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Section 8.8 This is a section in which we approximate a binomial distribution using a normal distribution. First we must “judge” whether the approximation is any good…and that is done by seeing whether BOTH of the following are true: - 3 > 0 and + 3 < 0 (remembering that the mean, , is equal to n*p and that the standard deviation, , is equal to the square root of n*p*q). Problem #232 n=10 and p=0.5, so = 10*0.5 = 5 and =10*0.5*0.5 = 1.5811 so - 3 = 5 – 3*1.5811 = 0.2565 which IS greater than zero, and + 3 = 5 + 3*1.5811 = 9.7434 which IS less than 10 (n), so this is a good approximation Problem #233 n=20 and p=0.4, so = 20*0.4 = 8 and =20*0.4*0.6 = 2.1909 so - 3 = 8 – 3*2.1909= 1.4273 which IS greater than zero, and + 3 = 8 + 3*2.1909= 14.5727 which IS less than 20 (n), so this is a good approximation Problem #234 n=30 and p=0.2, so = 30*0.2 = 6 and =30*0.2*0.8 = 2.1909 so - 3 = 6 – 3*2.1909= -0.5727 which IS NOT greater than zero, and + 3 = 6 + 3*2.1909= 12.5727 which IS less than 30 (n), so this is NOT a good approximation Problem #235 n=40 and p=0.1, so = 40*0.1 = 4 and =40*0.1*0.9 = 1.8974 so - 3 = 4 – 3*1.8974= -1.6921 which IS NOT greater than zero, and + 3 = 4 + 3*1.8974= 9.6921 which IS less than 40 (n), so this is NOT a good approximation Problems #236-239 For these problems, n=20 and p=0.4, so we will be using = 20*0.4 = 8 and =20*0.4*0.6 = 2.1909 Problem #236 The below picture describes the problem… =8 9 Since this is centered on 9, we must go up a half unit and down a half unit for the normal distribution, so from 8.5 to 9.5. Using the same processes that we did in section 8.7, we first get the z-scores: z8.5= (8.5-8)/2.1909 = 0.23 and z9.5= (9.5-8)/2.1909 = 0.68 …now from the table we get (respectively): 0.0910 and 0.2517; finally we subtract the two probabilities since they are on the same side of the mean, giving us: 0.2517 – 0.0910 = 0.1607 Using the table for the exact answer, you get 0.755 -0.596 = 0.169, which is an OK approximation, but not great Problem #237 The below picture describes the problem… =8 12 Since the edge of the shaded area is 12, we must go up a half unit away from the shaded area for the normal distribution, so 11.5. Using the same processes that we did in section 8.7, we first get the z-scores: z11.5= (11.5-8)/2.1909 = 1.60 …now from the table we get: 0.4452; finally we subtract that from 0.5 since it is a “tail” problem: 0.5 – 0.4452 = 0.0548 Using the table for the exact answer, you get 1 - 0.943 = 0.057, which is a pretty good approximation. Problem #238 The below picture describes the problem… 4 =8 9 Since there are two edges to the shaded area, you move a half unit away from the shaded area, for the normal distribution, so from 3.5 to 9.5. Using the same processes that we did in section 8.7, we first get the z-scores: z3.5= (3.5-8)/2.1909 = -2.05 and z9.5= (9.5-8)/2.1909 = 0.68 …now from the table we get (respectively): 0.4798 and 0.2517; finally we add the two probabilities since they are on the opposite sides of the mean, giving us: 0.2517 + 0.4798 = 0.7315 Using the table for the exact answer, you get 0.755 -0.016 = 0.739, which is an OK approximation, but not great Problem #239 The below picture describes the problem…and the reason for 7 as an end point is that it is the first number greater than 6… 7 =8 11 Since there are two edges to the shaded area, you move a half unit away from the shaded area, for the normal distribution, so from 6.5 to 11.5. Using the same processes that we did in section 8.7, we first get the z-scores: z6.5= (6.5-8)/2.1909 = -1.14 and z11.5= (11.5-8)/2.1909 = 1.60 …now from the table we get (respectively): 0.3729 and 0.4452; finally we add the two probabilities since they are on the opposite sides of the mean, giving us: 0.3729 + 0.4452 = 0.8181 Using the table for the exact answer, you get 0.943 - 0.250 = 0.687, which is NOT a good approximation Problem #240 “…fair coin…” means that p equals 0.5; and “…tossed 12 times…” means that n equals 12… (a) using the binomial equation: P(3) = C(12,3)*(0.5)3*(0.5)9 = 0.0537 (b) so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 12*0.5 = 6 and =12*0.5*0.5 = 1.7321 3 =6 Since this is centered on 3, we must go up a half unit and down a half unit for the normal distribution, so from 2.5 to 3.5. Using the same processes that we did in section 8.7, we first get the z-scores: z2.5= (2.5-6)/1.7321 = -2.02 and z3.5= (3.5-6)/1.7321 = -1.44 …now from the table we get (respectively): 0.4783 and 0.4251; finally we subtract the two probabilities since they are on the same side of the mean, giving us: 0.4783 – 0.4251 = 0.0532 Problem #241 “…In 84 tosses…” means that n equals 84; and “…points up about 30% of the time…” means that p equals 0.3… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 84*0.3 = 25.2 and =84*0.3*0.7 = 4.2 18 25 Since there are two edges to the shaded area, you move a half unit away from the shaded area, for the normal distribution, so from 17.5 to 25.5. Using the same processes that we did in section 8.7, we first get the z-scores: z17.5= (17.5-25.2)/4.2 = -1.83 and z25.5= (25.5-25.2)/4.2 = 0.07 …now from the table we get (respectively): 0.4664 and 0.0279; finally we add the two probabilities since they are on the opposite sides of the mean, giving us: 0.4664 + 0.0279 = 0.4943 (which is only off by 0.0116) Problem #242 “…6%...is color-blind…” means that p equals 0.06; and “…random sample of 250…” means that n equals 250… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 250*0.06 = 15 and =250*0.06*0.94 = 3.755 10 15 20 Since there are two edges to the shaded area, you move a half unit away from the shaded area, for the normal distribution, so from 9.5 to 20.5. Using the same processes that we did in section 8.7, we first get the z-scores: z9.5= (9.5-15)/3.755 = -1.46 and z20.5= (20.5-15)/3.755 = 1.46 …now from the table we get (respectively): 0.4279 and 0.4279; finally we add the two probabilities since they are on the opposite sides of the mean, giving us: 0.4279 + 0.4279 = 0.8558 Problem #243 “…25%...study…” means that p equals 0.25; and “…random sample of 48…” means that n equals 48… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 48*0.25 = 12 and =48*0.25*0.75 = 3 12 15 Fifteen is the first number less than 16 (which is one third of 48)… You move a half unit away from the shaded area, for the normal distribution, so 15.5. Using the same processes that we did in section 8.7, we first get the z-scores: z15.5= (15.5-12)/3= 1.17 …now from the table we get: 0.3790; finally we add the two probabilities since they are on the opposite sides of the mean, giving us: 0.5 + 0.3790 = 0.8790 Problem #244 “…60%...favors…” , which means that 40% oppose, so p equals 0.40; and “…random sample of 96…” means that n equals 96… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 96*0.40 = 38.4 and =96*0.40*0.60 = 4.8 38.4 49 49 is the first number greater than half of 96 (which is majority)… You move a half unit away from the shaded area, for the normal distribution, so 48.5. Using the same processes that we did in section 8.7, we first get the z-scores: z45.5= (45.5-38.4)/4.8= 2.10 …now from the table we get: 0.4821; finally since this is “tail” problem, we subtract the probability from 0.5: 0.5 - 0.4821 = 0.0179 Problem #245 “…in 60 trials…” , which means that n equals 60; and “…does not prefer either shape…” means that p equals 0.5… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 60*0.50 = 30 and =60*0.50*0.50 = 3.873 = 30 40 Since the edge of the shaded area is 40, we must go up a half unit away from the shaded area for the normal distribution, so 39.5. Using the same processes that we did in section 8.7, we first get the z-scores: z39.5= (39.5-30)/3.873 = 2.45 …now from the table we get: 0.4929; finally we subtract that from 0.5 since it is a “tail” problem: 0.5 – 0.4929 = 0.0071 Problem #246 This is a game problem…so I start with my standard matrix and fill it in… Event 15+ heads 15+ tails $ $20 -$10 P($) $*P($) The probabilities will have to calculated using a normal approximation… “…tossed 25 times…” , which means that n equals 25; and since nothing is stated about the coin’s fairness, it must be assumed to be fair, which means that p equals 0.5… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 25*0.50 = 12.5 and =25*0.50*0.50 = 2.5 = 12.5 15 Since the edge of the shaded area is 15, we must go up a half unit away from the shaded area for the normal distribution, so 14.5. Using the same processes that we did in section 8.7, we first get the z-scores: z14.5= (14.5-12.5)/2.5 = 0.80 …now from the table we get: 0.2881; finally we subtract that from 0.5 since it is a “tail” problem: 0.5 – 0.2881 = 0.2119 (which is the probability for 15 or more heads or 15 or more tails, since we assumed a fair coin)…so the game table becomes: Event $ P($) $*P($) 15+ heads $20 0.2119 $4.24 15+ tails -$10 0.2119 -$2.12 So the expected value of the game is $4.24 + (-$2.12) = $2.12 Problem #247 “…50 question multiple-choice test…” , which means that n equals 50; and “…each question having 5 possible answer choices (only one which is correct)…” means that p equals 1/5 or 0.2… so for the normal approximation, that gives us the following mean and standard deviation for this problem: = 50*0.20 = 10 and =50*0.20*0.80 = 2.8284 = 10 30 Since the edge of the shaded area is 30, we must go up a half unit away from the shaded area for the normal distribution, so 29.5. Using the same processes that we did in section 8.7, we first get the z-scores: z29.5= (29.5-10)/2.8284 = 6.89 …now from the table we get: 0.5; finally we subtract that from 0.5 since it is a “tail” problem: 0.5 – 0.5 = 0 (not possible)