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Binomial Distribution and Applications

Normal Distribution, Binomial Distribution, Poisson Distribution

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It's all about Normal Distribution, Binomial Distribution, and Poisson Distribution. In addition, theres example with answer!

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Page 1: Normal Distribution, Binomial Distribution, Poisson Distribution

Binomial Distribution and Applications

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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.

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Comparison between binomial and normal distributions

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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

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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?

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Solution : Method 1(Using Binomial Formula):

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Method 2(Using Binomial Table):

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From table of binomial distribution :

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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.

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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.

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Normal Distribution

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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.

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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.

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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.

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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:

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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

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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

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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

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Poisson distribution

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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.

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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...)

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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

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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

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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.

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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

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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