= 0.35 (or ˆp = We have 20 independent trials, each with probability of success (heads) equal to 0.5, so X has a B(20, 0.5) distribution.

Transcription

1 Chapter 5 Solutions 51 (a) n = 1500 (the sample size) (b) The Yes count seems like the most reasonable choice, but either count is defensible (c) X = 525 (or X = 975) (d) ˆp = = 035 (or ˆp = = 065) 52 n = 200 (the sample size), ˆp = 40% = 040, and X = n ˆp = We have 20 independent trials, each with probability of success (heads) equal to 05, so X has a B(20, 05) distribution 54 Assuming no multiple births (twins, triplets, quadruplets), we have four independent trials, each with probability of success (type O blood) equal to 025, so the number of children with type O blood has a B(4, 025) distribution 55 (a) For a B(4, 03) distribution, P(X = 0) = and P(X 3) = (b) For a B(4, 07) distribution, P(X = 4) = and P(X 1) = (c) The number of failures in a B(4, 03) distribution has a B(4, 07) distribution With 4 trials, 0 successes is equivalent to 4 failures, and 3 or more successes is equivalent to 1 or fewer failures p(1 p) 56 (a) For a B(100, 05) distribution, µˆp = p = 05 and σˆp = n = 1 20 = 005 (b) No; the mean and standard deviation of the sample count are both 100 times bigger (That is, ˆp = X/100, so µˆp = µ X /100 and σˆp = σ X /100) 57 (a) ˆp has approximately a distribution with mean 05 and standard deviation 0025, so P(04 < ˆp < 06) = P( 2 < Z < 2) = (b) P(045 < ˆp < 055) = P( 1 < Z < 1) = Note: The answers given are from A or software The rule would give the answers 095 and 068 quite reasonable answers, especially since this is an approximation anyway For comparison, the exact answers are P(04 < ˆp < 06) = or P(04 ˆp 06) = 09648, and P(045 < ˆp < 055) = or P(045 ˆp 055) = (Notice that the correct answer depends on our understanding of between ) 58 (a) P(X 3) = ( 4) ( 4 4) = (b) If the coin were fair, P(X 3) = ( 4) ( 4 4) 05 4 = (a) Separate flips are independent (coins have no memory, so they do not try to compensate for a lack of tails) (b) Separate flips are independent (coins have no memory, so they do not get on a streak of heads) (c) ˆp can vary from one set of observed data to another; it is not a parameter 169

2 170 Chapter 5 Sampling Distributions 510 (a) X is a count; ˆp is a proportion (b) The given formula is the standard deviation for a binomial proportion The variance for a binomial count is np(1 p) (c) The rule of thumb in the text is that np and n(1 p) should both be at least 10 If p is close to 0 (or close to 1), n = 1000 might not satisfy this rule of thumb (See also the solution to Exercise 518) 511 (a) A B(200, p) distribution seems reasonable for this setting (even though we do not know what p is) (b) This setting is not binomial; there is no fixed value of n (c) A B(500, 1/12) distribution seems appropriate for this setting 512 (a) This is not binomial; X is not a count of successes (b) A B(20, p) distribution seems reasonable, where p (unknown) is the probability of a defective pair (c) This should be (at least approximately) a B(n, p) distribution, where n is the number of students in our sample, and p is the probability that a randomly-chosen student eats at least five servings of fruits and vegetables 513 (a) C, the number caught, is B(10, 07) M, the number missed, is B(10, 03) (b) Referring to C, we find P(M 4) = = (software: 03504) 514 (a) X, the number of auction site visitors, is B(15, 05) (b) Symmetry tells us that this must be 05, which is confirmed in C: P(X 8) = = (a) The mean of C is (10)(07) = 7 errors caught, and for M the mean is (10)(03) = 3 errors missed (b) The standard deviation of C (or M) isσ = (10)(07)(03) = errors (c) With p = 09, σ = (10)(09)(01) = errors; with p = 099, σ = errors σ decreases toward 0 as p approaches (a) The mean of X is (15)(05) = 75; the mean of ˆp is 05 (b) The mean of X increases with n; itis75with n = 150, and 750 with n = 1500 The mean of ˆp is 05 for any value of n 517 m = 6: P(X 6) = and P(X 5) = (a) The population (the 75 members of the fraternity) is only 25 times the size of the sample Our rule of thumb says that this ratio should be at least 20 (b) Our rule of thumb for the approximation calls for np and n(1 p) to be at least 10; we have np = (1000)(0002) = The count of 5s among n random digits has a binomial distribution with p = 01 (a) P(at least one 5) = 1 P(no 5) = 1 (09) 5 = (Or take from C and subtract from 1) (b) µ = (40)(01) = 4

3 Solutions One sample of 15 flips is shown on the right Results will vary quite a bit; C shows that 995% of the time, there will be 4 or fewer bad records in a sample of 15 Out of 25 samples, most students should see 2 to 12 samples with no bad records That is, N, the number of samples with no bad records, has a B(25, 02683) distribution, and P(2 N 12) = (a) n = 4 and p = 1/4 = 025 (b) The distribution is below; the histogram is on the right (c) µ = np = 1 x P(X = x) µ For ˆp, µ = 049 and σ = p(1 p)/n = As ˆp is approximately ly distributed with this mean and standard deviation, we find: P(046 < ˆp < 052) = P( 191 < Z < 191) = ( computation of the probability gives Using a binomial distribution, we can also find P(468 X 428) = 09444) 523 Recall that ˆp is approximately ly distributed with mean µ = p and standard deviation p(1 p)/n (a) With p = 030, σ = , so P(028 < ˆp < 032) = P( 139 < Z < 139) = ( computation of the probability gives Using a binomial distribution, we can also find P(284 X 323) = 08301) (b) With p = 006, σ = , so P(004 < ˆp < 008) = P( 268 < Z < 268) = (Using a binomial distribution, we can also find P(41 X 80) = 09918) (c) P( 002 < ˆp p < 002) increases to 1 as p gets closer to 0 (This is because σ also gets close to 0, so that 002/σ grows) 524 When n = 300, the distribution of ˆp is approximately with mean 049 and standard deviation (nearly twice that in Exercise 522) When n = 5000, the standard deviation drops to (less than half as big as in Exercise 522) Therefore: n = 300 : P(046 < ˆp < 052) = P( 104 < Z < 104) = n = 5000 : P(046 < ˆp < 052) = P( 424 < Z < 424) = 1 Larger samples give a better probability that ˆp will be close to the true proportion p ( computation of the first probability gives 07014; using a binomial distribution, we can also find P(138 X 156) = These more accurate answers do not change our conclusion)

4 172 Chapter 5 Sampling Distributions 525 (a) ˆp = = 07 (b) We want Exact prob P(X 140) or P( ˆp 07) The first can be found exactly (using a binomial distribution), or we can compute either using a approximation (with or without the continuity correction) All possible answers are shown on the right (c) The sample results are higher than the national percentage, but the sample was so small that such a difference could arise by chance even if the true campus proportion is the same 526 As σˆp = p(1 p)/n, wehave = (049)(051)/n, son = 15,61875 round up to 15, (a) p = 1/4 = 025 (b) P(X 10) = (c) µ = np = 5 and σ = np(1 p) = 375 = successes (d) No: The trials would not be independent because the subject may alter his/her guessing strategy based on this information 528 (a) µ = (1200)(075) = 900 and σ = 225 = 15 students (b) P(X 951) = P(Z 34) = (c) With n = 1300, P(X 951) = P(Z 154) = Other answers are shown in the table on the right (a) X, the count of successes, has a binomial distribution with mean µ X = np = (900)(1/5) = 180 and σ X = (900)(02)(08) = 12 successes (b) For ˆp, the mean is µˆp = p = 02 and σˆp = (02)(08)/900 = (c) P( ˆp > 024) = P(Z > 3) = (d) From a standard distribution, P(Z > 2326) = 001, so the subject must score 2326 standard deviations above the mean: µˆp σ ˆp = This corresponds to 208 or more successes 530 (a) M has a binomial distribution with n = 30 and p = 06, so P(M = 20) = ( 30 20) (06) 20 (04) 10 = (b) P(1st woman is the 4th call) = (06) 3 (04) = (a) p = 23,772, ,128,094 = (b) If B is the number of blacks, then B has (approximately) a binomial distribution with parameters n = 1200 and p = 01137, so the mean is np = 1364 blacks (c) P(B 100) = P(Z < 331) = Note: In (b), the population is at least 20 times as large as the sample, so our rule of thumb for using a binomial distribution is satisfied In fact, the mean would be the same even if we could not use a binomial distribution, but we need to have a binomial distribution for part (c), so that we can approximate it with a distribution which we can safely do, because both np and n(1 p) are much greater than 10

5 Solutions 173 ( ) n 532 (a) n = n! = 1 The only way to distribute n successes among n observations is for n!0! ( ) n n! n (n 1)! all observations to be successes (b) n 1 = = = n Todistribute (n 1)!1! (n 1)! n 1 successes among ( ) n observations, the one failure must be either ( observation ) 1, 2, 3, n n!, n 1, or n (c) k = k! (n k)! = n! (n k)![n (n k)]! = n n k Distributing k successes is equivalent to distributing n k failures 533 (a) P( ˆp 08) = P(X 80) is on line 1 (b) P( ˆp 08) = P(X 200) is on line 2 (c) For atest with 400 questions, the standard deviation of ˆp would be half as big as the standard deviation of ˆp for a test with 100 questions: With n = 100, σ = (085)(015)/100 = ; and with n = 400, σ = (085)(015)/400 = (d) Yes: Regardless of p, n must be quadrupled to cut the standard deviation in half 534 (a) P(first appears on toss 2) = (b) P(first appears on toss 3) = (c) P(first appears on toss 4) = P(first appears on toss 5) = ( )( ( )( 6) )( ) ( ) 3 ( ) ( ) 4 ( ) = 5 36 = Y has possible values 1, 2, 3, P(first appears on toss k) = ( 5 6 ) k 1 ( 1 6) 536 With µ = 200, σ = 10, and n = 25, we have mean µ x = µ = 200 and standard deviation σ x = σ/ n = When n = 100, the mean is µ x = µ = 200 (unchanged), and the standard deviation is σ x = σ/ n = 1 Increasing n does not change µ x but decreases σ x (the variability of the sampling distribution) 538 When n = 100, σ x = σ/ 100 = 10/10 = 1 The sampling distribution of x is approximately N(200, 1), soabout 95% of the time, x is between 198 and When n = 400, σ x = σ/ 400 = 10/20 = 05 The sampling distribution of x is approximately N(200, 05), soabout 95% of the time, x is between 199 and With σ/ 70 = 01195, we have P(x < 11) = P(Z < 084) = (If we use the rounded value σ/ 70 = 012 given in the text, then P(x < 11) = P(Z < 083) = 07967) 541 (a) The given formula is the standard deviation of the mean The variance of the sample mean is 10 2 /20 (b) Standard deviation decreases with increasing sample size (c) µ x always equals µ, regardless of the sample size 542 (a) The standard deviation is σ/ 10 = seconds (b) In order to have σ/ n = 30 seconds, we need a sample of size n = 100

6 174 Chapter 5 Sampling Distributions 543 Mean µ = mm and standard deviation σ/ 4 = mm 544 (a) For this exercise, bear in mind that the actual distribution for a single song length is definitely not ; in particular, a distribution with mean 350 seconds and standard deviation 300 seconds extends well below 0 seconds The curve for x should be taller by a factor of 10 and skinnier by a factor of 1/ 10 (b) Using a N(350, 300) distribution, 1 P(331 < X < 369) = 1 P( 006 < Z < 006) = (c) Using a N(350, ) distribution, 1 P(331 < X < 369) = 1 P( 020 < Z < 020) = In Exercise 543, we found that σ x = mm, so x has a N(40135 mm, mm) distribution (a) On the right The curve for x should be taller by a factor of 2 and skinnier by a factor of 05 (twice as tall, half as wide) (b) The probability that an individual axle diameter differs from the target by at least 0006 mm two standard deviations is about 5% (by the rule), or (using A) (c) 0006 mm is four standard deviations for the distribution of x, sothis probability is P(Z < 4 orz > 4) < (a) If T is the total number of lightning strikes (in one year), then µ T = 10 6 = 60 strikes and σ T = = strikes (b) If x is the mean number of strikes per square kilometer (that is, x = T/10), then µ x = 6 strikes/km 2 and σ x = 24/ 10 = strikes/km (a) x is not systematically higher than or lower than µ; that is, it has no particular tendency to underestimate or overestimate µ (b) With large samples, x is more likely to be close to µ because with a larger sample comes more information (and therefore less uncertainty) 548 (a) P(X 23) = P(Z 48 ) = P(Z 046) = (with software: 03234) Because ACT scores are reported as whole numbers, we might instead compute P(X 225) = P(Z 035) = (software: 03616) (b) µ x = 208 and σ x = σ/ 25 = 096 (c) P(x 23) = P(Z 096 ) = P(Z 229) = (In this case, it is not appropriate to find P(x 225), unless x is rounded to the nearest whole number) (d) Because individual scores are only roughly, the answer to (a) is approximate The answer to (c) is also approximate but should be more accurate because x should have a distribution that is closer to 40139

7 Solutions (a) µ x = 05 and σ x = σ/ 50 = 07/ 50 = (b) Because this distribution is only approximately, it would be quite reasonable to use the rule to give a rough estimate: 06 is about one standard deviation above the mean, so the probability should be about 016 (half of the 32% that falls outside ±1 standard deviation) Alternatively, P(x > 06) = P(Z > ) = P(Z > 101) = (a) µ = (4)(031) + (3)(040) + (2)(020) + (1)(004) + (0)(005) = 288 and σ = = (b) µ x = µ = 288 and σ x = σ/ 50 = (c) P(X 3) = = 071, and P(x 3) = P(Z = P(Z 081) = (software value: 02098) 551 Let X be Sheila s measured glucose level (a) P(X > 140) = P(Z > 15) = (b) If x is the mean of three measurements (assumed to be independent), then x has a N(125, 10/ 3 ) or N(125 mg/dl, mg/dl) distribution, and P(x > 140) = P(Z > 260) = (a) µ X = ($500)(0001) = $050 and σ X = = $ (b) In the long run, Joe makes about 50 cents for each $1 ticket (c) If x is Joe s average payoff over a year, then µ x = µ = $050 and σ x = σ X / 104 = $15497 The central limit theorem says that x is approximately ly distributed (with this mean and standard deviation) (d) Using this approximation, P(x > $1) = P(Z > 032) = (software: 03735) Note: Joe comes out ahead if he wins at least once during the year This probability is easily computed as 1 (0999) 104 = The distribution of xisdifferent enough from a distribution so that answers given by the approximation are not as accurate in this case as they are in many others 553 The mean of three measurements has a N(125 mg/dl, mg/dl) distribution, and P(Z > 1645) = 005 if Z is N(0, 1), sol = = 1345 mg/dl 554 x is approximately with mean 15 and standard deviation 13/ 200 = flaws/yd 2,soP(x > 2) = P(Z > 544) = 0 (essentially) 555 If W is total weight, and x = W/25, then: P(W > 5200) = P(x > 208) = P(Z > / ) = P(Z > 257) = (a) Although the probability of having to pay for a total loss for one or more of the 12 policies is very small, if this were to happen, it would be financially disastrous On the other hand, for thousands of policies, the law of large numbers says that the average claim on many policies will be close to the mean, so the insurance company can be assured that the premiums they collect will (almost certainly) cover the claims (b) The central limit theorem says that, in spite of the skewness of the population distribution, the average loss among 10,000 policies will be approximately ly distributed with mean $250 and standard deviation σ/ 10,000 = $1000/100 = $10 Since $275 is 25 standard deviations above the mean, the probability of seeing an average loss over $275 is about 00062

8 176 Chapter 5 Sampling Distributions 557 Over 45 years, x (the mean return) is approximately with µ x = 92% and σ x = 206%/ 45 = 30709%, so P(x > 15%) = P(Z > 189) = and P(x < 5%) = P(Z < 137) = Note: We have to assume that returns in separate years are independent 558 Let D be the shaft diameter so that D has a N(245 cm, 001 cm) distribution (a) 9953% fit into the hole: P(D 2476) = P(Z 001 ) = P(Z 260) = (b) 9671%: If H is the hole diameter, then H D has a distribution with mean = 005 cm and standard deviation = cm Therefore, P(H D 0024) = P(Z ) = P(Z 184) = (software: 09670) 559 (a) The mean of six untreated specimens has a standard deviation of 22/ 6 = lbs, so P(x u > 50) = P(Z > ) = P(Z > 779), which is basically 1 (b) The mean of x u x t is = 27 lbs, and the standard deviation is 22 2 / /6 = lbs, so P(x u x t > 25) = P(Z > ) = P(Z > 180) = (a) The central limit theorem says that the sample means will be roughly Note that the distribution of individual scores cannot have extreme outliers because all scores are between 1 and 7 (b) For Journal scores, y has mean 48 and standard deviation 15/ 28 = For Enquirer scores, x has mean 24 and standard deviation 16/ 28 = (c) y x has (approximately) a distribution with mean 24 and standard deviation 15 2 / /28 = (d) P(y x 1) = P(Z ) = P(Z 338) = (a) y has a N(µ Y,σ Y / m ) distribution and x has a N(µ X,σ X / n ) distribution (b) y x has a distribution with mean µ Y µ X and standard deviation σ 2 Y /m + σ 2 X /n 562 We have been given µ X = 9%, σ X = 19%, µ Y = 11%, σ Y = 17%, and ρ = 06 (a) Linda s return R = 07X + 03Y has mean µ R = 07µ X + 03µ Y = 96% and standard deviation σ R = (07σ X ) 2 + (03σ Y ) 2 + 2ρ(07σ X )(03σ Y ) = % (b) R, the average return over 20 years, has approximately a distribution with mean 96% and standard deviation σ R / 20 = 37703%, so P(R < 5%) = P(Z < 122) = (c) After a 12% gain in the first year, Linda would have $1120; with a 6% gain in the second year, her portfolio would be worth $ By contrast, two years with a 9% return would make her portfolio worth $ Note: As the text suggests, the appropriate average for this situation is (a variation on) the geometric mean, computed as (112)(106) 1 = 89587% Generally, if the sequence of annual returns is r 1, r 2,, r k (expressed as decimals), the mean return is k (1 + r1 )(1 + r 2 ) (1 + r k ) 1 It can be shown that the geometric mean is always smaller than the arithmetic mean, unless all the returns are the same 563 The total height H of the four rows has a distribution with mean 4 8 = 32 inches and standard deviation 01 4 = 02 inch P(H < 315orH > 325) = 1 P(315 < H < 325) = 1 P( 250 < Z < 250) = = 00124

9 Solutions (a) The table of standard deviations is given below (b) The graph is below on the right; it is shown as a scatterplot, but in this situation it would be reasonable to connect the dots because the relationship between standard deviation and sample size holds for all n (c) As n increases, the standard deviation decreases at first quite rapidly, then more slowly (a demonstration of the law of diminishing returns) n σ/ n Sample size 565 Out of five independent drivers, the number A who have an accident in the same year has a binomial distribution with n = 5 and p = 02, so P(A 3) = (using C, software, or a calculator) Five roommates could not be considered independent, as they were not randomly chosen from the population of all drivers 566 (a) Julie s score is about the 65th percentile: P(X 1110) = P(Z 209 ) = P(Z 040) = (software: 06561) Note that a continuity correction would be reasonable here, as SAT scores are reported as whole numbers; that is, we could compute P(X 11105) However, this makes little difference: A gives the same answer, while software changes slightly to (b) x = 1110 is about the 9999th percentile: x is approximately with mean 1026 and standard deviation 209/ 80 = , so P(x 1110) = P(Z 359) = (The continuity correction would not be appropriate in this case) (c) The first answer is less accurate: The distribution of an individual s score (like Julie s) might not be, but the central limit theorem says that the distribution of x will be close to 567 (a) Out of 12 independent vehicles, the number X with one person has a binomial distribution with n = 12 and p = 07, so P(X 7) = (using C, software, or a calculator) (b) Y (the number of one-person cars in a sample of 80) has a binomial distribution with n = 80 and p = 07 Regardless of the approach used approximation, or exact computation using software or a calculator P(Y 41) = This would not be surprising: Assuming that all the authors are independent (for example, none were written by siblings or married couples), we can view the 12 names as being a random sample so that the number N of occurrences of the ten most common names would have a binomial distribution with n = 12 and p = 0056 Then P(N = 0) = (1 0056) 12 = Standard deviation

10 178 Chapter 5 Sampling Distributions 569 The probability that the first digit is 1, 2, or 3is = 0602, so the number of invoices for amounts beginning with these digits should have a binomial distribution with n = 1000 and p = 0602 More usefully, the proportion ˆp of such invoices should have approximately a distribution with mean p = 0602 and standard deviation p(1 p)/1000 = , so P( ˆp ) = P(Z 271) = Alternate answers shown on the right 570 (a) If R is the number of redblossomed plants out of a sample of 12, then P(R = 9) = 02581, using a binomial distribution with n = 12 and Exact prob p = 075 (For C, use p = 025 and find P(X = 3), where X = 12 R is the number of flowers with nonred blossoms) (b) With n = 120, the mean number of red-blossomed plants is np = 90 (c) If R 2 is the number of red-blossomed plants out of a sample of 120, then P(R 2 80) = P(Z 211) = (Other possible answers are given in the table on the right) 571 If x is the average weight of 12 eggs, then x has a N(65 g, 5/ 12 g) = N(65 g, g) distribution, and P( < x < ) = P( 144 < Z < 289) = (software: 09236) 572 The probability that an airman completes a tour of duty without being on a lost aircraft is = If ˆp is the sample proportion who have been on a diet, then ˆp has approximately a N(070, ) distribution, so P( ˆp 075) = P(Z 183) = Exact prob (software: 00339) Alternatively, as ˆp 075 is equivalent to 210 or more dieters in the sample, we can compute this probability using the binomial distribution; these answers are shown in the table 574 (a) The machine that makes the caps and the machine that applies the torque are not the same (b) T (torque) is N(70, 09) and S (cap strength) is N(101, 12), sot S is N(7 101, ) = N( 31 inch lb, 15 inch lb) The probability that the cap breaks is P(T > S) = P(T S > 0) = P(Z > 207) = (software: 00194) 575 The center line is µ x = µ = 425 and the control limits are µ ± 3σ/ 5 = to (a) x has a N(32, 6/ 25 ) = N(32, 12) distribution, and y has a N(29, 5/ 25 ) = N(29, 1) distribution (b) y x has a N(29 32, 5 2 / /25 ) = N( 3, 15620) distribution (c) P(y x) = P(y x 0) = P(Z 192) = 00274

11 Solutions (a) ˆp F is approximately N(082, ) and ˆp M is approximately N(088, ) (b) When we subtract two independent random variables, the difference is The new mean is the difference of the two means ( = 006), and the new variance is the sum of the variances ( = ), so ˆp M ˆp F is approximately N(006, ) (c) P( ˆp F > ˆp M ) = P( ˆp M ˆp F < 0) = P(Z < 238) = (software: 00085) 578 (a) Yes; this rule works for any random variables X and Y (b) No; this rule requires that X and Y be independent The incomes of two married people are certainly not independent, as they are likely to be similar in many characteristics that affect income (for example, educational background) 579 For each step of the random walk, the mean is µ = (1)(06) + ( 1)(04) = 02, the variance is σ 2 = (1 02) 2 (06) + ( 1 02) 2 (04) = 096, and the standard deviation is σ = 096 = Therefore, Y/500 has approximately a N(02, ) distribution, and P(Y 200) = P( Y ) = P(Z 456) = 0 Note: The number R of right-steps has a binomial distribution with n = 500 and p = 06 Y 200 is equivalent to taking at least 350 right-steps, so we can also compute this probability as P(R 350), for which software gives the exact value