1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. Source: http://en.wikipedia.org/wiki/Normal_distribution

8. λ The Poisson Distribution & The Normal Distribution

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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)

15. 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

16. 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

17. 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

18. 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

19. 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

21. μ = 0 f(x) +1.75 -1.75

22. 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

23. Look Up Standard Normal Tables μ = 0 f(x) +1.75 -1.75 4.01% (.0401)

24. 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

25. 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

26. 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

27. 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)

28. 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!)

29. 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

30. 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)

32. 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

33. 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

34. 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)

36. 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

37. 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

38. 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

39. Commonly Used Probabilities -3σ -2σ -1σ μ +1σ +2σ +3σ f(x) 68% 95% 99.7%

40. -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

41. -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

42. -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

43. Commonly Used Probabilities -3σ -2σ -1σ μ +1σ +2σ +3σ f(x) 68% 95% 99.7%