Outline

1. Outline • Two kinds of random variables • Discrete random variables • Continuous random variables • Symmetric distributions • Normal distributions • The standard normal distribution • Using the standard normal distribution • The normal approximation to the binomial Lecture 6

2. Discrete (DRV) Outcomes have countable values Possible values can be listed E.g., # of people in this room Possible values can be listed: might be …51 or 52 or 53… 1. Two kinds of random variables Lecture 6

3. Discrete RV Continuous RV Not countable Consists of points in an interval E.g., time till coffee break 1. Two kinds of random variables Lecture 6

4. The form of the probability distribution for a CRV is a smooth curve. Such a distribution may also be called a Frequency Distribution Probability Density Function 1. Two kinds of random variables Lecture 6

5. In the graph of a CRV, the X axis is whatever you are measuring E.g., exam scores, mood scores, # of widgets produced per hour. The Y axis measures the frequency of scores. 1. Two kinds of random variables Lecture 6

6. X The Y-axis measures frequency. It is usually not shown. Lecture 6

7. In a symmetric CRV, 50% of the area under the curve is in each half of the distribution. P(x ≤ ) = P(x ≥ ) = .5 Note: Because points on a CRV are infinitely thin, we can only measure the probability of intervals of X values We can’t measure or compute the probability of individual X values. 2. Symmetric Distributions Lecture 6

8. 50% of area on each side of µ μ A symmetric distribution which is not mound-shaped. The two sections (above and below the mean) each contain 50% of the observations. Lecture 6

9. 50% of area 50% of area µ A mound-shaped symmetric distribution (the normal distribution) Lecture 6

10. A very important set of CRVs has mound-shaped and symmetric probability distributions. These are called “normal distributions” Many naturally-occurring variables are normally distributed. 3. Normal Distributions Lecture 6

11. Are perfectly symmetrical around their mean, . Have standard deviation, , which measures the “spread” of a distribution  is an index of variability around the mean. 3. Normal Distributions Lecture 6

12.  µ Lecture 6

13. There are an infinite number of normal distributions They are distinguished on the basis of their mean (µ) and standard deviation (σ) 3. Normal distributions Lecture 6

14. The standard normal distribution is a special one produced by converting raw scores into Z scores Thus, µ = 0 and σ = 1 3. Standard Normal Distribution Lecture 6

15. The area under the curve between and some value X ≥  has been calculated for the standard normal distribution and is given in Table IV of the text. E.g., for Z = 1.62, area = .4474 Because distribution is symmetric, table values can also be used for scores below the mean (Z scores below 0). 3. Standard Normal Distribution Lecture 6

16. .4474 Z = 1.62 X Z = 0  Area gives the probability of finding a score between the mean and X when you make an observation Lecture 6

17. .4474  Z = -1.62 X For Z < 0, same values can be used as for Z > 0 Lecture 6

18. Suppose the average height for Canadian women is µ = 160 cm, with  = 15 cm. What is the probability that the next Canadian woman we meet is more than 175 cm tall? Note two things: 1. this is a question about a single case 2. it specifies an interval. 5. Using the Standard Normal Distribution Lecture 6

19. Using the Standard Normal Distribution Table gives this area We need this area 160 175 cm Lecture 6

20. µ Remember that area above the mean, , is half (.5) of the distribution. Lecture 6

21. Using the Standard Normal Distribution Call this shaded area P. We can get P from Table IV 160 175 Lecture 6

22. Using the Standard Normal Distribution Z = X -  = 175-160  15 = 1.00 Now, look up Z = 1.00 in the table. Corresponding area is .3413. Lecture 6

23. µ Z • Value of Z that marks one end of the interval – you want to find probability that a randomly selected case has a score that falls in this interval • Other end of the interval is at µ (= 0)

24. Using the Standard Normal Distribution This area is .3413 So this area must be .5 – .3413 = .1587 160 175 Lecture 6

25. Using the Standard Normal Distribution This area is .3413 So this area must be .5 – .3413 = .1587 Z = 0 Z = 1.0 Lecture 6

26. Using the Standard Normal Distribution • What is the probability that the next Canadian woman we meet is more than 175 cm tall? • Answer: .1587 Lecture 6

27. Binomial Random Variable – Method #3 When n is large and p is not too close to 0 or 1, we can use the normal approximation to the binomial probability distribution to work out binomial probabilities. How can you tell if the normal approximation is appropriate in a given case? Use the approximation if np ≥ 5 and nq ≥ 5 Lecture 6

28. BRV – Method #3 Normal curve Histogram X The histogram shows the probabilities of different values of the BRV X. Because area gives probability, we can use the area under a section of the normal curve to approximate the area of some part of the histogram Lecture 6

29. BRV – Method #3 In order to use the normal approximation, we have to be able to compute the mean and standard deviation for the BRV (in order to work out Z). μ = np (# of observations times P(Success)) σ = √npq There’s just one other issue to deal with… Lecture 6

30. BRV – Method #3 Notice how the normal curve misses one top corner of the rectangle for 7 and overstates the other top corner – these two errors cancel each other. 1 2 3 4 5 6 7 8 9 10 X Lecture 6

31. BRV – Method #3 1 2 3 4 5 6 7 8 9 10 X Notice how the rectangle for 7 runs from 6.5 to 7.5. The probability of X values up to and including 7 is given by the area to the left of 7.5. Lecture 6

32. BRV – Example 1 from last week Air Canada keeps telling us that arrival and departure times at Pearson International are improving. Right now, the statistics show that 60% of the Air Canada planes coming into Pearson do arrive on time. (This actually is an improvement over 10 years ago when only 42% of the Air Canada planes arrived on time at Pearson.) The problem is that when a plane arrives on time, it often has to circle the airport because there’s still a plane in its gate (a plane which didn’t leave on time). Statistics also show that 50% of the planes that arrive on time have to circle at least once, while only 35% of the planes that arrive late have to circle at least once. Lecture 6

33. BRV – Example 1 from last week c) Of the next 80 Air Canada planes that arrive at Pearson, what’s the probability that between 40 and 45 (inclusive) have to circle at least once? Lecture 6

34. BRV – Example 1 First, we check to see whether we can use the normal approximation. To do this, we need to know the probability that a plane has to circle at least once: P(C) = P(C ∩ Late) + P(C∩ Not Late) = [P(L) * P(C │L)] + [P(L) * P(C│L)] = .35 (.40) + .6 (.5) = .44 Lecture 6

35. BRV – Example 1 Now we can do the check: n = 80. p = .44 and q = (1 – p) = .56 np = 80 (.44) = 35.2 > 5 nq = 80 (.56) = 44.8 > 5 Thus, it’s OK to use the normal approximation. Lecture 6

36. BRV – Example 1 μ = np = 80 (.44) = 35.2 σ = √npq = √80(.44)(.56) = 4.44 Correction for continuity: to get area for 40 and up, we use X = 39.5. To get area for 45 and below, we use X = 45.5 Lecture 6

37. 40 35.2 45 40 and up starts at 39.5 45 and below starts at 45.5 Lecture 6

38. BRV – Example 1 Z39.5 = 39.5 – 35.2 = +.97 4.44 Z45.5 = 45.5 – 35.2 = +2.32 4.44 P(.97 ≤ Z ≤ 2.32) = .4894 - .3340 = .1554. That is the probability that between 40 and 45 (inclusive) of the next 80 planes have to circle at least once. Lecture 6

39. From here down = .5 + .4894 = .9894 40 35.2 45 From here down = .5 + .3340 = .8340

40. 40 35.2 45 Using the normal approximation, we estimate the combined area of all these rectangles to be .9894 – .8340 = .1554 Lecture 6

41. CRV Example 1 Wind speed in Windy City is normally-distributed and the middle 40% of wind speeds is bounded by 23.9 and 29.3. a. Wind speed would be expected to be lower than what value only 5% of the time? Lecture 6

42. .40 .20 23.9 µ 29.3 What value of Z is associated with p = .20? From Table, Z = 0.53 Since distribution is symmetric, µ = midpoint between 23.9 and 29.3. This is 26.6.

43. CRV Example 1a Since Z = X - µ then, σ = X - µ σ = 29.3 – 26.6 = 5.094 0.53 Z for p = .45 is 1.645 (from Table) Thus, required X = 26.6 – 1.645 (5.094) = 18.22 σ Z Lecture 6

44. 18.22 µ .05 .45 Lecture 6

45. CRV Example 1 Wind speed in Windy City is normally-distributed and the middle 40% of wind speeds is bounded by 23.9 and 29.3. b. UV radiation in Windy City is normally distributed with a mean of 10.4 and a variance of 6.25. Wind speed and UV radiation are independent of each other. A "bad day" in Windy City is any day on which either wind speed exceeds 35.0 or UV radiation exceeds 15. What is the probability of a bad day in Windy City? Lecture 6

46. 10.4 15 √6.25 = 2.5 CRV Example 1b Lecture 6

47. CRV Example 1b Z = 15 – 10.4 = 1.84 2.5 P(Z < 1.84) = .4671 (from Table) P(X > 15) = P(Z > 1.84) = .5 – .4671 = .0329 Lecture 6

48. 26.6 35 CRV Example 1b 5.094 Lecture 6

49. CRV Example 1b Z = 35 – 26.6 = 1.65 5.094 P(Z < 1.65) = .4505 (from Table) P(X > 35) = P(Z > 1.65) = .5 – .4505 = .0495 Lecture 6

50. CRV Example 1b P(Bad Day) = P[(UV > 15) or (Wind > 35)] = .0329 + .0495 – (.0329)(.0495) = .0808 (From Additive Rule of Probability) Lecture 6