Solution manual for statistics 2nd edition by lock

Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 492 Section P.1 Solutions P.1 We have P (not A) = − P (A) = − 0.4 = 0.6 P.2 We have P (not B) = − P (B) = − 0.3 = 0.7 P.3 By the additive rule, we have P (A or B) = P (A) + P (B) − P (A and B) = 0.4 + 0.3 − 0.1 = 0.6 P.4 We have P (A if B) = 0.1 P (A and B) = = 0.333 P (B) 0.3 P (B if A) = 0.1 P (A and B) = = 0.25 P (A) 0.4 P.5 We have P.6 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) = Since we are told that P (A and B) = 0.1, not zero, the events are not disjoint P.7 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.1 and P (A) · P (B) = 0.4 · 0.3 = 0.12, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.25 = P (B) = 0.3 P.8 We have P (not A) = − P (A) = − 0.8 = 0.2 P.9 We have P (not B) = − P (B) = − 0.4 = 0.6 P.10 By the additive rule, we have P (A or B) = P (A) + P (B) − P (A and B) = 0.8 + 0.4 − 0.25 = 0.95 P.11 We have P (A if B) = 0.25 P (A and B) = = 0.625 P (B) 0.4 P (B if A) = 0.25 P (A and B) = = 0.3125 P (A) 0.8 P.12 We have P.13 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) = Since we are told that P (A and B) = 0.25, not zero, the events are not disjoint P.14 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.25 and P (A) · P (B) = 0.8 · 0.4 = 0.32, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.3125 = P (B) = 0.4 P.15 Since A and B are independent, knowing that B occurs gives us no additional information about A, so P (A if B) = P (A) = 0.7 P.16 Since A and B are independent, knowing that A occurs gives us no additional information about B, so P (B if A) = P (B) = 0.6 P.17 Since A and B are independent, we have P (A and B) = P (A) · P (B) = 0.7 · 0.6 = 0.42 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 493 P.18 Since A and B are independent, we have P (A and B) = P (A)·P (B) = 0.7·0.6 = 0.42 By the additive rule, we have P (A or B) = P (A) + P (B) − P (A and B) = 0.7 + 0.6 − 0.42 = 0.88 P.19 There are two cells that are included as part of event A, so P (A) = 0.2 + 0.1 = 0.3 P.20 There are two cells that are included as part of event not B, so P (not B) = 0.1 + 0.3 = 0.4 P.21 There is one cell that is in both event A and event B, so we have P (A and B) = 0.2 P.22 There are three cells that represent being in event A or event B or both, so we have P (A or B) = 0.2 + 0.4 + 0.1 = 0.7 We could also use the additive rule P (A or B) = P (A) + P (B) − P (A and B) = 0.3 + 0.6 − 0.2 = 0.7 P.23 We have P (A if B) = P (A and B) 0.2 = = 0.333 P (B) 0.6 P (B if A) = P (A and B) 0.2 = = 0.667 P (A) 0.3 P.24 We have P.25 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) = We see in the table that P (A and B) = 0.2, not zero, so the events are not disjoint P.26 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.2 and P (A) · P (B) = 0.3 · 0.6 = 0.18, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.667 = P (B) = 0.6 P.27 The two events are disjoint, since if at least one skittle is red then all three can’t be green However, they are not independent or complements P.28 The two events are disjoint because both teams cannot win They are also complements because if Australia does not win, then South Africa wins However, they are not independent P.29 The two events are independent, as Australia winning their rugby match will not change the probability that Poland wins their chess match However, they are not disjoint or complements P.30 The two events are not disjoint or complements, as it is possible to have the rolls be {3,5} where the first die is a and the sum is To check independence we need to find P (A) = 1/6 and P (B) = P ({2, 6} or {3, 5} or {4, 4} or {5, 3} or {6, 3}) = 5/36 There is only one possibility for the intersection so P (A and B) = P ({3, 5}) = 1/36 We then check that P (A if B) = 1/36 P (A and B) = = 1/6 = P (B) = 5/36 P (B) 1/6 so A and B are not independent We can also verify that P (A and B) = 1/36 = P (A) · P (B) = (1/6) · (5/36) = 5/216 P.31 (a) It will not necessarily be the case that EXACTLY in 10 adults are left-handed for every sample We can only conclude that approximately 10% will be left-handed in the “long run” (for very large samples) Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 494 (b) The three outcomes each have probability for the results of baseball pitches only if they are equally likely This may not be the case (c) To find the probability of two consecutive 1’s on independent dice rolls we should multiply the probabilities instead of adding them Using the multiplicative rule, the probability that two consecutive rolls land with a is 16 × 16 = 36 (d) A probability that is not between and does not make sense P.32 (a) We are told that P (C) = 0.26, P (W ) = 0.13, and P (C and W ) = 0.03 (b) We use the additive rule for find the probability of one event or another P (C or W ) = P (C) + P (W ) − P (C and W ) = 0.26 + 0.13 − 0.03 = 0.36 The probability that a movie is either a comedy or produced by Warner Bros is 0.36 (c) We are finding the probability that a movie is a comedy if it is produced by Warner Bros, so we have P (C if W ) = P (C and W ) 0.03 = = 0.23 P (W ) 0.13 About 23% of movies produced by Warner Bros are comedies (d) We are finding the probability that a movie is produced by Warner Bros if it is a comedy, so we have P (W if C) = P (C and W ) 0.03 = = 0.115 P (C) 0.26 About 11.5% of comedies are produced by Warner Bros (e) To find the probability a movie is not a comedy we have P (not C) = − P (C) = − 0.26 = 0.74 (f) Saying C and W are disjoint would mean that Warner Bros never makes any comedies This is not true since we see that 3% of all the movies are comedies from Warner Bros It is clear that C and W are not disjoint (g) Saying C and W are independent would mean that knowing a movie comes from Warner Bros would give us no information about whether it is a comedy and knowing it is a comedy would give us no information about whether it came from Warner Bros This is almost true: P (C if W ) = 0.23 which is close to P (C) = 0.26 and P (W if C) = 0.115 which is close to P (W ) = 0.13, but neither result is exactly the same so the two events are not independent (at least for these approximated probability values.) P.33 (a) We are finding P (M P ) There are a total of 303 inductees and 206 of them are performers, so we have 206 P (M P ) = = 0.680 303 The probability that an inductee selected at random will be a performer is 0.680 (b) We are finding P (not F ) There are a total of 303 inductees and 256 of them not have any female members, so we have 256 P (not F ) = = 0.845 303 The probability that an inductee selected at random will not have any female members is 0.845 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 495 (c) In this case, we are interested only in inductees who are performers, and we want to know the probability they have female members, P (F if M P ) There are 206 performers and 38 of those have female members, so we have 38 P (F if M P ) = = 0.184 206 (d) In this case, we are interested only in inductees that not have any female members, and we want to know the probability of not being a performer, P (not M P if not F ) There are 256 inductees with no female members and 88 of them are not performers, so we have P (not M P if not F ) = 88 = 0.344 256 (e) We are finding P (M P and not F ) Of the 303 inductees, there are 168 that are performers with no female members, so we have 168 P (M P and not F ) = = 0.554 303 (f) We are finding P (not M P or F ) Of the 303 inductees, 38 + + 88 = 135 are either not performers or have female members (or both), so we have P (not M P or F ) = 135 = 0.446 303 Notice that this is the complement of the event found in part (e) P.34 (a) We are finding P (C) There are a total of 268 inductees and 233 of them were born in Canada, so we have 233 P (C) = = 0.869 268 The probability is 0.869 that an inductee selected at random is Canadian It is remarkable how much Canada has dominated the sport! (b) We are finding P (not D) There are a total of 268 inductees and 85 of them played defense while the other 183 not, so we have 183 P (not D) = = 0.683 268 The probability that an inductee selected at random will not be a defenseman is 0.683 (c) We are finding P (C and D) Of the 268 inductees, there are 74 that are defensemen born in Canada, so we have 74 P (Cand D) = = 0.276 268 (d) We are finding P (C or D) Of the 268 inductees, 233 were born in Canada and 85 are defensemen and 74 are both, so we have P (C or D) = P (C) + P (D) − P (C and D) = 233 85 74 244 + − = = 0.910 268 268 268 268 (e) In this case, we are interested only in inductees who are Canadian, and we want to know the probability that a Canadian inductee plays defense, P (D if C) There are 233 Canadians and 74 of them play defense, so we have 74 P (D if C) = = 0.318 233 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 496 (f) In this case, we are interested only in defensemen, and we want to know the probability that a defenseman inductee is also Canadian, P (C if D) There are 85 defensemen and 74 of them are Canadian, so we have 74 = 0.871 P (C if D) = 85 P.35 (a) There are 11 red ones out of a total of 80, so the probability that we pick a red one is 11/80 = 0.1375 (b) The probability that it is blue is 20/80 = 0.25 so the probability that it is not blue is − 0.25 = 0.75 (c) The single piece can be red or orange, but not both, so these are disjoint events The probability the randomly selected candy is red or orange is 11/80 + 12/80 = 23/80 = 0.2875 (d) The probability that the first one is blue is 20/80 = 0.25 When we put it back and mix them up, the probability that the next one is blue is also 0.25 By the multiplication rule, since the two selections are independent, the probability both selections are blue is 0.25 · 0.25 = 0.0625 (e) The probability that the first one is red is 11/80 Once that one is taken (since we don’t put it back and we eat it instead), there are only 79 pieces left and 11 of those are green By the multiplication rule, the probability of a red then a green is (11/80) · (11/79) = 0.191 P.36 (a) There are 18 yellow ones out of a total of 80, so the probability that we pick a yellow one is 18/80 = 0.225 (b) The probability that it is brown is 8/80 = 0.10, so the probability that it is not brown is 1−0.10 = 0.90 (c) The single piece can be blue or green, but not both, so these are disjoint events The probability the randomly selected candy is blue or green is 20/80 + 11/80 = 31/80 = 0.3875 (d) The probability that the first one is red is 11/80 = 0.1375 When we put it back and mix them up, the probability that the next one is red is also 0.1375 By the multiplication rule, since the two selections are independent, the probability both selections are red is 0.1375 · 0.1375 = 0.0189 (e) The probability that the first one is yellow is 18/80 Once that one is taken (since we don’t put it back and we eat it instead), there are only 79 pieces left and 20 of those are blue By the multiplication rule, the probability is (18/80) · (20/79) = 0.0570 P.37 Let S denote successfully making a free throw and F denote missing it (a) As free throws are independent, we can multiply the probabilities P (Makes two) = P (S1 and S2 ) = P (S1 ) · P (S2 ) = 0.908 × 0.908 = 0.824 (b) The probability of missing one free throw is P (F ) = − 0.908 = 0.092 So, P (Misses two) = P (F1 and F2 ) = P (F1 ) · P (F2 ) = 0.092 × 0.092 = 0.008 (c) He can either miss the first and make the second shot, or make the first and miss the second So, P (Makes exactly one) = P (S1 and F2 ) + P (F1 and S2 ) = 0.908 × 0.092 + 0.092 × 0.908 = 0.167 P.38 Let CBM and CBW denote the events that a man or a woman is colorblind, respectively (a) As 7% of men are colorblind, P (CBM ) = 0.07 (b) As 0.4% of women are colorblind, P (not CBW ) = − P (CBW ) = − 0.004 = 0.996 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock CHAPTER P 497 (c) The probability the woman is not colorblind is 0.996, and the probability that the man is not colorblind is − 0.07 = 0.93 As the man and woman are selected independently, we can multiply their probabilities: P (Neither is Colorblind) = P (not CBM ) · P (not CBW ) = 0.93 × 0.996 = 0.926 (d) The event that “At least one is colorblind” is the complement of part (d) that “Neither is Colorblind” so we have P (At least one is Colorblind) = − P (Neither is Colorblind) = − 0.926 = 0.074 We could also this part as P (CBM or CBW ) = P (CBM )+P (CBW )−P (CBM and CBW ) = 0.07+0.004−(0.07)(0.004) = 0.074 P.39 (a) The probability that a women is not color-blind is − 0.04 = 0.996, and the probability that a man is not colorblind is − 0.07 = 0.93 As all events are independent, we can multiply their probabilities: P (Nobody is Colorblind) = 0.99625 × 0.9315 = 0.305 (b) The event “At least one is Colorblind” is the complement of the event “Nobody is Colorblind” which has its probability computed in part (a), so P (At least one is Colorblind) = − P (Nobody is Colorblind) = − 0.305 = 0.695 (c) The probability that the randomly selected student is a man is 15/40 = 0.375 and the probability that it is a women is 25/40 = 0.625 Using the additive rule for disjoint events, P (Colorblind) = P (Colorblind Man OR Colorblind Woman) = P (Colorblind and Man) + P (Colorblind and Woman) By the multiplicative rule, P (Colorblind and Man) = P (Man) · P (Colorblind if Man) = 0.375 × 0.07 = 0.02625 Similarly, P (Colorblind and Woman) = 0.625 × 0.004 = 0.0025 So, P (Colorblind) = 0.02625 + 0.0025 = 0.029 P.40 85995 100000 = 0.860 73548 100000 = 0.735 (a) As 85,995 out of 100,000 men live to age 60, the probability is (b) As 73.548 out of 100,000 men live to age 70, the probability is − (c) As 17,429 out of 100,000 men live to age 90 and 14,493 out of 100,000 live to age 91, the probability that a man dies at age 90 is 17429−14493 = 0.0294 100000 (d) Use conditional probability: P (dies at 90 if lives till 90) = P (dies at 90 and lives till 90) P (dies at 90) 0.0294 = = = 0.169 P (lives till 90) P (lives till 90) 17429/100000 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 498 (e) Use conditional probability: P (dies at 90 if lives till 80) = P (dies at 90) 0.0294 P (dies at 90 and lives till 80) = = = 0.058 P (lives till 80) P (lives till 80) 50344/100000 (f) As 85,995 out of 100000 men live to age 60 and 17,429 out of 100000 live to age 90, the probability that a man dies between the ages of 60 and 89 is 85995−17429 = 0.686 100000 (g) Use conditional probability: P (lives till 90 if lives till 60) = P.41 P (lives till 90) 17429/100000 P (lives till 90 and lives till 60) = = = 0.203 P (lives till 60) P (lives till 60) 85995/100000 (a) The probability that the S&P 500 increased on a randomly selected day is 423/756 = 0.5595 (b) Assuming independence, the probability that the S&P 500 increases for two consecutive days is 0.5595× 0.5595 = 0.3130 (using the multiplicative rule) The probability that the S&P 500 increases on a day given that it increased the day before remains 0.5595 if the events are independent (c) The probability that the S&P 500 increases for two consecutive days is 234/755 = 0.3099 The probability that the S&P 500 increases on a day, given that it increased on the previous day is P (Increase both days) 0.3099 = = 0.5539 P (Increase first day) 0.5595 (d) The difference between the results in part (b) and part (c) is very small and insignificant, so we have little evidence that daily changes are not independent (However, since the question does not ask for a formal hypothesis test, other answers are acceptable.) P.42 If you are served one of the pancakes at random, let A be the event that the side facing you is burned and B be the event that the other side is burned We want to find P (B if A) As only one of three pancakes is burned on both sides, P (A and B) = 1/3 As out of total sides are burned, P (A) = 3/6 = 1/2 So, P (B if A) = P (A and B) 1/3 = = P (A) 1/2 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 499 Section P.2 Solutions P.43 Here is the tree with the missing probabilities filled in Note that the probabilities of all branches arising from a common point must sum to one, so P (I) + P (II) + P (III) = =⇒ P (I) = − 0.43 − 0.31 = 0.26 P (A if II) + P (B if II) = =⇒ P (A if II) = − 0.24 = 0.76 We obtain the probabilities at the end of each pair of branches with the multiplicative rule, so P (II and B) = P (II) · P (B if II) = 0.43(0.24) = 0.1032 P (III and A) = P (III) · P (A if III) = 0.31(0.80) = 0.248 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 500 P.44 Here is the tree with the missing probabilities filled in We use the conditional probability rule to find P (A if I) = 0.09 P (I and A) = = 0.5 P (I) 0.18 We use the complement rule (sum of a branches from one point equals one) to find P (B if I) = − P (A if I) = − 0.5 = 0.5 We use the multiplicative rule to find P (I and B) = P (I) · P (B if I) = 0.18(0.5) = 0.09 From the multiplicative rule for III and A we have P (III and A) = P (III) · P (A if III) =⇒ 0.268 = P (III) · 0.67 =⇒ P (III) = 0.268 = 0.4 0.67 Using the complement rule several more times P (II) = − P (I) − P (III)) = − 0.18 − 0.4 = 0.42 P (A if II) = − P (B if II) = − 0.45 = 0.55 P (B if III) = − P (A if III) = − 0.67 = 0.33 Finally, several more multiplicative rules along pairs of branches give P (II and A) = P (II) · P (A if II) = 0.42(0.55) = 0.231 P (II and B) = P (II) · P (B if II) = 0.42(0.45) = 0.189 P (III and B) = P (III) · P (B if III) = 0.40(0.33) = 0.132 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 501 P.45 Here is the tree with the missing probabilities filled in First,the sum of all the joint probabilities for all pairs of branches must be one so we have P (A and I) = − (0.225 + 0.16 + 0.45 + 0.025 + 0.025) = 0.115 Using the total probability rule we have P (I) = P (I and A) + P (I and B) + P (I and C) = 0.115 + 0.225 + 0.16 = 0.5 P (II) = P (II and A) + P (II and B) + P (II and C) = 0.45 + 0.025 + 0.025 = 0.5 The remaining six probabilities all come from the conditional probability rule For example, P (A if I) = P (A and I) 0.115 = = 0.23 P (I) 0.5 P.46 Here is the tree with the missing probabilities filled in Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 522 P.141 The plots below show the three required regions as areas on the appropriate normal curve Using technology, we can find the areas directly, and we see that the areas are as follows (If you are using a paper table, you will need to first convert the values to the standard normal using z-scores.) For additional help, see the online supplements (a) 0.691 (b) 0.202 (c) 0.643 (a) below 80 on N(75,10) (b) above 25 on N(20,6) (c) between 11 and 14 on N(12.2, 1.6) P.142 The plots below show the three required regions as areas on the appropriate normal curve Using technology, we can find the areas directly, and we see that the areas are as follows (If you are using a paper table, you will need to first convert the values to the standard normal using z-scores.) For additional help, see the online supplements (a) 0.252 (b) 0.048 (c) 0.452 (a) above on N(5,1.5) (b) below 15 on N(20,3) (c) between 90 and 100 on N(100,6) P.143 The plots below show the three required regions as areas on the appropriate normal curve Using technology, we can find the areas directly, and we see that the areas are as follows (If you are using a paper table, you will need to first convert the values to the standard normal using z-scores.) For additional help, see the online supplements (a) 0.023 (b) 0.006 (c) 0.700 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P (a) above 200 on N(120,40) 523 (b) below 49.5 on N(50,0.2) (c) between 0.8 and 1.5 on N(1,0.3) P.144 The plots below show the three required regions as areas on the appropriate normal curve Using technology, we can find the areas directly, and we see that the areas are as follows (If you are using a paper table, you will need to first convert the values to the standard normal using z-scores.) For additional help, see the online supplements (a) 0.0122 (b) 0.869 (c) 0.0807 (a) below 0.21 on N(0.3,0.04) (b) above 472 on N(500,25) (c) between and 10 on N(15,6) P.145 The plots below show the required endpoint(s) for the given normal distribution Using technology, we can find the endpoints directly, and we see that the requested endpoints are as follows (If you are using a paper table, you will need to find the endpoints on a standard normal table and then convert them back to the requested normal.) For additional help, see the online supplements (a) 59.3 (b) 2.03 (c) 60.8 and 139.2 Notice that this is very close to our rough rule that about 95% of a normal distribution is within standard deviations of the mean Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 524 (a) 0.01 above on N(50,4) (b) 0.70 below on N(2,0.05) (c) 95% between on N(100,20) P.146 The plots below show the required endpoint(s) for the given normal distribution Using technology, we can find the endpoints directly, and we see that the requested endpoints are as follows (If you are using a paper table, you will need to find the endpoints on a standard normal table and then convert them back to the requested normal.) For additional help, see the online supplements (a) 30.4 (b) 335.7 (c) 4.1 and 15.9 Notice that this is very close to our rough rule that about 95% of a normal distribution is within standard deviations of the mean (a) 0.25 above on N(25,8) (b) 0.02 below on N(500,80) (c) 95% between on N(10,3) P.147 The plots below show the required endpoint(s) for the given normal distribution Using technology, we can find the endpoints directly, and we see that the requested endpoints are as follows (If you are using a paper table, you will need to find the endpoints on a standard normal table and then convert them back to the requested normal.) For additional help, see the online supplements (a) 110 (b) 9.88 (a) 0.75 below on N(100,15) (b) 0.03 above on N(8,1) Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 525 P.148 The plots below show the required endpoint(s) for the given normal distribution Using technology, we can find the endpoints directly, and we see that the requested endpoints are as follows (If you are using a paper table, you will need to find the endpoints on a standard normal table and then convert them back to the requested normal.) For additional help, see the online supplements (a) 2.44 (b) 541 (a) 0.10 below on N(5,2) (b) 0.05 above on N(500,25) P.149 We standardize the endpoint of 40 using a mean of 48 and standard deviation of to get z= x−μ 40 − 48 = = −1.6 σ The graphs below show the lower tail region on each normal density The shaded area in both curves is 0.0548 P.150 We use technology to find the endpoint on a standard normal curve that has 30% above it The graph below shows this point is z = 0.5244 We then convert this to a N(48,5) endpoint with x = μ + z · σ = 48 + 0.5244 · = 50.62 Note: We could also use technology to find the N(48,5) endpoint directly The graphs below show the upper 30% region on each normal density Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 526 P.151 We use technology to find the endpoint on a standard normal curve that has 5% above it The graph below shows this point is z = 1.64 We then convert this to a N(10,2) endpoint with x = μ + z · σ = 10 + 1.64 · = 13.3 Note: We could also use technology to find the N(10,2) endpoint directly The graphs below show the upper 5% region on each normal density P.152 We standardize the endpoint of 13.4 using a mean of 10 and standard deviation of to get z= 13.4 − 10 x−μ = = 1.7 σ The graphs below show the upper tail region on each normal density We see that the areas are identical, and both are 0.0446 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 527 P.153 Using technology to find the endpoint for a standard normal density, we see that it is z = −1.28 (as shown in the figure) We convert this to N(500,80) using x = μ + z · σ = 500 − 1.28 · 80 = 397.6 Note: We could also use technology to find the N(500,80) endpoint directly The graphs show the lower 10% region on each normal density P.154 We convert the standard normal endpoint to N(500,80) using x = μ + z · σ = 500 + 2.1 · 80 = 668 The graphs below show the region on each normal density We see that the area is identical in both and is 0.018 P.155 We convert the standard normal endpoints to N(100,15) using x = 100 + · 15 = 115 and x = 100 + · 15 = 130 The graphs below show the region on each normal density We see that the area is identical in both and is 0.1359 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 528 P.156 We use technology to find the endpoints for a standard normal density, the 10th and 90th percentiles are z = −1.282 and z = −1.282 (as shown in the figure below) We convert these to N(100,15) using x = 100 − 1.282 · 15 = 80.77 and x = 100 + 1.282 · 15 = 119.23 Note: We could also use technology to find the N(100,15) endpoints directly The graphs below show the middle 80% region on each normal density P.157 For Critical Reading, we have a N (495, 116) curve, which will be centered at its mean, 495 We label the points that are one standard deviation away from the mean (379 and 611) and two standard deviations ways (263 and 727) so that approximately 95% of the area falls between the values two standard deviations away Those points should be out in the tails, with only about 2.5% of the distribution beyond them on each side See the figure Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 529 P.158 The plots below show the required endpoint(s) and/or probabilities for the given normal distributions Note that a percentile always means the area to the left Using technology, we can find the endpoints and areas directly, and we obtain the answers below (Alternately, we could convert to a standard normal and use the standard normal to find the equivalent area.) (a) The area below 700 is 0.961, so that point is the 96th percentile of a N(495,116) distribution (b) The point where 30% of the scores are below it is a score of 434 (a) 700 on Critical Reading (b) 30th percentile Critical Reading P.159 The plots below show the required endpoint(s) and/or probabilities for the given normal distributions Note that a percentile always means the area to the left Using technology, we can find the endpoints and areas directly, and we obtain the answers below (Alternately, we could convert to a standard normal and use the standard normal to find the equivalent area.) (a) The area below 450 is 0.384, so that point is about the 38th percentile of a N(484,115) distribution (b) The point where 90% of the scores are below it is a score of 631 (a) 450 on Writing (b) 90th percentile Writing Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 530 P.160 (a) The N (55.5, 2.7) curve is centered at its mean, 55.5 inches The points two standard deviations away from the mean on either side, 50.1 and 60.9, enclose 95% of the area These values should be near the tails, with only about 2.5% of the heights beyond them on each side See the figure (b) Using technology we find the area between 4’4” (52 inches) and 5’ (60 inches) for a N (55.5, 2.7) distribution We see that the areas is 0.8548 Alternately, we can compute z-scores z= 52 − 55.5 = −1.30 2.7 z= 60 − 55.5 = 1.67 2.7 Using technology or a table the area between −1.30 and 1.67 on a standard normal curve is 0.855, matching what we see directly About 85.5% of 10 year old boys will be between 52 and 60 inches tall Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 531 (c) Using technology we find an endpoint for a N (55.5, 2.7) distribution that has an area of 0.99 below it Alternately, we can use technology or a table to find the 99%-tile for a standard normal distribution, z = 2.326 and convert this value to the N (55.5, 2.7) scale, Height = 55.5 + 2.326 · 2.7 = 61.78 Again, of course, we arrive at the same answer using either approach A 10 year old boy who is taller than 99% of the other boys his age is 61.8 inches tall or about 5’2” P.161 (a) Using technology we find the area between 68 inches and 72 inches in a N (70, 3) distribution We see in the figure that the area is 0.495 Alternately, we can compute z-scores: z= 68 − 70 = −0.667 z= 72 − 70 = 0.667 Using technology or a table the area between −0.667 and 0.667 on a standard normal curve is 0.495, matching what we see directly About 49.5% (or almost exactly half) of US men are between 68 and 72 inches tall Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock 532 CHAPTER P (b) Using technology we find an endpoint for a N (70, 3) distribution that has an area of 0.10 below it We see in the figure that the height is about 66.2 inches Alternately, we can use technology or a table to find the 10%-tile for a standard normal distribution, z = −1.28 and convert this value to the N (70, 3) scale, Height = 70 − 1.28(3) = 66.2 Again, of course, we arrive at the same answer using either approach A US man whose height puts him at the 10th percentile is 66.2 inches tall or about 5’6” P.162 A N (0.325, 0.021) curve is centered at its mean, 0.325 The values two standard deviations away, 0.283 and 0.367, are labeled so that approximately 95% of the area falls between these two values They should be out in the tails, with only about 2.5% of the distribution beyond them on each side See the figure P.163 We use technology to find the points on a N(3.16, 0.40) curve that have 25% and 75%, respectively, of the distribution below them Note: for the upper quartile we can also look for 25% above the value In the figure below we see that the two quartiles are at Q1 = 2.89 and Q3 = 3.43 Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 533 Alternately, we can use technology or a table to find the 25% and 75% endpoints for a standard normal distribution, z = ±0.674 We then convert these endpoints to corresponding points on a N (3.16, 0.40) distribution Q1 = 3.16 − 0.674 · 0.40 = 2.89 Q3 = 3.16 + 0.674 · 0.40 = 3.43 P.164 Using technology we can find the area above 0.35 on a N (0.325, 0.021) curve directly We see in the figure that the area is 0.1169 Alternately, we can compute a z-score z= 0.35 − 0.325 = 1.19 0.021 Using technology or a table the area above 1.19 on a standard normal curve is 0.1169, the same as the area we found above Thus about 11.7% of samples of 500 US adults will contain more than 35% with at least a bachelor’s degree P.165 The plots below show the three required regions as areas in a N(21.97,0.65) distribution We see that the areas are 0.0565, 0.0012, and 0.5578, respectively Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 534 (a) more than 23 (b) less than 20 (c) between 21.5 and 22.5 If converting to a standard normal, the relevant z-scores and areas are shown below (a) z = (b) z = 23−21.97 = 0.65 20−21.97 = 0.65 21.5−21.97 0.65 1.585 The area above 1.585 for N (0, 1) is 0.0565 −3.031 The area below -3.031 for N (0, 1) is 0.0012 (c) z = = −0.7231 and z = N (0, 1) is 0.5578 P.166 22.5−21.97 0.65 = 0.8154 The area between −0.7231 and 0.8154 for (a) Here is a sketch of a normal curve with center at and standard deviation of 2.5 (b) Using technology, we can find the area above 3.0 for a N (0, 2.5) density Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 535 Alternately, we can compute a z-score z= 3−0 = 1.20 2.5 Using technology or a table the area above 1.20 on a standard normal curve is 0.1151 Thus about 11.5% of randomization samples would have slopes of 3.0 (c) We use technology to find the point on a N(0,2.5) curve that has 5% of the distribution below it This point is at a slope of −4.112 Alternately, we can use technology or a table to find the 5%-tile for a standard normal distribution, z = −1.645 We then convert this point to the corresponding slope on a N (0, 2.5) distribution, − 1.645 · 2.5 = −4.112 P.167 We use technology to determine the answers We see in the figure that the results are: (a) 0.0509 or 5.09% of students scored above a 90 (b) 0.138 or 13.8% of students scored below a 60 (c) Students with grades below 53.9 will be required to attend the extra sessions (d) Students with grades above 86.1 will receive a grade of A Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock CHAPTER P 536 P.168 We standardize the original scores using the N (62, 18) distribution z1 = 47 − 62 = −0.833 18 and z2 = 90 − 62 = 1.556 18 We then convert these z-scores to a N (75, 10) distribution x1 = 75 − 0.833 · 10 = 66.67 and x2 = 75 + 1.556 · 10 = 90.56 Rounding to integers the new exam scores would be 67 and 91 P.169 (a) For any N (μ, σ) distribution, when we standardize μ − 2σ and μ + 2σ we must get z = −2 and z = +2 For example, if we use N (100, 20) the interval within two standard deviations of the mean goes from 60 to 140 z= 60 − 100 = −2 20 and z= 140 − 100 = +2 20 Using technology, the area between −2 and +2 on a standard normal curve is 0.954 (b) Similar to part (a), if we go just one standard deviation in either direction the standardized z-scores will be z = −1 and z = +1 Using technology, the area between −1 and +1 on a standard normal curve is 0.683 (c) Similar to part (a), if we go three standard deviations in either direction the standardized z-scores will be z = −3 and z = +3 Using technology, the area between −3 and +3 on a standard normal curve is 0.997 (d) The percentages within one, two, or three standard deviations of the mean, roughly 68% between μ ± σ, roughly 95% between μ ± 2σ, and roughly 99.7% between μ ± 3σ, should hold for any normal distribution since the standardized z-scores will always be z = ±1 or z = ±2 or z = ±3, respectively Full file at https://TestbankDirect.eu/Solution-Manual-for-Statistics-2nd-Edition-by-Lock ... https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock. .. https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock. .. https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock Solution Manual for Statistics 2nd Edition by Lock Full file at https://TestbankDirect.eu /Solution- Manual- for- Statistics- 2nd- Edition- by- Lock