standardized (mean 0 and variance 1) normal curve is shown.

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In this graph, we have indicated the areas between the regions as follows:

-1 Z 168.27%

-2 Z 2 95.45%

-3 Z 399.73%

This means that 68.27% of the scores lie within 1 standard deviation of the mean.

This comes from:

Also, 95.45% of the scores lie within 2 standard deviations of the mean.

This comes from:

Finally, 99.73% of the scores lie within 3 standard deviations of the mean.

This comes from:

The total area from - < z< is 1.

The z-Table

The areas under the curve bounded by the ordinates z= 0 and any positive value

of zare found in the z-Table. From this table the area under the standard normal

curve between any two ordinates can be found by using the symmetry of the curve

about z= 0.

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EXAMPLE 1

Find the area under the standard normal curve for the following, using the z-table.

Sketch each one.

(a) between z= 0 and z= 0.78

(b) between z= -0.56 and z= 0

(c) between z= -0.43 and z= 0.78

(d) between z= 0.44 and z= 1.50

(e) to the right of z= -1.33.

Answer

From the z-table:

(a) 0.2823

(b) 0.2123

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(c) 0.1664 + 0.2823 = 0.4487

(d) 0.4332 - 0.1700 = 0.2632

(e) 0.4082 + 0.5 = 0.9082

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EXAMPLE 2

Find the following probabilities:

(a) P(Z> 1.06)

(b) P(Z< -2.15)

(c) P(1.06

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This time, we need to take the area of the whole left side (0.5) and subtract the

area from z= 0 to z= 2.15 (which is actually on the right side, but the z-table is

assuming it is the right hand side.)

(c) This is the same as asking "What is the area between 1.06 and 4.00 under the

standard normal curve?"

(d) This is the same as asking "What is the area between -1.06 and 4.00 under the

standard normal curve?"

We find the area on the left side from z= -1.06 to z= 0 (which is the same as z= 0

to z= 1.06), then add the area between z= 0 to z= 1.04 (on the right side):

EXAMPLE 3

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It was found that the mean length of 100 parts produced by a lathe was 20.05 mm

with a standard deviation of 0.02 mm. Find the probability that a part selected at

random would have a length

(a) between 20.03 mm and 20.08 mm

(b) between 20.06 mm and 20.07 mm

(c) less than 20.01 mm

(d) greater than 20.09 mm.

Answer

X= length of part

(a) 20.03 is 1 standard deviation below the mean;

20.08 is standard deviations above the mean

So the probability is 0.7745.

(b) 20.06 is 0.5 standard deviations above the mean;

20.07 is 1 standard deviation above the mean

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(c) 20.01 is 2 s.d. below the mean.


(d) 20.09 is 2 s.d. above the mean, so the answer will be the same as (c),

P(X> 20.09) = 0.0228.

EXAMPLE 4

A company pays its employees an average wage of $3.25 an hour with a standard

deviation of 60 cents. If the wages are approximately normally distributed,

determine

a. the proportion of the workers getting wages between $2.75 and $3.69 anhour;

b. the minimum wage of the highest 5%.Answer

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X= wage

(a)

So about 56.6% of the workers have wages between $2.75 and $3.69 an hour.

(b) W= minimum wage of highest 5%

x= 1.645 (from table)

Solving gives:x= 4.237

So the minimum wage of the top 5% of salaries is $4.24.

EXAMPLE 5

The average life of a certain type of motor is 10 years, with a standard deviation of

2 years. If the manufacturer is willing to replace only 3% of the motors that fail,

how long a guarantee should he offer? Assume that the lives of the motors follow

a normal distribution.

Answer

X= life of motor

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x= guarantee period

Normal Curve: = 10, = 2

We need to find the value (in years) that will give us the bottom 3% of thedistribution. These are the motors that we are willing to replace under the

guarantee.

P(X < x) = 0.03

The area that we can find from the z-table is

0.5 - 0.03 = 0.47

The corresponding z-score is z= -1.88.

Since , we can write:

Solving this givesx= 6.24.

So the guarantee period should be 6.24 years.

Example 6

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In the standard normal curve to the right, the mean is 0 and the standard

deviation is 1.

The green shaded area in the diagram represents the area that is within 1.45

standard deviations from the mean. The area of this shaded portion is 0.4265 (or

42.65% of the total area under the curve).

To get this area of 0.4265, we read down the left side of the table for the standard

deviation's first 2 digits (the whole number and the first number after the decimal

point, in this case 1.4), then we read across the table for the "0.05" part (the toprow represents the 2nd decimal place of the standard deviation that we are

interested in.)

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279

We have:

(left column) 1.4 + (top row) 0.05 = 1.45 standard deviations

The area represented by 1.45 standard deviations to the right of the mean is

shaded in green in the standard normal curve above.

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You can see how to find the value of 0.4265 in the full z-table below. Follow the

"1.4" row across and the "0.05" column down until they meet at 0.4265.

Exercises

1. In a Binomial distribution containing of 5 independent trials the probability of 1

and

2 successes are 0.49 and 0.2048 respectively. Find parameter p of the

distribution.

2.Ten coins are thrown simultaneously. Find the probability of getting 7 heads.

3. The probability of a man hitting a target is 1/3.

(i) If he fires 5 times, what is the probability of his hitting the target at least

twice.

(ii) How many times must he fires so that the probability of his hitting the target

at

least once is more than 90%.

4. Find mean and variance of Poisson distribution.

5.The probability of a Poisson variate taking the values 1 and 2 are equal.

Find (i) (ii) P (X1) (iii) P (1

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(ii) how many students get more than 40

(iii) how many students get below 20

(iv) how many students get more than 50

****

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