Contents. The Binomial Distribution. The Binomial Distribution The Normal Approximation to the Binomial Left hander example

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1 Contents The Binomial Distribution The Normal Approximation to the Binomial Left hander example The Binomial Distribution When you flip a coin there are only two possible outcomes - heads or tails. This is an example of a dichotomous event. Other examples are getting an answer right vs. wrong on a test, catching vs. missing a bus, or eating vs. not eating your vegetables. A roll of a dice, on other hand, is not a dichotomous event since there are six possible outcomes. If you flip a coin repeatedly, say 1 times, and count up the number of heads, this number is drawn from what s called a binomial distribution. Other examples are counting the number of correct answers on an exam, or counting the number of days that your eight year old eats his vegetables at dinner. Importantly, each event has to be independent, so that the outcome of one event does not depend on the outcomes of other events in the sequence. We can define a binomial distribution with three parameters: P is the of a successful event. That is the event type that you re counting up - lie heads or correct answers or did eat vegetables. For a coin flip, P =.5. For guessing on a 4-option multiple choice test, P = 1/4 =.25. For my eight year old eating his vegetables, P =.5. n is the number of repeated events. is the number of successful events out of n. The of obtaining successful events out of n, with P is: n!!(n )! P (1 P ) n where n! = n(n 1)(n 2)..., or n factorial. For example, if you flip a fair coin (P=.5) 5 times, the of getting 2 heads is: P r( = 5) = 5! 2!(5 2)! (.5)2 (1.5) (5 2) = (1)(.5 2 )(.5) 3 =.3125 Table B in our textboo and our Excel spreadsheet gives you this number, where the columns are for different values of P and the rows are different values of : n

2 We can plot this binomial frequency distribution as a bar graph: The table in the boo goes up to n=15, and the Excel spreadsheet goes up to n=2. But what about higher values of n? The shape of the distribution for n=5 should loo familiar. It loos normal! The Normal Approximation to the Binomial Here s the distribution for n = 2: There s that familiar bell curve. It turns out that the discrete binomial distribution can be approximated by the continuous normal distribution with a nown mean and standard deviation. The mean of the normal distribution is intuitive. If you have 2 coin flips, each with.5, then the average number of heads should be (2)(.5) = 1. So: µ = np = 1 The standard deviation is: σ = (n)(p )(1 P ) Which is for n = 2 is: 2

3 (2)(.5)(1.5) = 2.24 Here s that normal distribution drawn on top of the binomial distribution for n=2: You can see that it s a pretty good fit. The fit gets better with larger values of n, and for values of P that don t get too far away from 1 or -1. Since the fit is good, we can use Table A to estimate probabilities from binomial distribution problems for appropriate values of n and P. For example, if you flip a coin 2 times, what is the of obtaining 13 or more heads? Let s zoom in on the figure above, coloring the events that we are counting in green: Using table B, the actual of obtaining 13 or more heads is: Pr(=13) + Pr(=14) Pr( = 2) = = 316 Looing at the figure above, notice that the widths of each bar is 1 unit. The area of each bar (height times width) is therefore equal to the of that event. That means that Pr(>=13) is equal to the sum of the areas of the green bars. 3

4 So, to approximate the area of the green bars with the normal distribution (the red curve), we need to find the area under the red curve that covers the same range as the green bars. Loo closely, the green bar at = 13 covers the range from 12.5 to It follows that to approximate the area of the green bars, we need to find the area under the normal distribution above = Since we now the mean and standard deviation of this normal distribution, we can find the z-score: z = x µ σ = = 1.12 Using Table A: Pr(z>1.12) = 314 This is pretty close to the actual answer of 316. Left hander example Of the 9 students in our class that too the survey, 8 reported themselves as left-handed. If 1% of the population is left-handed, what is the that 8 or fewer people in will be left handed in a random sample of 9 people? The normal distribution that best approximates the distribution of left-handed people in a sample size of 9 will have a mean of: µ = np = (9)() = 9 The standard deviation of: σ = (n)(p )(1 P ) = (9)()(1 ) = To convert our value of 8 left-handers to a z-score, we need to include the bar that ranges from 7.5 to 8.5. So this time we need add.5 to 8 and calculate Pr(x<=8.5). z = x µ σ = =.37 Pr(z < -.37) =

5 It s question time. The following 1 questions are all binomial distribution problems. Use the binomial table (Table B or Excel spreasheet) if n is less than or equal to 2, and use the normal approximation to the binomial if n = 2. 1) For P =.5 and n = 14, find P r( >= 2) 2) For P =.85 and n = 32, find P r( <= 29) 3) For P =.75 and n = 26, find P r( <= 19) 4) For P =.5 and n = 7, find P r( >= 4) 5) For P =.35 and n = 5, find P r( <= 1) 6) For P =.65 and n = 22, find P r( >= 21) 7) For P =.8 and n = 42, find P r( >= 33) 8) For P =.6 and n = 9, find P r( <= 8) 9) For P =.85 and n = 33, find P r( >= 29) 1) For P =.5 and n = 15, find P r( >= 2) 5

6 Answers 1) For P =.5 and n = 14, find P r( >= 2) P r( >= 2) = = 529 2) For P =.85 and n = 32, find P r( <= 29) µ = np = (32)(.85) = 27.2 σ = (32)(.85)(1.85) = z = ( ) = 1.14 P r( <= 29.5) = P r(z <= 1.14) = ) For P =.75 and n = 26, find P r( <= 19) µ = np = (26)(.75) = 19.5 σ = (26)(.75)(1.75) = z = ( ) = P r( <= 19.5) = P r(z <= ) =.5 4) For P =.5 and n = 7, find P r( >= 4) P r( >= 4) = =.5 5) For P =.35 and n = 5, find P r( <= 1) P r( <= 1) = = ) For P =.65 and n = 22, find P r( >= 21) µ = np = (22)(.65) = 14.3 σ = (22)(.65)(1.65) = z = ( ) = 2.77 P r( >= 2.5) = P r(z >= 2.77) =.28 6

7 7) For P =.8 and n = 42, find P r( >= 33) µ = np = (42)(.8) = 33.6 σ = (42)(.8)(1.8) = z = ( ) =.42 P r( >= 32.5) = P r(z >=.42) = ) For P =.6 and n = 9, find P r( <= 8) With P >.5 we need to switch the problem to P = 1-.6 =.4, n = 9, P r( >= 9 8) = P r( >= 1) P r( >= 1) = = ) For P =.85 and n = 33, find P r( >= 29) µ = np = (33)(.85) = 28.5 σ = (33)(.85)(1.85) = z = ( ) =.22 P r( >= 28.5) = P r(z >=.22) = ) For P =.5 and n = 15, find P r( >= 2) P r( >= 2) = = 71 7