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COMMONLY USED PROBABILITY DISTRIBUTION

COMMONLY USED PROBABILITY DISTRIBUTION. CHAPTER 2 BCT2053. Introduction. Probability – chance of an event occurring Distribution – a function which assigns to each possible value of the random variable

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COMMONLY USED PROBABILITY DISTRIBUTION

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  1. COMMONLY USED PROBABILITY DISTRIBUTION CHAPTER 2 BCT2053

  2. Introduction • Probability – chance of an event occurring • Distribution – a function which assigns to each possible value of the random variable • Probability Distribution – the values/function that a random variable can assume and the corresponding probabilities of the values • Types of probability distribution: • Discrete – describe discrete random variable • Continuous – describe continuous random variable

  3. CONTENT • 2.1 Review on Binomial and Poisson Distributions • 2.2 Poisson Approximation for Binomial Distribution • 2.3 Review on Normal Distribution • 2.4 Central Limit Theorem • 2.5 Normal Approximation to the Binomial Distribution • 2.6 Normal Approximation to the Poisson Distribution • 2.7 Normal Probability Plots

  4. OBJECTIVEAt the end of this chapter, you should be able to:1. Explain what a Binomial Distribution, identify Binomial experiments and compute Binomial probabilities2. Find the expected value (mean), variance, and standard deviation of a Binomial experiment. 2.1 Binomial Distribution

  5. Binomial Distribution • A Binomial distribution results from a procedure that meets all the following requirements • The procedure has a fixed number of trials ( the same trial is repeated) • The trials must be independent • Each trial must have outcomes classified into 2 relevant categories only (success & failure) • The probability of success remains the same in all trials • Example: toss a coin, Baby is born, True/false question, product, etc ...

  6. Notation for the Binomial Distribution Then, X has the Binomial distribution with parameters n and p denoted by X ~ Bin (n, p) which read as ‘‘X is Binomial distributed with number of trials n and probability of success p’’

  7. Binomial Experiment or not ? • An advertisement for Vantin claims a 77% end of treatment clinical success rate for flu sufferers. Vantin is given to 15 flu patients who are later checked to see if the treatment was a success. • A study showed that 83% of the patients receiving liver transplants survived at least 3 years. The files of 6 liver recipients were selected at random to see if each patients was still alive. • In a study of frequent fliers (those who made at least 3 domestic trips or one foreign trip per year), it was found that 67% had an annual income over RM35000. 12 frequent fliers are selected at random and their income level is determined. For each problem, state what are X, n, p, and q.

  8. Binomial Probability Formula

  9. EXERCISE 2.1 • A fair coin is tossed 10 times. Let X be the number of heads that appear. What is the distribution of X? • A lot contains several thousand components. 10 % of the components are defective. 7 components are sampled from the lot. Let X represents the number of defective components in the sample. What is the distribution of X ?

  10. Solves problems involving linear inequalities • At least, minimum of, no less than • At most, maximum of, no more than • Is greater than, more than • Is less than, smaller than, fewer than

  11. EXERCISE 2.1 • Find the probability distribution of the random variable X if X ~ Bin (10, 0.4). Find also P(X = 5) and P(X < 2). Then find the mean and variance for X. • A fair die is rolled 8 times. Find the probability that no more than 2 sixes comes up. Then find the mean and variance for X.

  12. EXERCISE 2.1 • A survey found that, one out of five Malaysians say he or she has visited a doctor in any given month. If 10 people are selected at random, find the probability that exactly 3 will have visited a doctor last month. • A survey found that 30% of teenage consumers receive their spending money from part time jobs. If 5 teenagers are selected at random, find the probability that at least 3 of them will have part time jobs.

  13. Solve Binomial problems by statistics table • Use Cumulative Binomials Probabilities Table • n number of trials • p probability of success • k number of successes in n trials – X • It give P (X ≤ k) for various values of n and p • Example: n = 2 , p = 0.3 • Then P (X ≤ 1) = 0.9100 • Then P (X = 1) = P (X ≤ 1) - P (X ≤ 0) = 0.9100 – 0.4900 = 0.4200 • Then P (X ≥ 1) = 1 - P (X <1) = 1 - P (X ≤ 0) = 1 – 0.4900 = 0.5100 • Then P (X < 1) = P (X ≤ 0) = 0.4900 • Then P (X > 1) = 1 - P (X ≤ 1) = 1- 0.9100 = 0.0900

  14. Using symmetry properties to read Binomial tables • In general, • P (X = k | X ~ Bin (n, p)) = P (X = n - k | X ~ Bin (n,1 - p)) • P (X ≤ k | X ~ Bin (n, p)) = P (X ≥ n - k | X ~ Bin (n,1 - p)) • P (X ≥ k | X ~ Bin (n, p)) = P (X ≤ n - k | X ~ Bin (n,1 - p)) • Example: n = 8 , p = 0.6 • Then P (X ≤ 1) = P (X ≥ 7 | p = 0.4) = P ( 1 - X ≤ 6 | p = 0.4) = 1 – 0.9915 = 0.0085 • Then P (X = 1) = P (X = 7 | p = 0.4) = P (X ≤ 7 | p = 0.4) - P (X ≤ 6 | p = 0.4) = 0.9935 – 0.9915 = 0.0020 • Then P (X ≥ 1) = P (X ≤ 7 | p = 0.4) = 0.9935 • Then P (X < 1) = P (X > 7 | p = 0.4) = P ( 1 - X ≤ 7 | p = 0.4) = 1 – 0.9935 = 0.0065 • Then P (X > 1) = P (X < 7 | p = 0.4) = P (X ≤ 6 | p = 0.4) = 0.9915

  15. EXERCISE 2.1 • Given that n = 12 , p = 0.25. Then find • P (X ≤ 3) • P (X = 7) • P (X ≥ 5) • P (X < 2) • P (X > 10) • Given that n = 9 , p = 0.7. Then find • P (X ≤ 4) • P (X = 8) • P (X ≥ 3) • P (X < 5) • P (X > 6)

  16. EXERCISE 2.1 • A large industrial firm allows a discount on any invoice that is paid within 30 days. Of all invoices, 10% receive the discount. In a company audit, 12 invoices are sampled at random. • What is probability that fewer than 4 of 12 sampled invoices receive the discount? • Then, what is probability that more than 1 of the 12 sampled invoices received a discount.

  17. EXERCISE 2.1 • A report shows that 5% of Americans are afraid being alone in a house at night. If a random sample of 20 Americans is selected, find the probability that • There are exactly 5 people in the sample who are afraid of being alone at night • There are at most 3 people in the sample who are afraid of being alone at night • There are at least 4 people in the sample who are afraid of being alone at night

  18. OBJECTIVEAt the end of this chapter, you should be able to:1. Explain what a Poisson Distribution, identify Poisson experiments and compute Poisson probabilities.2. Find the expected value (mean), variance, and standard deviation of a Poisson experiment. 2.1 Poisson Distribution

  19. Poisson Distribution • The Poisson distribution is a discrete probability distribution that applies to occurrences of some event over a specified interval ( time, volume, area etc..) • The random variable X is the number of occurrences of an event over some interval • The occurrences must be random • The occurrences must be independent of each other • The occurrences must be uniformly distributed over the interval being used • Example of Poisson distribution • The number of emergency call received by an ambulance control in an hour. • The number of vehicle approaching a bus stop in a 5 minutes interval. • 3. The number of flaws in a meter length of material

  20. Poisson Probability Formula • λ, mean number of occurrences in the given interval is known and finite • Then the variable X is said to be ‘Poisson distributed with mean λ’ • X ~ Po (λ)

  21. EXERCISE 2.2 • A student finds that the average number of amoebas in 10 ml of ponds water from a particular pond is 4. Assuming that the number of amoebas follows a Poisson distribution, find the probability that in a 10 ml sample, • there are exactly 5 amoebas • there are no amoebas • there are fewer than three amoebas

  22. EXERCISE 2.2 • On average, the school photocopier breaks down 8 times during the school week (Monday - Friday). Assume that the number of breakdowns can be modeled by a Poisson distribution. Find the probability that it breakdowns, • 5 times in a given week • Once on Monday • 8 times in a fortnight (2 week)

  23. EXERCISE 2.2Solve Poisson problems by statistics table • Given that X ~ Po (1.6). Use cumulative Poisson probabilities table to find • P (X ≤ 6) • P (X = 5) • P (X ≥ 3) • P (X < 1) • P (X > 10) Find also the smallest integer n such that P ( X > n) < 0.01

  24. EXERCISE 2.2 • A sales firm receives, on the average, three calls per hour on its toll-free number. For any given hour, find the probability that it will receive the following: • At most three calls • At least three calls • 5 or more calls

  25. EXERCISE 2.2 • The number of accidents occurring in a weak in a certain factory follows a Poisson distribution with variance 3.2. Find the probability that in a given fortnight, • exactly seven accidents happen. • More than 5 accidents happen.

  26. 2.2 Using the Poisson distribution as an approximation to the Binomial distribution • When n is large (n > 50) and p is small (p < 0.1), the Binomial distribution X ~ Bin (n, p) can be approximated using a Poisson distribution with X ~ Po (λ) where mean, λ = np < 5. • The larger the value of n and the smaller the value of p, the better the approximation.

  27. EXERCISE 2.2 • Eggs are packed into boxes of 500. On average 0.7 % of the eggs are found to be broken when the eggs are unpacked. Find the probability that in a box of 500 eggs, • Exactly three are broken • At least two are broken

  28. EXERCISE 2.2 • If 2% of the people in a room of 200 people are left-handed, find the probability that • exactly five people are left-handed. • At least two people are left-handed. • At most seven people are left-handed.

  29. OBJECTIVEAt the end of this chapter, you should be able to:1. Identify the properties of the normal distribution and find the area under the standard normal distribution, given various Z values.3. Find probabilities for a normally distributed variable by transforming it into a standard normal variable.4. Find specific data values for given percentages, using the standard normal distribution. 2.3 Normal Distribution

  30. Continuous Distribution • A discrete variable cannot assume all values between any two given values of the variables. • A continuous variable can assume all values between any two given values of the variables. • Examples of continuous variables are the heights of adult men, body temperatures of rats, and cholesterol levels of adults. • Many continuous variables, such as the examples just mentioned, have distributions that are bell-shaped, and these are called approximately normally distributed variables.

  31. Properties of Normal Distribution • Also known as the bell curveor the Gaussian distribution, named for the German mathematician Carl Friedrich Gauss (1777–1855), who derived its equation. • X is continuous where and

  32. Example: Histograms for the Distribution of Heights of Adult Women

  33. Observation • The larger the data size, then the distribution of the data will approximately bell shape (normal). • No variable fits normal distribution perfectly, since a normal distribution is a theoretical distribution. • However, a normal distribution can be used to describe many variables, because the deviations from normal distribution are very small.

  34. The Normal Probability Curve • The Curve is bell-shaped • The mean, median, and mode are equal and located at the center of the distribution. • The curve is unimodal (i.e., it has only one mode). • The curve is symmetric about the mean, (its shape is the same on both sides of a vertical line passing through the center. • The curve is continuous, (there are no gaps or holes) For each value of X, there is a corresponding value of Y.

  35. The Normal Probability Curve • The curve never touches the x axis. Theoretically, no matter how far in either direction the curve extends, it never meets the x axis—but it gets increasingly closer. • The total area under the normal distribution curve is equal to 1.00, or 100%. • A Normal Distribution is a continuous, symmetric, bell shaped distribution of a variable.

  36. Area Under a Normal Distribution Curve • The area under the part of the normal curve that lies • within 1 standard deviation of the mean is approximately 0.68, or 68%; • within 2 standard deviations, about 0.95, or 95% • within 3 standard deviations, about 0.997, or 99.7%.

  37. Other Characteristics • Finding the probability • Area under curve Example Given , Find the value of a and b if

  38. Shapes of Normal Distributions

  39. The Standard Normal Distribution • The standard normal distributionis a normal distribution with a mean of 0 and a standard deviation of 1. TIPS

  40. Different between 2 curves Area Under the Normal Distribution Curve Area Under the Standard Normal Distribution Curve

  41. Finding Area under the Standard Normal Distribution • STEP 1Draw a picture. • STEP 2Shade the area desired. • STEP 3Find the correct figure in the following Procedure Table (the figure that is similar to the one you’ve drawn). • STEP 4Follow the directions given in the appropriate block of the Procedure Table to get the desired area. GENERAL PROCEDURE • EXAMPLE 1 • P (0 < Z < 2.34) = 0.4904 • P (-2.34 < Z < 0) = 0.4904 • P (0 < Z < 0.156) = 0.062 • P (-1.738 < Z < 0) = 0.4589

  42. Finding Area under the Standard Normal Distribution • EXAMPLE 2 • P (Z >1.25) = 0.1056 • P (Z <-2.13) = 0.0166 • P (Z >2.099) = 0.0179 • P (Z <-0.087) = 0.4653 • EXAMPLE 3 • P (0.21 < Z < 2.34) = 0.4072 • P (-2.134 < Z < -0.21) = 0.4004 • P (0.67 < Z < 1.156) = 0.1276 • P (-1.738 < Z < -0.79) = 0.1737

  43. Finding Area under the Standard Normal Distribution • EXAMPLE 4 • P (-0.21 < Z < 2.34) = 0.5736 • P (-2.134 < Z < 0.21) = 0.5688 • P (-0.67 < Z < 1.156) = 0.6248 • P (Z < |0.79|) = 0.5704 • EXAMPLE 5 • P (Z < 1.21) = 0.8869 • P (Z < 2.099) = 0.9821 • P (Z < 0.512) = 0.6957

  44. Finding Area under the Standard Normal Distribution • EXAMPLE 6 • P (Z >-1.25) = 0.8944 • P (Z >-2.13) = 0.9834 • P (Z >-0.087) = 0.5347 • EXAMPLE 7 • P (Z >|2.34|) = 0.0192 • P (Z >|0.147|) = 0.8832

  45. EXERCISE 2.3 • Given X ~ N(110,144), find • P (110 < X < 128) (d) P (X > 170) • P (X < 150) (e) P (98 < X < 128) • P (X > 130) (f) P (X < 60) Transform the original variable X where to a standard normal distribution variable Z where TIPS

  46. EXERCISE 2.3 • If Z ~ N(0,1), find the value of a if • P(Z < a) = 0.9693 • P(Z < a) = 0.3802 • P(Z < a) = 0.7367 • P(Z < a) = 0.0793 • If X ~ N(μ,36) and P ( X > 82) = 0.0478, find μ. • If X ~ N(100, σ²) and P ( X < 82) = 0.0478, find σ. TIPS

  47. EXERCISE 2.3 Applications of the Normal Distribution 5. The mean number of hours an American worker spends on the computer is 3.1 hours per workday. Assume the standard deviation is 0.5 hour. Find the percentage of workers who spend less than 3.5 hours on the computer. Assume the variable is normally distributed. 6. Length of metal strips produced by a machine are normally distributed with mean length of 150 cm and a standard deviation of 10cm. Find the probability that the length of a randomly selected is a) Shorter than 165 cm b) within 5cm of the mean

  48. EXERCISE 2.3 Applications of the Normal Distribution 7. Time taken by the Milkman to deliver to the Jalan Indah is normally distributed with mean of 12 minutes and standard deviation of 2 minutes. He delivers milk everyday. Estimate the numbers of days during the year when he takes a) longer than 17 minutes b) less than ten minutes c) between 9 and 13 minutes 8. To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.

  49. EXERCISE 2.3Applications of the Normal Distribution 9. The heights of female student at a particular college are normally distributed with a mean of 169cm and a standard deviation of 9 cm. a) Given that 80% of these female students have a height less than h cm. Find the value of h. b) Given that 60% of these female students have a height greater than y cm. Find the value of y. 10. For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study.

  50. OBJECTIVEAt the end of this chapter, you should be able to:1. Use the central limit theorem to solve problems involving sample means for large samples (probability of mean) 2.4 Central Limit Theorem

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