Solutions for practice questions: Chapter 15, Probability Distributions If you find any errors, please let me know at mailto:msfrisbie@pfrisbie.com. 1. Let X represent the savings of a resident; X ~ N(3000, 500 2 ). a) P(X > 3200) 0.345, so about 34.5% of the townspeople have savings greater than $3200. That uses a normal CDF command. b) The probability that one person has a savings between $2300 and $3300 is P(2300 < X < 3300) 0.645. If the two people are chosen at random from a normally distributed set of data, then they should be independent, and the probability of both of them having that income should be 0.645 2 0.416. 1 c) If P(X < d) = 74.22% = 0.7422, we can use an inverse normal command. The area requested by the calculator is an area to the left of d, which is the 0.7422 given. The calculator says that d $3325.07. Probably it s safest to give this to three significant figures, although you might be asked for the nearest dollar or nearest cent. There s not going to be an exact answer. So I ll call it $3330. 2. Since the discs are being replaced, the repeated trials are independent of each other. There are precisely two possible outcomes (red and black), so this is a binomial experiment. The questions are about selecting black discs, and the 1 I d like to point out that the hypothesis, that the residents of a small town should have savings that are normally distributed, is pretty silly. probability of getting a black is 5 out of 40, or 0.125. a) If there are eight selections, we can define X as the number of black discs selected. X ~ B(8, 0.125) i) P(X = 1) 0.393. That s a binomial PDF command, with n = 8, p = 0.125, and X = 1. ii) This one uses the binomial CDF command; the C is for cumulative. At least once means the minimum number of successes is 1. The maximum is, of course, 8 that s the total number of discs selected. P(1 X 8) 0.656. b) If the probability that a black disc is selected is 0.125, and there are 400 trials, one would expect 0.125 400 = 50. 3. a) If P(X 12) = 0.1, then the area of A is also 0.1. Duh. b) If P(X 8) is also 0.1, then the mean must be 8 + 12 = 10, 2 because the normal distribution is symmetric. c) Ah, show that. Since we know that P(X 8) = 0.1, we can find its z-score using the inverse normal command on the calculator. z 1.28155 Note that you get z-scores from a calculator by using the default mean, 0, and standard deviation, 1, in the inverse normal dialog.
The z-score represents the number of standard deviations that a particular data point is from the mean, so we get the equation z =. Substituting the known values of z, x, and µ and solving should do the trick. 8 10 2 1.28155 = = 2 = 1.56061 1.28155 1.56. Q. E. D. d) So now we have that X ~ N(10, 1.56). P(X 11) 0.739. 4. This is a classic binomial experiment. Let X be the number of heads obtained; X ~ B(8, 0.5). a) P(X = 4) 0.273. This is done with the binomial PDF. 2 b) P(X = 3) 0.219. Same thing. c) Getting 3, 4, or 5 heads is a binomial CDF thing. P(3 X 5) 0.711. 5. Let X be the lifespan of the insect. Then X ~ N(57, 4.4 2 ). a) Since the graph is a standard normal curve, the numbers on the horizontal axis are z-scores. The value of a corresponds to an x- value of 55. 55 57 z = 0.455 a 4.4 And the value of b corresponds to 60 hours. 60 57 z = 0.682 b 4.4 b) i) P(X > 55) 0.675 Note that since I ve already said how X is distributed, there s really no more work 2 Yes, you can also use combinations. to show. This question was written before command terms were standardized and when the normal tables were still in the formula packet. You would not be expected to do any more than this. ii) P(55 < X < 60) 0.428 c) i) If after t hours, 90% of the insects are dead, that must be a long time substantially to the right of the mean. Since it s supposed to be a standard normal curve and t is in hours, I ll need to convert it t 57 into a z-score: z =. 4.4 ii) To find the value of t requires an inverse normal command, with the area to the left as 0.9. P(X < t) < 0.9 t 62.6 hours 6. Let X represent the speed of a randomly selected vehicle. We know that X ~ N(µ, 10 2 ), and that P(X > 50) is 0.3. a) Using an inverse normal command with an area to the left of 1 0.3 = 0.7 gives a z-score for 50 km/h of 0.524. 50 µ Then 0.524 =. So 10 50 µ = 5.24, and µ 44.76 44.8 km/h. Q.E.D. When your book here says, The following part is optional and is
currently not in the syllabus of Maths SL, what it means is that nowhere in this unit has there been any mention of this topic, and there is no reason to believe that you have any idea how to approach this question. It is absurd that the question has been included in the text, and I will not attempt to teach you how to do it here. 7. Let X represent the IQ of an individual in the described population. X ~ N(100, 15 2 ) a) P(90 < X < 125) 0.700, using a normal CDF command. b) The chances that two people selected at random both have IQs in a particular range is the product of their individual probabilities the randomness of the selection from a group that is normally distributed guarantees their independence. P(X > 125) 0.04779, and 0.04779 2 0.00228, to three s. f. c) Hey, it s another problem you haven t been taught how to do. Ignoring it. 8. Look, another normal distribution problem. While I could use a variable other than X, if I do that, my chances of making typos (and your chances of getting candy) go up significantly. We can t have that. Also, I might need two different variables at some point. When that happens, I ll use Y. Let X represent the weight of a randomly selected bag of cement. X ~ N(25.7, 0.50 2 ) a) P(X < 25.0) 0.0808. That means about one bag in 12 would be under the advertised mass of 25 kg. (I read ahead.) b) The mean is increased to some value so that the probability of a bag being under 25.0 kg is down to 0.025. But the standard deviation (which probably amounts to a measure of how good the bag-filling machine is at its job, or maybe a measure of the accuracy of the scales used in filling) is unchanged. The z-score that goes with 0.025 being the area to the left of the mean is approximately 1.95996, using an inverse normal command. Then, using the definition of a z-score: z = 25 µ 1.95996 = 0.50 25 µ = 0.50( 1.95996) µ = 25 0.50( 1.95996) 25.9800 kg, or about 26.0 kg, to 3 s. f. Q.E.D. c) Now there will be 2.5% of the bags under 25 kg and 2.5% over 26 kg. Because of the symmetry of the normal distribution, the new mean is 25.5 kg. We know from part (b) that the z-score corresponding to a left-end area of 0.025 is 1.95996, so here s the work for. z = 25 25.5 1.95996 = 25 25.5 = 0.255 kg. 1.95996 d) Because the mean of the new machine is set at 25.5 kg rather than the old 26 kg, each bag s contents cost the company $0.40 less (half of the $0.80 per kg price I guess we re assuming that this is cost to the company; it s not entirely clear).
$5000 12500 $0.40 = bags 9. This time X will represent the mass of a randomly selected packet of breakfast cereal; X ~ N(750, 25 2 ). a) i) P(X < 740) 0.345 ii) P(X 780) 0.115. Notice the equals? At least includes the endpoint, but the normal CDF command is typed exactly the same way in either case. iii) P(740 < X < 780) 0.540. b) This choose two thing seems pretty popular. The random choice thing makes them independent, so this is just the square of the answer to (a)(i). 0.345 2 0.119. c) If 70% of the packets are more than x grams, then 30% are less; that s the area to the left. P(X < x) = 0.30 gives x 737 g. 10. Tallopia. Not a real country, in case you were wondering. Here, X will represent the height of a randomly selected adult; X ~ N(187.5, 9.5 2 ). a) P(X > 197) 0.159, or 15.9%. b) Okay, the wording here is a little difficult. First, we need to find the height of the person at the 99th percentile, and then we add 17 cm clearance for that guy. 3 P(X < x) = 0.99 gives x 209.6 cm for the one-percenters in terms of height. Then 17 cm more than that is 226.6 cm. To the nearest centimeter, that s 227 cm. 11. Let s define X as the mass of a randomly selected lion. It is claimed that X ~ N(310, 30 2 ). 4 a) P(X > 350) 0.0912 b) Okay, so the area between a and b is 0.95, and those are symmetric about the mean. That gives an extra 0.05 to be shared by both tails, and symmetry tells us that it will be 0.025 on each end. That means P(X < a) = 0.025, and P(X > b) = 1 0.025 = 0.975. Judicious use of the inverse normal command will take care of this: a 251 kg, b 369 kg. 12. Look, it s yet another normal distribution question let X be the reaction time of a human being; X ~ N(0.76, 0.06 2 ). a) Once again we re looking at a standard normal curve with mean of 0 and standard deviation of 1. That makes the values of a and b z-scores. 0.70 0.76 a = = 1 0.06 0.79 0.76 b = = 0.5 0.06 b) i) P(X > 0.70) 0.841 ii) P(0.70 < X < 0.79) 0.533 Both of those are normal CDF commands. c) i) If we know that 3% of the population has a reaction time less than c seconds, the 3% must be on the left end of the graph. Just like the last time we did this (in #5), it s necessary to change c seconds into a z- 3 It s gotta be a guy, right? No human population after eighth grade has the tallest one percent of people being female, does it? 4 I like that they ve worded it this way. Of course, it was claimed by the authors of the question, not by scientists or something.
ii) score, too. You can see that below the point. If P(X < c) = 0.03, then using an inverse normal command tells us that c 0.647 seconds. 5 13. Over a long period here means that this probability of being faulty is unchanging, and therefore the trials (of selecting a calculator to be tested) are independent. This is a binomial situation. a) One would expect 2% of the 100 calculators to be faulty. That s two. b) I m going to do this two ways. The old-school binomial coefficient way goes like this: Of the 100 calculators, we are choosing three to be faulty. The probability of being faulty is 0.02, and of being functional is 0.98. There are three faulty and 97 functional. 100 0.02 3 0.98 97 0.182 3 The other way to do this is with the binomial PDF command on the calculator. To do that, first I d have to define X as the number of faulty calculators. 5 If you were wondering, the standard normal curve diagrams are mathematically accurate. I typed in the formula for a standard normal curve and used an inverse normal command to figure out where to draw the vertical and shade. I m all about the attention to detail here (as opposed to when I just scrawl on the screen in class). X ~ B(100, 0.02), and then P(X = 3) 0.182. c) While both of those methods worked fairly easily in part (b), in part (c), the binomial coefficient method is more annoying if there are more than one outcome to account for. So I m just going with the calculator. The statement about the distribution of X is still true here. P(X > 1) 0.597. That s a binomial CDF command, with a lower bound of 2 and upper bound of 100. 14. Here s a diagram. The inverse normal command can give a z-score for each of the two x-values. Since the formula for a z-score, z =, contains both µ and, two different z-scores will give a system of equations to solve. P(X < 40) = 0.12 gives a z-score of 40 µ 1.17499, so 1.17499 =. P(X > 90) = 1 0.15 = 0.85 gives a z- 90 µ score of 1.03643, so 1.03643 =. You can solve this by hand, but if you multiply both sides of each equation by, they turn into linear equations that can be solved with the linear system solver on your calculator. 1.17499 = 40 µ 1.03643 = 90 µ The calculator says µ 66.6 km/h and 22.6 km/h.
15. a) Expected value is the sum of each payoff times its probability. 3 6 1 8 0 + 1 + 2 =. 10 10 10 10 b) i) Here s the tree. ii) I ll make a table to match the first one. The first row is the number of red balls. y 0 1 2 2 16 12 P(Y = y) 30 30 30 c) I started to try to explain this in words, but I think a tree is a better way to organize it. There are two ways to get two reds; I ll add those up. 2 1 4 12 2 48 + = + 6 10 6 30 60 180 1 8 9 3 = + = = 30 30 30 10 d) Rolling a 1 or a 6 means bag A. P A two red P ( A two red) = P two red 2 1 2 6 10 60 2 1 = = = = 3 18 18 9 10 60 ( ) ( ) 16. This is a binomial situation, because there are repeated trials (fifty of them) with a constant probability of success (4%), so those trials are independent. In this case, success actually means that a defective bearing has been identified. a) Let X be the number of defective bearings in the sample; X ~ B(50, 0.04). Clear of defects means X = 0. 50 0 50 P(X = 0) = ( 0.04 ) ( 0.96 ) 0 0.130. b) To be accepted, there must be two or fewer defective bearings. P(X 2) 0.677. I used the binomial CDF command here; the alternative is to add up the probabilities that X is 0, 1 and 2. c) E(X) = np = 50 0.04 = 2. 17. Once again, this is a binomial probability question. Let X be the number of correctly functioning CDs; X ~ B(10, 0.98). a) A package is eligible to be returned if any of the discs fail. That means it s returned if the number of correctly functioning discs is less than 10. P(X < 10) 0.183 This can also be done with the complement: 1 P(X = 10). b) Three packages are purchased, and each, independently, has the probability of being returned of
0.182. It s a binomial situation inside another one. This time Y will be the number of returnworthy packages, and Y ~ B(3, 0.183). P(Y = 1) 0.366 18. a) The sum of the probabilities in a probability distribution is 1. 2k + 2k 2 + (k 2 + k) + (2k 2 + k) = 1 5k 2 + 4k = 1 5k 2 + 4k 1 = 0 (5k 1)(k + 1) = 0 1 k =, k = 1 5 But only one of these makes any sense; a k-value of 1 gives a negative probability for X = 0. 1 Therefore k =. 5 b) E(X) is the expected value, calculated as x P ( X = x) 3. x= 0 Using the value of k from above, the table looks like this: x 0 1 2 3 2 2 6 P(X = x) 5 25 25 2 2 6 7 0 + 1 + 2 + 3 5 25 25 25 35 7 = = 25 5 7 25 ii) P(X 4) 0.000119 b) This time it s a normal distribution. Let Y be the size (I m guessing diameter, although it doesn t actually say); Y ~ N(3, 0.5 2 ). P(Y > 2.5) 0.841, or 84.1%. 19. Domestic consumption at first I thought that was as opposed to foreign, but that would still be commercial, from which I conclude that domestic means at home. So lots of salads, I guess. a) The probability is constant and there are 12 trials, so this is binomial. We ll let X be the number of fruits that are not large enough to sell; X ~ B(12, 0.023). i) P(X = 3) 0.00217