The Binomial and Poisson Distributions |

**Introduction**

The Binomial and Poisson are **discrete** distributions -- for they are based

on **counting successes** rather than **measuring data** like a **continuous **distribution does.

In other words, the data values are frequencies rather than measures. For example,

if we're counting the number of customers in a Metro Store who buy the ice cream that's

on special, we would label a "buy" as a success and we'd build our probability distribution

from the frequencies recorded. Obviously, frequencies are **integers** since there's

no such thing as "half a buy". With a continuous distribution like the normal distribution,

our data comes from **measuring**, say the lengths of a school of fish or the weight of a

sample of bags of flour -- so the data can take on real number values, not just integers.

**The Binomial Distribution**

The Binomial Distribution counts the **number of successes in a fixed number of trials**.

For example, we could count the number of 5's obtained in 20 rolls of a single die,

or the number of Heads that come up in 200 coin tosses. These are both **binomial events**.

The word "*binomial*" in algebra means an expression with 2 terms. Here is means the

experiment has exactly **2 possible outcomes** -- either a **success** or **failure** -- for each trial.

These 4 characteristics identify the **Binomial Distribution**

- 1) each outcome is classified as either a

2) it is a

3) Each trial is

4) The

If the value of P(*x*) cannot be found in a table, we use this formula

**P( x) =_{ }**

where

When we analyze the formula we see that:

_{n}**C**** _{ x }**is the number of ways to get

Since

**Illustration**

Say we want to know the probability of getting **3 Heads in 5** coin **tosses**.

In this case, ** n** = 5,

so P(3H) =

Let's discuss the

so we need to multiply the probabilities of 3 successes and 2 failures by the number

of ways this can happen. For instance,we could get {1H, 2T, 2H}, or {2H, 2T, 1H}

or {2H, 1H, 2T} etc. The

get the desired outcome.

Had we wanted the probability of getting **at most** 3 Heads in 5 coin tosses,

we would** sum** the probabilities of getting 0, 1, 2 or 3 Heads in 5 tosses.

That is, we would find P(0) + P(1) + P(2) + P(3), using the correct values for ** x, n **and

Had we wanted the probability of getting

we would find 1 - [ P(0) + P(1) + P(2) + P(3) ] or we would sum P(4) + P(5).

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**3 more characteristics of the Binomial Distribution**

1) if **n** stays constant while ** p** approaches 0.5, the shape of the Binomial

Distribution becomes more symmetric.

2) if ** p** stays constant while

Distribution becomes more symmetric.

3) the mean **µ = np** and the variance ** ****r**** ² = np(1 - p)** (*formulas 4 and 5 on list*).

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Since the outcomes of a coin toss forms a binomial population, we'll do some coin tossing.

With a coin toss, there are exactly 2 possible outcomes -- Heads or Tails.

Each outcome is independent of the last, since the coin has no idea

what side came up on the last toss.

The probability of a success -- we'll choose Heads (H) as a success -- is always ½.

Though we used the formula to find P(3H) above, there are of course **tables
of Binomial Probabilities**. Problem is, most tables offer only a limited number

of values for

between 1 and 20. So if

either the formula or our calculators to find the needed probability.

Let's repeat the illustration only this time we'll use a Tables of Binomial

Probabilities instead of the formula to find our answers.

**Example 1**

Using Binomial Probability tables, find:

a) the probability of getting exactly 3 heads in 5 coin tosses.

b) the probability of getting at most 3 heads in 5 coin tosses.

The table entries for ** n = 5 **and

n |
x |
P(x) |

5 | 0 | 0.031 |

1 | 0.156 | |

2 | 0.312 | |

3 | 0.312 | |

4 | 0.156 | |

5 | 0.031 |

a) the probability of getting exactly 3 heads in 5 tosses = 0.312

b) the probability of getting at most 3 heads in 5 tosses = .031 + .156 + 2(.312) = 0.811

We could have found{ 1 - [P(4) + P(5)]} = 1 - (.156 + .031) = 0.813. The small

difference is due to rounding to 3 decimal places.

**Calculator Notes**

To use the TI-83 calculator to compute these probabilities, we use either *binompdf*

or *binomcdf* from the DISTR menu (2nd + VARS). For a question like (a) above,

to find the probability of **exactly** *x* successes in *n* trials, we use *binompdf*.

For (b) part type questions, we use *binomcdf*, since *pdf* stands for *probability density function* and *cdf* stands for *cumulative density function*. The order of data entry is {*n, p, x*} for

both types of questions. So for this questions we'd enter {5, 0.5, 3} to find the values.

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**Example 2**

If the probability is 0.60 that a car stolen in Montreal will be recovered, find the

probability that:

a) at most 3 of 10 stolen cars will be recovered.

b) at least 7 of 10 stolen cars will be recovered.

Since neither of these asks for **exactly** 3 or 7 cars will be recovered, we will use

the *binomcdf* function with *n = 10*, *p = 0.60*. For a) we set *x = 3* and for b) *x = 6*.

Also in b) part, we'll subtract the calculator result from 1 since the question says

*at least 7.*

a) P(*x **—** 3*) is found with *binomcdf {10, 0.60, 3}* = 0.05476 or 0.055.

b) P(*x **¥** 7*) is found with 1 - (*binomcdf {10, 0.60, 6})* = 1 - 0.61722 = 0.38228

**Note:** we could also just add up the values we get from *binompdf* with *x = 7, 8 , 9 *and* 10*.

Either way we get the same answer.

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**The Poisson Distribution**

When *n* is large, and *p* is small for a Binomial population or sample, we often approximate

the probabilities using the Poisson Distribution. The general rule of thumb is to do this

when *n **¥** 100* and *np < 10*.

In general use however, The Poisson Distribution **counts** the number of **successes**

over a **fixed interval** of time or within a specified region.

For example: the number of murders recorded in Detroit in a month, or the

number of cars parked at the airport during a 48 hour period.

__To apply the Poisson distribution 2 conditions must be met__:

1) The number of successes that occur in any interval is independent

of those that occur in other non-overlapping intervals.

2) The probability of a success in an interval is proportional to the size of the interval.

The formula for calculating Poisson probabilities is:

where µ = the mean or expected number of successes,

*e* = the constant *e*, and *x* = the number of successes observed or counted.

The standard deviation of the Poisson Distr. =

**Note:** many statistics books use k (lambda) unstead of l for the expected mean value.

This is most often done when using the Poisson Distribution with no connection to the

Binomial Distribution.

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**Example 3**

If a bank receives an average of 6 bad checks per day, what is the probability

that on any given day, it will receive 4 bad checks?

**Solution**

So, l or k = 6, and* x = 4*. Substituting into the formula we get:

Like the Binomial and other distributions, we don't often use the formula because we

use tables or our calculators. Like the Binomial, the TI-83 also produces Poisson

probability values. The correct order for data entry is *poissonpdf {**l**, x}*.

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**Practice**

1) If incompatibility is given as the reason for separating in 55% of all divorces,

find the probability that this will be the reason given for 4 of the next 6 divorces.

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2) A food distributor claims that 80% of the cans of mixed nuts she sells contain

at least 3 pecans. If 8 of the cans are tested, what is the probability that:

- a) none of them contain more than 3 pecans?

- b) exactly 1 of them contains more than 3 pecans?

- c) more than 6 of them contain more than 3 pecans?

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3) It is known that 20% of all patients taking a certain drug will become drowsy within

2 minutes, what is the probability that of 14 randomly chosen patients on this drug:

- a) at most 2 of them will become drowsy within 2 minutes

b) at least 5 of them will become drowsy within 2 minutes

c) 2, 3 or 4 of them will become drowsy within 2 minutes.

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4) The number of monthly breakdowns of a computer is a random variable

having the Poisson distribution with k = 1.8.

Find the probabilities that this computer will:

- a) function for a month with no breakdowns

b) have only 1 breakdown in a month.

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5) The number of emergency calls received per day by the Montreal 911 service

is a random variable having the Poisson distribution with k = 5.5.

What is the probability that on any given day, there will be only 4 emergency calls?

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**Solutions**

1) *n = 6, p = 0.55 *and* x = 4*

Using tables, we find that *p = 0.55* is not included, so we use the formula or calculator.

*binompdf(6, .55, 4) = 0.278*

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2) *n = 8, p = 0.80*

a) *P(0) = binompdf(8, .80, 0) = 0.00*. We could use tables to find this since 0.8 is listed.

b) *P(1) = binompdf(8, .80, 1) = 0.001*.

c) *P(x > 6) = 1 - binomcdf(8, .80, 6) = 0.50332*.

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3) *n = 14, p = 0.20*

a) *P(x **[** 2) = **binomcdf(14, .20, 2) = 0.448*

b) *P(x **¥** 5) = 1 - **binomcdf(14, .20, 4) = 0.12984*

c) *P(x **= 2, 3 or 4) = **P(x = 2) + P(x = 3) + P(x = 4) = 0.672*

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4) Using Poisson Distribution with k = 1.8.

a) *x = 0 **t** *

b) *x = 1 **t** *

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5) Using Poisson Distribution with k = 5.5.

*x = 4 **t*

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