Probability distribution

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By Abhishek Darge 70 Punit raut 98 Shriya singh 109 Priti Shrivastav 111 Probability Distribution

Transcript of Probability distribution

Page 1: Probability distribution

ByAbhishek Darge 70Punit raut 98Shriya singh 109Priti Shrivastav 111Sameer surve 112Chetan Vinjuda 116

Probability Distribution

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Probability Distribution• A probability distribution describes how the outcome of an

experiment are expected to vary• Since such distribution deals with expectation, they provide

useful models in making inferences and decision in the face of uncertainty

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Q1• The owner of a bakery may be considering how much one-

kg cakes he can sell in a day. He has kept record of the sale of this type of cake made over last 100 days as given below:

• Based on these historical data, develop probability distribution of demand for the cake in question.• Answer:

No of cakes sold (x): 0 1 2 3 4 5 TotalNo of days (f): 10 20 20 35 10 5 100

No of cakes(x) 0 1 2 3 4 5 TotalProbability(p) 0.10 0.20 0.20 0.35 0.10 0.05 1

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Types of Probability Distribution

Discrete Probability Distribution

1. Binomial Distribution2. Poisson Distribution

Continuous Probability Distribution

1. Uniform Probability Distribution2. Exponential Probability

Distribution3. Normal Probability Distribution4. Student’s t Distribution5. Chi-Square Distribution6. F Distribution

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Discrete Probability Distribution• In a probability distribution of random variable X, in which

X can only take the values of discrete integers, then it is called discrete probability distribution.• Finite number of outcome values

– Suppose one toss a coin 3 times then sample space consist of 8 equally likely events: HHH, HHT, HTH, HTT,THH, THT, TTH, TTT

– Now random variable X counts no of heads appearing in 3 tosses then we will get following four situations

– P(X) is probability of seeing exactly X heads

X 0 1 2 3P(X) 1/8 3/

83/8 1/8

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Continuous Probability Distribution• The probability distribution of a random variable is called

Continuous Probability Distribution if the given random variable is continuous.• Large number of outcomes• Suppose that the Indian air force sets the qualification that

all pilots must weight between 55kg and 65kg. Then the weight of pilot would be an example of continuous variable. Since pilot’s weight could take any value between 55kg and 65kg

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Binomial Distribution• Binomial distribution takes place when there are only two

mutually exclusive possible outcomes.• E.g.: Flipping a coin

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Formula for Binomial Distribution

– r: the no. of successes that result from the binomial experiment– n: no. of trials in binomials– p: probability of success of an individual trial– q: probability of failure of an individual trial– b(x; n, p): binomial probability– : no of combinations of n things, taken r at a time

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Q2• Fit the binomial distribution to following distribution of 156

samples

• Answer:

No. of defective items 7 6 5 4 3 2 1 0No of samples 1 6 32 36 48 24 7 2

No. of defective items 7 6 5 4 3 2 1 0Expected frequencies 1.22 8.53 25.6 42.65 48.6

525.6

8.53

1.22

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Poisson Distribution• Poisson distribution is discrete random variable distribution

that expresses probability of given number of event in a fix interval of time, if these event occur with a known average rate and independent of the time since the last event.• It can be used as an alternative to binominal distribution in

case of very large sample• E.g.

– No of accidents per year in a district of Maharashtra– No of typing error per page– No of vehicles passing a certain point per minute

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Formula for Poisson Distribution

• e: A constant = approx. 2.71828.• m: mean number of successes that occur in a specific

region• x: actual number of successes that occur in a specific

region• P(x; m): The Poisson probability that exactly x successes

occur in a Poisson experiment, when the mean no of successes is µ.

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Q3• If 4% of the electric geysers manufactured by a company

are defective, use Poisson distribution to find the probability in a sample of 100 geysers when:1) None is defective 2) 5 geysers are defective(Given: = 0.018)

• Answer:– P(0)= 0.018– P(5)= 0.154

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Normal Probability Distribution• A family of continuous probability distributions described

by the normal equation is called the normal distribution• Normal distribution is defined by following equation

• Where, – e: a mathematical constant equal to 2.7183– μ: expected value of mean– σ: Standard deviation– x: a particular value of the random variable, and -∞<x<+∞

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Standard Normal Distribution• The normal distribution with σ = 1 and μ = 0 is called

standard normal distribution• In fact, it is possible to convert any normal random variable

x into a standardized normal variable z. This is called z-transformation. And this is done by following formula

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Q4• 2000 applicant appeared in an interview. Distribution of

marks is assumed to be normal with mean(μ) = 30 and σ = 6.25. How many applicants are expected to get marks:– Between 20 and 40– Less than 35 – Above 50?

• Answer:1) 1422 2) 424 3) 1

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Homework QuestionsThe bolts produced by a certain machine were checked by examining samples if 12. The following table shows the distribution of 130 samples to according to the no of defective they contain

Fit a binomial distribution and find the expected frequencies if the chances of a bolt being defective is ½. Find the mean and variance of the fitted distribution.Assuming that on an average 3% of output in a factory manufacturing water filters are defective in a package what is probability that None is defective At the most 4 defective (Given = 0.0001234)

Q5

Q6

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Thank You.