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Topics Covered Discrete probability distributions –The Uniform Distribution –The Binomial Distribution –The Poisson Distribution Each is appropriately applied in certain situations and to particular phenomena
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Continuous Random Variables Continuous random variable can assume all real number values within an interval (e.g., rainfall, pH) Some random variables that are technically discrete exhibit such a tremendous range of values, that is it desirable to treat them as if they were continuous variables, e.g. population Continuous random variables are described by probability density functions
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Probability density functions are defined using the same rules required of probability mass functions, with some additional requirements: The function must have a non-negative value throughout the interval a to b, i.e. f(x) >= 0 for a <= x <= b The area under the curve defined by f(x), within the interval a to b, must equal 1 Probability Density Functions x area=1 f(x) a b
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The Normal Distribution The most common probability distribution is the normal distribution The normal distribution is a continuous distribution that is symmetric and bell-shaped Source: http://mathworld.wolfram.com/NormalDistribution.html
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The Normal Distribution Most naturally occurring variables are distributed normally (e.g. heights, weights, annual temperature variations, test scores, IQ scores, etc.) A normal distribution can also be produced by tracking the errors made in repeated measurements of the same thing; Karl Friedrich Gauss was a 19th century astronomer who found that the distribution of the repeated errors of determining the position of the same star formed a normal (or Gaussian) distribution
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The probability density function of the normal distribution: You can see how the value of the distribution at x is a f(x) of the mean and standard deviation The Normal Distribution
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Source: http://en.wikipedia.org/wiki/Normal_distribution
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λ The Poisson Distribution & The Normal Distribution
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The probability density function of a normal distribution approximating the probability mass function of a binomial distribution Source: http://en.wikipedia.org/wiki/Normal_Distribution
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The Normal Distribution As with all frequency distributions, the area under the curve between any two x values corresponds to the probability of obtaining an x value in that range The total area under the curve is equal to one Source: http://en.wikipedia.org/wiki/Normal_Distribution
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The Normal Distribution The way we use the normal distribution is to find the probability of a continuous random variable falling within a range of values ([c, d]) The probability of the variable between c and d is the area under the curve The areas under normal curves are given in tables such as that found in Table A.2 in Appendix A. Textbook (Rogerson) x f(x) cd
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Normal Tables Variables with normal distributions may have an infinite number of possible means and standard deviations Source: http://en.wikipedia.org/wiki/Normal_distribution
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Normal Tables Variables with normal distributions may have an infinite number of possible means and standard deviations Normal tables are standardized (a standard normal distribution with a mean of zero and a standard deviation of one) Before using a normal table, we must transform our data to a standardized normal distribution
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Standardization of Normal Distributions The standardization is achieved by converting the data into z-scores z-score The z-score is the means that is used to transform our normal distribution into a standard normal distribution ( = 0 & = 1)
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Standardization MonthT (°F)Z-score J39.53-1.56 F46.36-1.03 M46.42-1.02 A60.320.05 M66.340.51 J75.491.22 J75.391.21 A77.291.36 S68.640.69 O57.57-0.16 N54.88-0.37 D48.2-0.89 Mean = 59.70 Standard deviation = 12.97 Original data:Z-score:Mean = 0 Standard deviation = 1
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Example I – A data set with = 55 and = 20: Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>1.75) or P(X<1.75)) x - Z-score = = x - 55 20 If one of our data values x = 90 then: x - Z-score = = 90 - 55 20 = 35 20 = 1.75 Standardization of Normal Distributions
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Look Up Standard Normal Tables Using our example z-score of 1.75, we find the position of 1.75 in the table and use the value found there
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Look Up Standard Normal Tables μ = 0 f(x) +1.75.0401 P(Z > 1.75) = 0.0401 P(Z <= 1.75) = 0.9599 μ = 55 f(x).0401 +90 P(X > 90) = 0.0401 P(X <= 90) = 0.9599
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Example II – If we have a data set with = 55 and = 20, we calculate z-scores using: Using z-scores in conjunction with standard normal tables (like Table A.2 on page 214 of Rogerson) we can look up areas under the curve associated with intervals, and thus probabilities (P(X>20) or P(X<=20)) x - Z-score = = x - 55 20 If one of our data values x = 20 then: x - Z-score = = 20 - 55 20 = -35 20 = -1.75 Standardization of Normal Distributions
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μ = 0 f(x) +1.75 -1.75
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Look Up Standard Normal Tables Using our example z-score of -1.75, we find the position of 1.75 in the table and use the value found there; because the normal distribution is symmetric the table does not need to repeat positive and negative values
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Look Up Standard Normal Tables μ = 0 f(x) +1.75 -1.75 4.01% (.0401)
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Look Up Standard Normal Tables μ = 0 f(x) μ = 55 f(x) cd -1.75 4.01% (.0401) P(Z <= -1.75) = 0.0401 P(Z > -1.75) = 0.9599 P(X <= 20) = 0.0401 P(X > 20) = 0.9599
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Finding the P(x) for Various Intervals a 3. a 2. a 1. P(0 Z a) = [0.5 – (table value)] Total Area under the curve = 1, thus the area above x is equal to 0.5, and we subtract the area of the tail P(Z a) = [1 – (table value)] Total Area under the curve = 1, and we subtract the area of the tail P(Z a) = (table value) Table gives the value of P(x) in the tail above a
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Finding the P(x) for Various Intervals a 6. a 5. a 4. P(Z a) = [1 – (table value)] This is equivalent to P(Z a) when a is positive P(Z a) = (table value) Table gives the value of P(x) in the tail below a, equivalent to P(Z a) when a is positive P(a Z 0) = [0.5 – (table value)] This is equivalent to P(0 Z a) when a is positive
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Finding the P(x) for Various Intervals P(a Z b) if a 0 a 7. b With this set of building blocks, you should be able to calculate the probability for any interval using a standard normal table or = [0.5 – (table value for a)] + [0.5 – (table value for b)] = [1 – {(table value for a) + (table value for b)}] = (0.5 – P(Z b)) = 1 – P(Z b)
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Finding the P(x) – Example Suppose we are in charge of buying stock for a shoe store. We will assume that the distribution of shoe sizes is normally distributed for a gender. Let’s say the mean of women’s shoe sizes is 20 cm and the standard deviation is 5 cm If our store sells 300 pairs of a popular style each week, we can make some projections about how many pairs we need to stock of a given size, assuming shoes fit feet that are +/- 0.5 cm of the length of the shoe (because of course we must use intervals here!)
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Finding the P(x) – Example 1. How many pairs should we stock of the 25 cm size? We first need to convert this to a range according to the assumption specified above, since we can only evaluate P(x) over an interval P(24.5 cm ≤ x ≤ 25.5 cm) is what we need to find Now we need to convert the bounds of our interval into z-scores: x - Z lower = = 24.5 - 20 5 = 4.5 5 = 0.9 x - Z upper = = 25.5 - 20 5 = 5.5 5 = 1.1
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Finding the P(x) – Example 1. How many pairs should we stock of the 25 cm size? We now have our interval expressed in terms of z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables 1.1 0.9 We can calculate this area by finding P(0.9 Z 1.1) = P(Z 0.9) - P(Z 1.1)
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Finding the P(x) – Example 1. How many pairs should we stock of the 25 cm size? We now have our interval expressed in terms of z-scores as P(0.9 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables 1.1 0.9 We can calculate this area by finding P(0.9 Z 1.1) = P(Z 0.9) - P(Z 1.1) = 0.1841 – 0.1357 = 0.0484 Now multiply our P(0.9 ≤ Z ≤ 1.1) by total sales per week to get the number of shoes we should stock in that size: 300 х 0.0484 = 14.52 ≈ 15
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Finding the P(x) – Example 2. How many pairs should we stock of the 18 cm size? (assuming shoes fit feet that are +/- 0.5 cm of the length of the shoe) Again, we convert this to a range according to the assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 18.5 cm) is what we need to find Now we need to convert the bounds of our interval into z-scores: x - Z lower = = 17.5 - 20 5 = -2.5 5 = -0.5 x - Z upper = = 18.5 - 20 5 = -1.5 5 = -0.3
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Finding the P(x) – Example 2. How many pairs should we stock of the 18 cm size? We now have our interval expressed in terms of z- scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables -0.5 -0.3 We can calculate this area by finding P(-0.5 Z -0.3) = P(Z -0.3) - P(Z -0.5)
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Finding the P(x) – Example 2. How many pairs should we stock of the 18 cm size? We now have our interval expressed in terms of z- scores as P(-0.5 ≤ Z ≤ -0.3) for use in a standard normal distribution, which we can evaluate using our standard normal tables -0.5 -0.3 We can calculate this area by finding P(-0.5 Z -0.3) = P(Z -0.3) - P(Z -0.5) = 0.3821 – 0.3085 = 0.0736 Now multiply our P(-0.5 ≤ Z ≤ -0.3) by total sales per week to get the number of shoes we should stock in that size: 300 х 0.0736 = 22.08 ≈ 22
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Finding the P(x) – Example 3. How many pairs should we stock in the 18 to 25 cm size range? As always, we convert this to a range according to the assumption specified above, since we can only evaluate P(x) over an interval P(17.5 cm ≤ x ≤ 25.5 cm) is what we need to find We have already found the appropriate Z-scores x - Z lower = = 17.5 - 20 5 = -2.5 5 = -0.5 x - Z upper = = 25.5 - 20 5 = 5.5 5 = 1.1
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Finding the P(x) – Example 3. How many pairs should we stock in the 18 to 25 cm size range? We now have our interval expressed in terms of z- scores as P(-0.5 ≤ Z ≤ 1.1) for use in a standard normal distribution, which we can evaluate using our standard normal tables -0.5 1.1 We can calculate this area by finding P(-0.5 Z 1.1) = 1 – [P(Z -0.5) + P(Z 1.1)] = 1 – [0.3085 + 0.1357] = 1 – 0.4442 = 0.5558 Now multiply our P(-0.5 ≤ Z ≤ 1.1) by total sales per week to get the # of shoes we should stock in that range: 300 х 0.5558 = 166.74 ≈ 167
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Commonly Used Probabilities -3σ -2σ -1σ μ +1σ +2σ +3σ f(x) 68% 95% 99.7%
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-σ μ +σ f(x) Commonly Used Probabilities Standard normal distribution: P(z > 1) = 0.1587 p(-1 ≤ z ≤ 1) = 1 – 2*0.1587 = 0.6826 Normal distribution as illustrated in the Figure: P(μ-σ ≤ x ≤ μ-σ) = 0.6826 ≈ 0.68 For a normal distribution, 68% of the observations lie within about one standard deviations of the mean
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-2σ μ +2σ f(x) Commonly Used Probabilities Standard normal distribution: P(z > 2) = 0.0228 p(-2 ≤ z ≤ 2) = 1 – 2*0.0228 = 0.9543 Normal distribution as illustrated in the Figure: P(μ-2σ ≤ x ≤ μ-2σ) = 0.9543 ≈ 0.95 For a normal distribution, 95% of the observations lie within about two standard deviations of the mean
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-3σ μ +2σ f(x) Commonly Used Probabilities Standard normal distribution: P(z > 3) = 0.0013 p(-3 ≤ z ≤ 3) = 1 – 2*0.0013 = 0.9974 Normal distribution as illustrated in the Figure: P(μ-3σ ≤ x ≤ μ-3σ) = 0.9974 ≈ 0.997 For a normal distribution, 99.7% of the observations lie within about two standard deviations of the mean
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Commonly Used Probabilities -3σ -2σ -1σ μ +1σ +2σ +3σ f(x) 68% 95% 99.7%
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