06-Random Variables complete

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/25/2014 for the course STATS 250 at University of Michigan.

Ask a homework question - tutors are online