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06-Random Variables complete

# So now x has a binn120 p010 model the mean or expected

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Unformatted text preview: who actually makes a purchase? 10 P(X 1) = 1 – P(X = 0) = 1 – [ (0.25)0(0.75)10] 0 = 1 – [1(1)(0.0563)] = 0.9437 d. How many shoppers in your random sample of size 10 would you expect to actually make a purchase? 25% of 10 is 2.5, so 2.5 on average 49 In the previous question (part d), you just computed the mean of a binomial distribution. If X has the binomial distribution Bin(n, p) then Mean of X is = E(X) = np Standard Deviation of X, is = np1 p Try it! More Work with the Binomial Suppose that about 10% of Americans are left‐handed. Let X represent the number of left‐handed Americans in a random sample of 12 Americans. Then X has a Bin(n =12, p = 0.10) distribution (be as specific as you can). Note that the mean or expected number of left‐handed Americans in such a random sample would be = np = 12(0.10) = 1.2. The standard deviation (reflecting the variability in the results from the mean across many such random samples) is = np(1 p) 12(0.10)(0.90) = 1.04. a. What is the probability that the sample contains 2 or fewer left‐handed Americans? P(X 2) = P(X = 0) + P(X = 1) + P(X = 2) 12 12 12 (0.10)0(0.90)12 + (0.10)1(0.90)11 + (0.10)2(0.90)10 1 2 0 = = 1(1)(0.2824) + 12(0.10)(0.3138) + 66(0.01)(0.3487) = 0.2824 + 0.3756 + 0.2301 = 0.8891 (so it is quite likely) b. Suppose a random sample of 120 Americans had been taken instead of just 12. So now X has a Bin(n=120, p=0.10) model. The mean or expected number of left‐handed Americans in the sample would be = np = 120(0.10) = 12. The standard deviation would be = np(1 p) 120(0.10)(0.90) = 3.29. So how might you try to find the probability that a random sample of 120 Americans would result in 20 or fewer left‐handed Americans? Note that 2 out of 12 is 16.67% and that 20 out of 120 is also equal to 16.67%. P(X 20) = P(X = 0) + P(X = 1) + … + P(X = 19) + P(X = 20) Could do it but – yuk! We will soon see an approximation that we can use that will make this computation much more reasonable. 50 8.5 General Continuous Random Variables A continuous random variable, X , takes on all possible values in an interval (or a collection of intervals). The way that we determine probabilities for continuous random variables differs in one important respect from how we determine probabilities for discrete random variables. For a discrete random variable, we can find the probability that the variable X exactly equals a specified value. We can’t do this for a continuous random variable. Instead, we are only able to find the probability that X could take on values in an interval. We do this by determining the corresponding area under a curve called the probability density function of the random variable. In Chapter 2, we summarized the general shapes of distributions of a quantitative response that often arise with real data. The shape of a distribution was found by drawing a smooth curve that traces out the overall pattern that is displayed in a histogram. With a histogram, the area of each rectangle is proportional to the frequency or count for e...
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