# 2 px0 px1 px2 016203 029489 026835 0725 c px n n xx p

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2) = P(x=0) + P(x=1) + P(x=2) = 0.16203 + 0.29489 + 0.26835 = 0.725 c. P(x) = n!(n-x)!x!pxqn-xP(x=0) = [1300!/(1300!0!)](0.0014)0(0.9986)1300= [1](0.0014)0(0.9986)700= 0.16182 P(x=1) = [1300!/(1299!1!)](0.0014)1(0.9986)1299= [1300](0.0014)1(0.9986)699= 0.29492 P(x=2) = [1300!/(1298!2!)](0.0014)2(0.9986)1298= [844350](0.0014)2(0.9986)698= 0.26855 P(x2) = P(x=0) + P(x=1) + P(x=2) = 0.16182 + 0.29492 + 0.26855 = 0.725 To the usual level of accuracy, the answers agree. The Poison is a good approximation to the binomial. 17. As illustrated by the table at the right, the Poisson approximations are not acceptable. The binomial probabilities [taken from Table A-1 using n=10 and p=0.5] are symmetric around x = 5. The Poisson probabilities [calculated from P(x) = μxe-μ/x! using μ= np = (10)(0.5) = 5] are positively skewed around x = 4.5. For x = 4, for example, the relative error is approximately (0.205 – 0.1755)/0.205 = 14%. In addition, the Poisson shows that the impossible x11 has probability 1 – 0.9863 = 0.0137. binomial Poisson x P(x) P(x) 0 0.001 0.0067 1 0.010 0.0337 2 0.044 0.0842 3 0.117 0.1404 4 0.205 0.1755 5 0.246 0.1755 6 0.205 0.1462 7 0.117 0.1044 8 0.044 0.0653 9 0.010 0.0363 10 0.001 0.01811.000 0.9863
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Statistical Literacy and Critical Thinking 131 Statistical Literacy and Critical Thinking1. A random variable is a variable that has a single value, determined by chance, for each outcome of a procedure. Yes, a discrete random variable can have an infinite number of possible values – e.g., the number of coin tosses required to get the first head can be 1,2,3,4,… 2. A discrete random variable has a countable number of possible values – i.e., its possible values are either finite, or they can by placed in a well-defined 1-to-1 correspondence with the positive integers. A continuous random variable has an infinite number of possible values that are associated with measurements on a continuous scale – i.e., a scale in which there are generally no gaps or interruptions between the possible values. 3. In a binomial probability distribution, there are exactly two possible outcomes: p is the probability of one of the outcomes (usually called “success”), and q is the probability of the other outcome (usually called “failure”). Since the two outcomes are mutually exclusive and exhaustive, p+q = 1. This means that q = 1–p and that p = 1–q. 4. No. Other named discrete probability distributions identified in the exercises are the geometric distribution (Section 5-3, Exercise 46) and the hypergeometric distribution (Section 5-3, Exercise 47). There are also general unnamed discrete probability distributions (e.g., Section 5-2, Exercise 11. Chapter Quick Quiz 1. No. Since P(x=0)+P(x=1) = 0.8+0.8 = 1.6 violates the requirement that ΣP(x) = 1, the given scenario does not define a probability distribution. 2. As shown by the table at the right, μ= Σ[x·P(x)] = 0.7. 3. The scenario describes a binomial probability distribution with n = 400 and p = 0.5. μ= np = (400)(0.5) = 200 4. The scenario describes a binomial probability distribution with n = 400 and p = 0.5. σ2= npq = (400)(0.5)(0.5) = 100; σ= 10 5. Yes. Any value more than two standard deviations from the mean is generally considered unusual, and 35 is (35-20.0)/4.0 = 3.75 standard deviations from the mean. 6. Yes. The scenario describes a binomial probability distribution with n = 5 and p = 0.2 A check of the given P(x) values indicates they were generated by the binomial formula P(x) = n!(n-x)!x!pxqn-x7. P(x1) = 1 – P(x=0) = 1 – 0.4096 = 0.5904 8. P(x4) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) = 0.4096 + 0.4096 + 0.1536 + 0.0256 + 0.0016 = 0.9984
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