WEEK #23: Statistics for Spread; Binomial Distribution

Transcription

1 WEEK #23: Statistics for Spread; Binomial Distribution Goals: Study measures of central spread, such interquartile range, variance, and standard deviation. Introduce standard distributions, including the binomial distribution. Textbook reading for Week #23: Study Adler Section 6.9, 7.4

2 Standard Distributions We have seen distributions, both discrete and continous, that originate with data from experiments, or with formulas we made up to fit an example. Many common and important classes of experiments have well-understood distributions. We will start with the binomial distribution. Coin Flipping The canonical example for the binomial distribution is a sequence of coin flips. Compute the probability distribution for the number of heads flipped, if we flip a coin three times. The coin is loaded, so the probability of a heads each time is only 0.4 (tails probability is 0.6). The Binomial Distribution Consider the scenario of repeatedly running a two-outcome experiment. e.g. coin flipping, where each outcome is either heads or tails, or selecting alleles from parents, where each outcome is either B or b 1 We arbitrarily label one of the outcomes as a success (e.g. heads, B allele). If the probability of a successful outcome in each experiment is p, then what is the probability of getting exactly k successes in n trials? 1 These two-outcome experiments are sometimes called Bernoulli trials.

3 We note that there is an essential counting step in computing these probabilities: How many ways can there be k successful outcomes out of n trials? Fortunately, this is an well-known question, with a well-understood solution, and the calculation is built into most scientific calculators: Number of ways to choose k successes in n tries = n choose k = ( ) n = k n! k!(n k)! On a Casio 991 model calculator, you can compute this with the (<SHIFT> ) or ncr button. Use the binomial distribution to find the probability that you will roll exactly 3 ones while rolling a fair six-sided die 10 times. Consider a multi-genic phenotype, for which the visible effect of the genotype depends on the number of B copies a plant has in total over 12 different loci (24 possible B copies in total). Find the probability that a new offspring has a total of 16 different B alleles out of the possible 24, given that B alleles are distributed with p = 0.5 in the population at all loci.

4 We can now formalize these calculations by defining the binomial distribution. Binomial Distribution If a trial has two outcomes, and each trial is independent, the probability of k successful outcomes in n trials is given by b(k;n,p) = where p is the probability of success in each trial. ( ) n p k (1 p) n k k Properties of the Binomial Distribution If you flip 100 coins, intuitively how many do you expect to come up heads? If the coins were loaded, so that the probability of a heads was only 0.1 instead of 0.5, intuitively how many heads would you expect out of 100 tosses? Mean of the Binomial Distribution The mean of a binomial distribution, b(k;n,p), is given by E(b) = b(k;n,p) = n p Variance of the Binomial Distribution variance = σ 2 = Var(b(k; n, p)) = np(1 p) Standard Deviation of the Binomial Distribution Standard dev. = σ = Std. Dev(b(k; n, p)) = np(1 p)

5 Histograms of the Binomial Distribution Knowing the binomial distribution function, it straightforward to compute the probability of each number of successes, and so to draw a graph of the entire distribution. What is the range of the number of successes in n trials? How many possible outcomes is does that entail? Consider the distribution of the number of heads turning up in 10 flips of a fair coin. Sketch the distribution you would expect for the total number of heads out of Consider the distribution of the number of heads turning up in 10 flips of a loaded coin, where heads have only a 0.1 probability each flip. Sketch the distribution you would expect for the total number of heads out of

6 Here are some other distributions of the binomial distribution, as we change the probability of each individual success (p), and the number of trials (n). p = 0.1, n = 10 p = 0.3, n = 10 p = 0.5, n = 10 ] ] ] ] p = 0.1, n = 20 p = 0.3, n = 20 p = 0.5, n = 20 ] ] ] ] ] ] p = 0.1, n = 50 p = 0.3, n = 50 p = 0.5, n = 50 ] ] ] ] ] ] p = 0.1, n = 100 p = 0.3, n = 100 p = 0.5, n = 100 ] ] ] ] ] ] Comment on the patterns you see in the distributions.

7 Normal Approximation to the Binomial Distribution From the histograms of the binomial distribution, it seems that for large n values the binomial distribution starts to look a lot like a normal, gaussian, or bell-curve distribution. One commonly referenced rule of thumb is: A binomial distribution will be approximately normal in shape if both np and n(1 p) are above 10. Relate this observation back to the previous histograms. Beyond its mathematical interest (why does the binomial look like the normal distribution?), we can take advantage of well-understood properties of the normal distribution in analyzing binomial data. Theorem: For a normal distribution, the probability of an outcome within ±2 standard deviations of the mean is 95% (rounded) Under the assumption that some binomial distributions approximate the normal distribution, express this theorem as it applies to binomial distributions.

8 Example: A drug is undergoing re-evaluation by Health Canada for effectiveness as an anti-fungal treatment. The manufacturer claims the drug is effective 60% of the time in killing off the fungus. Health Canada tracks 100 patients who are treated. Sketch the probability distribution for the number of patients who are cured by the treatment, assuming the manufacturer s claims are true In this trial, only 53% of the patients are cured by the drug. Comment on how much you can trust the claimed 60% cure rate.

9 Doubts have been raised in other countries about the same drug, so a larger trial is commissioned with 1,000 patients. Sketch the probability distribution for the number of patients cured in this trial, assuming again a 60% curative probability for each patient Again, only 53% of the patients are cured by the drug. Is your conclusion the same or different than in the last example, and why?