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Elementary Statistics and Inference 22S:025 or 7P:025 Lecture 23 1 Elementary Statistics and Inference 22S:025 or 7P:025 Chapter 18 2 Chapter 18 A. Introduction When we toss a coin several times, the tosses are independent – the probability (chance) of tossing a head is the same for each toss: P( H ) = 1 2 Possible patterns of outcomes for 5 tosses of a coin. 3 1 Chapter 18 (cont.) HHHHH – 5 HHTTT HHHHT HTTHT HHHTH HTTTH HTHTT HHTHH – 4 THTTH – 2 HTHHH TTHTH THHHH TTTHH THHTT TTHHH TTHHT THTHH HTTHT THHTH THHHT TTTTH HTHHT – 3 TTTHT HHTHT TTHTT – 1 HHHTT THTTT HHTTH HTTTT HTHTH HTTHH TTTTT - 0 4 Chapter 18 (cont.) Toss a coin 5 times – outcomes Number of Heads Patterns 0 1 1 5 2 10 3 10 4 5 5 1 2 5 = 32 possible pattern 5 Chapter 18 (cont.) The number of heads can be determined mathematically – in “n” tosses of the coin the number of heads and tails is: n! n ⋅ (n − 1) ⋅ (n − 2) − − − 1 = X !(n − X !) X ( X − 1) − − − 1 ⋅ (n − X )(n − X − 1) − − − 1 If n=5, n=5 X=3: 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = = 10 patterns for 3 heads 3!(2)! (3 ⋅ 2 ⋅1)(2 ⋅1) If n=5, X=2: 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = = 5 patterns for 2 heads 4!1! (4 ⋅ 3 ⋅ 2 ⋅1)(1) 6 2 Chapter 18 (cont.) We can approximate the probability of obtaining the number of heads in “n” repetitions of tossing the coin (a binomial event), using the Normal Curve approximation. 7 Chapter 18 (cont.) B. Probability Histograms 8 Chapter 18 (cont.) 9 3 Chapter 18 (cont.) 10 Chapter 18 (cont.) As the number of tosses increases the probability histogram approaches a normal distribution. The sum on the pair of dice approaches a normal distribution histogram as the number of repetitions increases. The product of the numbers on a pair of dice does not approach a normal distribution as the number of tosses increases. 11 Chapter 18 (cont.) 12 4 Chapter 18 (cont.) 13 Chapter 18 (cont.) Exercise Set A – (pp. 312-315) #1, 2, 3, 4, 5, 6 14 Chapter 18 (cont.) 15 5 Chapter 18 (cont.) C. Probability Histograms and The Normal Curve In tossing a coin, the count of heads (tails) probability histogram approaches a Normal probability histogram. 16 Chapter 18 (cont.) 17 Chapter 18 (cont.) 18 6 Chapter 18 (cont.) 19 Chapter 18 (cont.) D. Normal Approximation The Normal Curve represents the probability distribution of a continuous variable and can be used to approximate the probability of obtaining sums/counts in a chance process. Example: A coin is tossed 100 times. Estimate the chance of obtaining a) exactly 50 heads b) between 45 and 55 heads inclusive (include endpoints) c) between 45 and 55 heads exclusive (exclude endpoints) 20 Chapter 18 (cont.) a) avg of the box = ½ SD of the box = ½ ⎛1⎞ E (count of heads) = n ⋅ avg of the box = 100⎜ ⎟ = 50 ⎝ 2⎠ ⎛1⎞ SE (count of heads) = n ⋅ SD of the box = 100 ⎜ ⎟ = 5 ⎝2⎠ 7.97% 49.5 SE=5 50.5 X M=50 Z= Z=-.10 Z=.10 X − Mean SD About 7.97% of scores are between Z±.10. The probability of exactly 50 heads ≈ 7.97%. 21 7 Chapter 18 (cont.) b) SE=5 72.87% 55.5 44.5 Inclusive – include end points X M=50 Z= − 5.5 = −1.10 5 Z= Z= 5.5 = 1.10 5 X −M SE Percent of scores between ±Z=1.10 is 72.87%. Probability of scores between 44.5 and 55.5 is .7287, or 72.87% of scores. 22 Chapter 18 (cont.) c) SE=5 63.19% 54.5 45.5 Exclusive – exclude end points X M=50 Z= 45.5 − 50 = −.90 5 Z= 54.5 − 50 = +.90 5 Z Percent of scores between ±Z=.90 is 63.19%. Probability of scores between 45 and 55 is .6319 or 63.19% of scores. 23 Chapter 18 (cont.) Note: Ordinarily we would find such probabilities by finding area between 45 and 55. This amounts to replacing the area under the histogram between 45 and 55 by the area under the Normal Curve between the corresponding values in standard score units (Zscores). ) Keeping p g track of the end p points has an official name – the continuity connection. See a more detailed comment on page 318 of text. Exercise Set B – (pp. 318-319) #1, 2, 3, 4, 5, 6 24 8 Chapter 18 (cont.) #3. A coin is tossed 100 times. Estimate the chance of obtaining 60 heads. SE=5 M=50 59.5 Z= 59.5 − 50 9.5 = = 1.8 5 5 X 60.5 Z= 60.5 − 50 10.5 = = 2.1 5 5 Z 25 Chapter 18 (cont.) Area between mean and 2.1 standard score units = 48.22. Area between mean and 1.8 standard score units = 46.40. The area (percent) between 1.8 and 2.1 standard score units is 1.8%. Probability of obtaining exactly 60 heads is about 1.82% or .0182. 26 27 9 28 10