# Statistics 100 Binomial and Normal Random Variables

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1 Statistics 100 Binomial and Normal Random Variables Three different random variables with common characteristics: 1. Flip a fair coin 10 times. Let X = number of heads out of 10 flips. 2. Poll a random sample of 100 Massachusetts voters. Let X = number of voters that preferring Elizabeth Warren over Scott Brown. 3. Randomly sample 50 Starbucks customers. Let X = number of customers preferring lattes to cappuccinos Characteristics in common: Each random variable counts number of successes in fixed number of trials. Each trial has equal probability of success. The outcome of each trial is independent. A random variable possessing these characteristics has a binomial distribution, and is referred to as a binomial random variable. Two important parameters of a binomial distribution: n fixed number of trials p probability of success on each trial Once you are able to identify a random variable as binomial, knowing n and p are sufficient for calculating all the probabilities of each value. Examples of random variables that are not binomial: Stand at the Bank of America in Harvard Square. Let X = number of cars passing by in the next 10 minutes (no fixed number of trials) Gather a random sample of five men and five women. Let X = number of persons out of 10 weighing more than 150 lbs. (successes on trials do not have equal probability) 1

2 Draw 4 cards (without replacement) from a deck of 52 cards. Let (trials are not independent) X = number of aces among the four Conventional notation: X B(n, p) (which translates to X has a binomial distribution with n trials and probability of success p ) To obtain probabilities for binomial random variables, use Table C in the textbook: Identify n and p for the binomial random variable Go to page in Table C corresponding to the appropriate n Go to column corresponding to the appropriate p The values inside the table are the probabilities for each value of a binomial random variable Can also use the dbinom command in R. Example: ESP Experiment Experiment consists of trying to guess one of 5 symbols (square, circle, triangle, wavy lines, plus) on a single trial. Procedure is repeated 20 times. Assume ESP does not exist, so that guesses are random. Let X = Number of correct guesses out of 20 X has a binomial distribution. More specifically, X B(20, 1 5 ). P(X = 7) = P(X 7) = Binomial distribution: Mean, variance and standard deviation: If X B(n, p), then it can be shown that µ X = np σx 2 = np(1 p) σ X = σx 2 = np(1 p) A related random variable: ˆp 2

3 Define ˆp = X/n = proportion of successes out of n trials Probabilities of each value ˆp takes on are the same as for X B(n, p). Mean, variance, and standard deviation of ˆp: µˆp = p σ 2ˆp = p(1 p)/n σˆp = p(1 p)/n Note that as n gets large, σˆp becomes smaller. Therefore, ˆp becomes close to its mean, namely p, for large sample sizes. ESP example again: We have X B(20, 1 5 ). Also, define ˆp to be X/20, the proportion of successes out of 20. What is the probability of observing 40% or more correct guesses? P(ˆp 0.4) = P(X/20 0.4) = P(X 8) = = (from Table C) What are the mean, variance and standard deviation of X, and of ˆp? µ X = np = 20( 1 5 ) = 4 σ 2 X = np(1 p) = 20( 1 5 )(4 5 ) = 3.2 σ X = 3.2 = 1.79 µˆp = p = 1 5 σ 2ˆp = p(1 p)/n = ( 1 5 )(4 )/20 = σˆp = =

4 Another example: Suppose that about 40% of all households own DVD players. A random sample of 12 households is selected. Find 1. the probability that at most 5 of the households own DVD players 2. the probability that between 50% and 70% of the households in the sample own DVD players 3. the mean and standard deviation of the number of households owning DVD players 4. the probability that the number of households owning DVD players is within one standard deviation of the mean Solution to each question: Let X be the number of households out of 12 that own DVD players. Then X B(12, 0.4). Also, let ˆp = X/12 be the proportion of households out of 12 that own DVD players. With X B(12, 0.4), 1. P(X 5) = 2. P(0.50 ˆp 0.70) = 3. µ X = np = 12(0.4) = 4.8 σ X = np(1 p) = 12(0.4)(0.6) = P( X ) = P(3.103 X 6.497) = P(X = 4 or X = 5 or X = 6) = Normal distribution: A random variable X that has a normal distribution is a continuous random variable that can take on all possible real values, has values near the mean occurring more often than values away from the mean, i.e., is bell-shaped, is symmetrically distributed around its mean. Usually think of a random variable as being approximately normally distribution. 4

5 Probability density function of the Normal distribution: p(x) = 1 2πσ 2 e (x µ)2 /2σ 2 where µ is the mean, and σ is the standard deviation. Notice that once µ and σ are specified, we know the density function exactly, which means we can calculate probabilities associated with X. You ll never need to know the formula for this density function! Density function of a normal distribution: Can write X N(µ, σ) This translates to: X has a normal distribution with mean µ and standard deviation σ Special case: When µ = 0 and σ = 1, a random variable with a normal distribution has a standard normal distribution. Write Z N(0, 1) 5

6 Usually reserve Z to denote a random variable with a standard normal distribution. Density function of a standard normal distribution: Computing probabilities involving standard normal random variables: Table A in textbook reports probabilities of the form P(Z < z) for given z. Examples: P(Z < 1.23) = P(Z > 2.84) = P( 1.61 Z 3.44) = P(Z < 5.98) = Two important and useful mathematical facts: 6

7 1. If X N(µ, σ), then the random variable X µ σ N(0, 1) This fact will allow us to compute probabilities involving normal random variables with arbitrary means and variances. 2. If X 1, X 2,..., X n are iid random variables with X i N(µ, σ), then X N(µ, σ/ n). In combination with the above fact, this will help to compute probabilities involving sample means. Probabilities from normally distributed random variables with arbitrary means and standard deviations: Suppose X is the blood sugar level (mg/dl) of a randomly selected adult after eating watermelon. Assume X N(170, 30). What is the probability a randomly selected adult will have a blood sugar level less than 150 after eating watermelon? P(X < 150) = P( X 170 < 30 = P(Z < 0.67) = ) 30 What is the probability a randomly selected adult will have a blood sugar level between 170 and 210 after eating watermelon? P(170 < X 210) = P( < X = P(0 < Z 1.33) = P(Z 1.33) P(Z 0) = = ) 30 A reverse problem: One fourth of adults after eating watermelon have blood sugar levels above what value? Solution: Let x be the desired value. Then 0.25 = P(X > x) = P( X > x 170 ) 30 7

8 = P(Z > x 170 ) 30 so that 0.75 = P(Z x 170 ) 30 From Table A, we know that P(Z 0.68) Therefore x = 0.68 x = A slightly more challenging (but relevant) problem: Suppose we select four adults at random who have eaten watermelon. What is the probability their average blood sugar level is less than 150? Let X i, i = 1, 2, 3, 4 be the blood sugar level for the i-th adult. Know that X i N(170, 30) and that we can assume the X i are independent. Define X = X 1 + X 2 + X 3 + X 4 4 Then we have E( X) = 170, SD( X) = 30/ 4 = 15, and that X is normally distributed, so that X N(170, 15) Distribution of an individual s blood sugar level: X N(170, 30) 8

9 Density Blood sugar level Distribution of the mean of four blood sugar measurements: X N(170, 15) 9

10 Density Blood sugar level So because X N(170, 15), P( X < 150) = P( X < ) = P(Z < 1.33) = Back to data summaries: How do you tell if data follows a normal distribution? 10

11 Is this Normal Data? Or is this Normal Data? Number of Observations Number of Observations Data N= Data N=500 Use normal probability plot. Constructing a normal probability plot from n observations (on the computer): 1. Sort the n observations in ascending order. 2. Find n values on a standard normal distribution ( normal scores ) that section off equal areas. 3. Produce a scatter plot of the sorted data on the vertical axis and the normal scores on the horizontal axis. If the plot is approximately linear, then the data are approximately normal. Otherwise, the plot shows departures from normality. Now which one is a normal distribution? 11

12 Is this Normal Data? Or is this Normal Data? Data Data Normal Scores N= Normal Scores N=500 The most remarkable result in the entire field of probability and statistics: Suppose X 1,..., X n are iid random variables with arbitrary probability distributions (not necessarily normal!), each with mean µ and standard deviation σ. The Central Limit Theorem (CLT): As n becomes large, X = X X n n N(µ, σ n ) Also, n X i = X X n N(nµ, nσ) i=1 This is INCREDIBLE!! Why you should be falling out of your seats: It s plausible that the distribution of X (or n i=1 X i ) is symmetric as n becomes large, but 12

13 why specifically normal?? (there are many probability density functions that are symmetric and bell-shaped) The only two parameters of the distribution of the X i that are relevant to averages (or sums) are µ and σ. You may have no clue about the probability distribution of any of the X i, but you do know the distribution of the average (or sum) of the X i (as n becomes large). Implications: 1. Can compute probabilities associated with X (or n i=1 X i ) using the normal distribution (when n is large). 2. More problems you can solve on Stat 100 homework, exams, etc. Rule of thumb: If n 30, and the distribution of the X i are not very skewed, then the distribution of X can be treated as approximately normal. (if n < 30, consult a statistician) Example: Find the probability that the average of 1000 independent dice rolls will be greater than Let X i be the value of the i-th die roll, i = 1,..., Can easily compute E(X i ) = 3.5, and SD(X i ) = for every i. Also, µ X = 3.5 σ X = 1.708/ 1000 = Therefore, by the CLT, approximately X N(3.5, 0.054) P( X > 3.65) = P( X > ) = P(Z > 2.78) = 1 P(Z < 2.78) = = Another application: Normal approximation to the Binomial distribution If X B(n, p), then for large enough n, the distribution of X is well approximated by X N(np, np(1 p)) 13

14 Why does the CLT apply to this situation? Because X is counting the number of successes on n independent and equi-probable trials, we can consider each trial separately: Y i = { 1 if trial i results in success 0 if trial i results in failure. Then X = Y 1 + Y Y n, which is the sum of n iid random variables. So by the CLT, as n becomes large, X is well-approximated by a normal distribution. A similar rule applies for the random variable ˆp = X/n when n is large: because (as in the preceding demonstration) ˆp N(p, p(1 p)/n) ˆp = Y 1 + Y Y n. n For binomial random variables, how large is large? To use the normal approximation to the binomial distribution, should have both np 10, and n(1 p) 10 as a useful rule of thumb. Example: Let X be the number of males in a randomly selected group of 15 people. What is the probability that more than 75% are male? Assume X B(15, 1 2 ). Because np = 15( 1 2 ) = 7.5 < 10, we should not use the normal approximation, so we use the binomial probabilities from Table C: P(X > 0.75(15)) = P(X > 11.25) = P(X = 12) + + P(X = 15) = =

15 Now let X be the number of males in a randomly selected group of 150 people. Here X B(150, 1/2). This time, both np = 150(1/2) = 75 and n(1 p) = 150(1/2) = 75 are greater than 10, so the normal approximation is okay. Probability histogram of X B(150, 1/2): Probability Values of X Normal approximation of of X B(150, 1/2): 15

16 Probability Values of X With X B(150, 1/2), we have E(X) = np = 150( 1 2 ) = 75 SD(X) = np(1 p) = 150( 1 2 )(1 2 ) = 6.12 Can approximate X N(75, 6.12) So the probability that more than 75% are male in a sample of 150 is: P(X > 0.75(150)) = P(X > 112.5) = P( X > ) = P(Z > 6.13) 0 16

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