Binomial Distribution and Applications

Binomial Probability DistributionIs the binomial distribution is a continuous

distribution?Why?

Notation: X ~ B(n,p)There are 4 conditions need to be satisfied for a

binomial experiment:1. There is a fixed number of n trials carried out.2. The outcome of a given trial is either a

“success” or “failure”.

3. The probability of success (p) remains constant from trial to trial.

4. The trials are independent, the outcome of a trial is not affected by the outcome of any other trial.

Comparison between binomial and normal distributions

Binomial Distribution If X ~ B(n, p), then

where

success of

trials.in successes ofnumber r

1 1! and 1 0! also ,1...)2()1(!

yprobabilitP

n

nnnn

.,...,1,0r )1()!(!

! )1()( npp

rnr

nppcrXP rnrrnr

n

r

Exam Question Ten percent of computer parts produced by a

certain supplier are defective. What is the probability that a sample of 10 parts contains more than 3 defective ones?

Solution : Method 1(Using Binomial Formula):

Method 2(Using Binomial Table):

From table of binomial distribution :

Example 2If X is binomially distributed with 6 trials and a probability of success equal to ¼ at each attempt. What is the probability of

a)exactly 4 succes.

b)at least one success.

Example 3Jeremy sells a magazine which is produced in order to raise money for homeless people. The probability of making a sale is, independently, 0.50 for each person he approaches. Given that he approaches 12 people, find the probability that he will make:

(a)2 or fewer sales;(b)exactly 4 sales;(c)more than 5 sales.

Normal Distribution

Normal Distribution In general, when we gather data, we expect to see

a particular pattern tothe data, called a normal distribution. A normal distribution is onewhere the data is evenly distributed around the mean, which when plotted as ahistogram will result in a bell curve also known as a Gaussian distribution.

thus, things tend towards the mean – the closer a value is to the mean, the more you’ll see it; and the number of values oneither side of the mean at any particular distance are equal or in symmetry.

To speak specifically of any normal distribution, two quantities have to be specified: the mean , where the peak of the density occurs, and the standard deviation , which indicates the spread or girth of the bell curve.

Z-score with mean and standard deviation of a set of

scores which are normally distributed, we can standardize each "raw" score, x, by converting it into a z score by using the following formula on each individual score:

Example 1a) Find the z-score corresponding to a raw score of 132 from a normal distribution with

mean 100 and standard deviation 15.

b) A z-score of 1.7 was found from an observation coming from a normal distribution with mean 14 and standard deviation 3.  Find the raw score.

Solutiona)We compute                    132  -  100        z    =   __________         =  2.133                          15b) We have                       x  -  14        1.7    =   ________                                           3To solve this we just multiply both sides by the denominator 3,        (1.7)(3)  =  x - 14        5.1  =  x - 14        x  =  19.1

Example 2Find a) P(z < 2.37)

b) P(z > 1.82)

Solutiona)We use the table. Notice the picture on the table has shaded region corresponding to the area to the left (below) a z-score. This is exactly what we want.Hence P(z < 2.37) = .9911

b) In this case, we want the area to the right of 1.82. This is not what is given in the table. We can use the identity P(z > 1.82) = 1 - P(z < 1.82)reading the table gives P(z < 1.82) = .9656Our answer is P(z > 1.82) = 1 - .9656 = .0344

Example 3Find P(-1.18 < z < 2.1)

SolutionOnce again, the table does not exactly handle this type of area. However, the area between -1.18 and 2.1 is equal to the area to the left of 2.1 minus the area to the left of -1.18. That is P(-1.18 < z < 2.1) = P(z < 2.1) - P(z < -1.18)To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table to get P(z < 2.10) = .9821. The table also tells us that P(z < -1.18) = .1190Now subtract to get P(-1.18 < z < 2.1) = .9821 - .1190 = .8631

Poisson distribution

Definitions a discrete probability distribution for the count of

events that occur randomly in a given time. a discrete frequency distribution which gives the

probability of a number of independent events occurring in a fixed time.

Poisson distribution only apply one formula:

Where: X = the number of events λ = mean of the event per interval

Where e is the constant, Euler's number (e = 2.71828...)

Example:Births rate in a hospital occur randomly at an average rate of 1.8 births per hour.What is the probability of observing 4 births in a given hour at the hospital?AssumingX = No. of births in a given houri) Events occur randomlyii) Mean rate λ = 1.8Using the poisson formula, we cam simply calculate the distribution.

P(X = 4) =( e^-1.8)(1.8^4)/(4!)

Ans: 0.0723

If the probability of an item failing is 0.001, what is the probability of 3 failing out of a population of 2000?

Λ = n * p = 2000 * 0.001 = 2Hence, use the Poisson formula X = 3,

P(X = 3) =

Ans: 0.1804

Example:A small life insurance company has determined that on the average it receives 6 death claims per day. Find the probability that the company receives at least seven death claims on a randomly selected day.

Analysis method 1st: analyse the given data. 2nd: label the value of x, λ At least 7 days, means the probability must be ≥ 7. but

the value will be to the infinity. Hence, must apply the probability rule which is

P(X ≥ 7) = 1 – P(X ≤ 6) P(X ≤ 6) means that the value of x must be from 0, 1, 2, 3,

4, 5, 6. Total them up using Poisson, then 1 subtract the answer. Ans = 0.3938

Example:The number of traffic accidents that occurs on a particular stretch of road during a month follows a Poisson distribution with a mean of 9.4. Find the probability that less than two accidents will occur on this stretch of road during a randomly selected month.

P(x < 2) = P(x = 0) + P(x = 1)

Ans: 0.000860