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Solutions to Practice Exam 3 Math 203 March 16, 2012 1. Suppose that on any given day, the probability you will like one of the meals at the “Explore” station in Anderson Commons is 53 . During next week, Sunday through Saturday, what is the probability you will like an “Explore” meal on (a) exactly 5 of those days? Solution. This is a binomial trial question. We are asking for exactly 5 successes out of 7 days, where the probability of success is 35 . So n = 7, k = 5, p = 53 , and q = 25 , and the answer is 5 2 3 2 C(7, 5) = 0.2612736. 5 5 (b) at least 5 of those days? Solution. Now we want at least 5 successes; in other words, we want either 5, 6, or 7 successes. So our answer is 5 2 6 1 7 0 3 3 3 2 2 2 C(7, 5) + C(7, 6) + C(7, 7) = 0.419904 5 5 5 5 5 5 2. Let X be the random variable corresponding to how many pounds of candy a random trick-or-treater in Moorhead gets on Halloween. Suppose the probability distribution of X is given by the following table. x Pr(X = x) 0.5 1 1.5 2 0.3 0.4 0.2 0.1 (a) Find the expected value E(X), also known as µ. Solution. For a probability distribution, the formula for µ is µ = x1 p 1 + x2 p 2 + · · · + xn p n . Thus the answer is µ = (0.5)(0.3) + (1)(0.4) + (1.5)(0.2) + (2)(0.1) = 1.05. (b) Find the standard deviation σ. Solution. The formula for the standard deviation of a probability distribution is p σ = (x1 )2 p1 + (x2 )2 p2 + · · · + (xn )2 pn − µ2 , so the answer is p √ σ = (0.5)2 (0.3) + (1)2 (0.4) + (1.5)2 (0.2) + (2)2 (0.1) − 1.052 = 0.2225 ≈ 0.4717. Page 2 of 4 Practice Exam 3 Solutions Math 203, Spring 2012 3. Your friend Joe is a big film fan, and on average he likes 75% of the films he sees. During the Fargo Film Festival, Joe skips his classes and abandons his homework to watch 48 films in a row. Let the random variable X denote how many of those films he will like. (a) What is the formula giving the probability distribution of X? In other words, what is Pr(X = k)? Solution. This is a binomial distribution with n = 48, p = 0.75, and q = 1 − p = 0.25. So the probability distribution is Pr(X = k) = C(48, k)(0.75)k (0.25)48−k . (b) How many of the 48 movies do you expect that Joe will like? Solution. For a binomial distribution, µ = E(X) = np = 48 · 0.75 = 36. (c) What is the standard deviation of the random variable X? √ √ √ Solution. For a binomial distribution, σ = npq = 48 · 0.75 · 0.25 = 9 = 3. (d) Use the normal distribution to approximate the probability that Joe will like at least 30 of the 48 movies. Solution. We will pretend that X, instead of a binomial random variable, is instead a normal random variable with µ = 36 and σ = 3. Since we want at least 30, the cutoff should be 29.5. So 29.5 − 36 Pr(X ≥ 30) = Pr(X ≥ 29.5) ≈ Pr Z ≥ = Pr(Z ≥ −2.1667) 3 = 1 − Pr(Z ≤ −2.1667) ≈ 1 − 0.0158 = 0.9842. 4. A publisher of textbooks has 1 misprint per 50 pages. The misprints occur randomly. In 100 pages, what is the probability of (a) no misprints? Solution. Since the misprints occur randomly, this is a Poisson distribution. We are looking at 100 pages, so 1 misprint λ misprints = ; 50 pages 100 pages thus λ = 2. Using the formula Pr(X = k) = λk −λ e , k! k Pr(X = k) 0 0.13534 1 0.27067 2 0.27067 .. .. . . Thus Pr(X = 0) = 0.13534. we fill in the table Practice Exam 3 Solutions Page 3 of 4 Math 203, Spring 2012 (b) at most 2 misprints? Solution. Using the table above, we see Pr(X ≤ 2) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 0.13534 + 0.27067 + 0.27067 = 0.67668. (c) more than 2 misprints? Solution. Pr(X > 2) = 1 − Pr(X ≤ 2) = 1 − 0.67668 = 0.32332. 5. Suppose the weights of pumpkins in Hornbacher’s are normally distributed with mean µ = 6 lb. and standard deviation σ = 2 lb. (a) What fraction of the pumpkins weigh at least 10 pounds? Solution. Our random variable X is normally distributed with µ = 6 and σ = 2. The fraction of pumpkins over 10 pounds equals the probability that a randomly selected pumpkin weights more than 10 pounds, so the fraction we want is 10 − 6 = Pr(Z ≥ 2) = 1−Pr(Z ≤ 2) ≈ 1−0.9772 = 0.0228. Pr(X ≥ 10) = Pr Z ≥ 2 (b) What fraction of the pumpkins weigh between 4 and 8 pounds? Solution. Again we use the table (which would be given at the back of the test) to compute that 8−6 4−6 ≤Z≤ = Pr(−1 ≤ Z ≤ 1) = 1 − 2 Pr(Z ≤ −1) Pr(4 ≤ X ≤ 8) = Pr 2 2 ≈ 1 − 2(0.1587) = 0.6826. 6. The grade point average (GPA) of graduating seniors at Tippecanoe College is normally distributed with a mean of 2.7 and a standard deviation of 0.5. The top 8% of the class (by GPA) will graduate cum laude. What is the minimum GPA required to graduate cum laude? Solution. Let m stand for the minimum GPA. Then we know that 8% of the population earns a GPA greater than or equal to m; in other words, 0.08 = Pr(X ≥ m). Applying our rules, then, we see that 0.08 = 1 − Pr(X ≤ m) = 1 − Pr(Z ≤ m − 2.7 ) 0.5 so Pr(Z ≤ m−2.7 ) = 0.92. Looking in the table, we find that Pr(Z ≤ 1.41) = 0.9207, 0.5 so we want m − 2.7 = 1.41. 0.5 Solving for m, we find m = 2.7 + 0.5 · 1.41 = 3.405. Practice Exam 3 Solutions Page 4 of 4 Math 203, Spring 2012 7. If a cow gets a certain disease, there is only a 25% probability that it will recover. If 200 cows get the disease, what is the probability that at least 60 will recover? Give an approximate answer using a normal approximation. Solution. This is a binomial distribution, with n = 200, p = 0.25, and q = 0.75. √ Thus the mean is µ = np = 200(0.25) = 50 and the standard deviation is σ = npq = p 200(0.25)(0.75) ≈ 6.1237. Because we are approximating a binomial distribution with a normal distribution, we need to adjust by 0.5. Thus 59.5 − 50 Pr(X ≥ 60) = Pr(X ≥ 59.5) ≈ Pr Z ≥ ≈ Pr(Z ≥ 1.55). 6.1237 Then using the table, we look up and find Pr(Z ≥ 1.55) = 1 − Pr(Z ≤ 1.55) = 1 − 0.9394 = 0.0606.