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# Solucionario Walpole 8 ED

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```prospects for bankruptcy.
Also, the probability that the first success appears in the 11th drill is pq10 = 0.014
which is even smaller.
5.95 It is a negative binomial distribution.
(
x\u22121
k\u22121
)
pkqx\u2212k =
(
6\u22121
2\u22121
)
(0.25)2(0.75)4 = 0.0989.
5.96 It is a negative binomial distribution.
(
x\u22121
k\u22121
)
pkqx\u2212k =
(
4\u22121
2\u22121
)
(0.5)2(0.5)2 = 0.1875.
5.97 n = 1000 and p = 0.01, with µ = (1000)(0.01) = 10. P (X < 7) = P (X \u2264 6) = 0.1301.
5.98 n = 500;
(a) If p = 0.01,
P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212
14\u2211
x=0
(
500
x
)
(0.01)x(0.99)500\u2212x = 0.00021.
This is a very rare probability and thus the original claim that p = 0.01 is ques-
tionable.
(b) P (X = 3) =
(
500
3
)
(0.01)3(0.99)497 = 0.1402.
(c) For (a), if p = 0.01, µ = (500)(0.01) = 5. So
P (X \u2265 15) = 1\u2212 P (X \u2264 14) = 1\u2212 0.9998 = 0.0002.
For (b),
P (X = 3) = 0.2650\u2212 0.1247 = 0.1403.
5.99 N = 50 and n = 10.
70 Chapter 5 Some Discrete Probability Distributions
(a) k = 2; P (X \u2265 1) = 1\u2212 P (X = 0) = 1\u2212 (
2
0)(
48
10)
(5010)
= 1\u2212 0.6367 = 0.3633.
(b) Even though the lot contains 2 defectives, the probability of reject the lot is not
very high. Perhaps more items should be sampled.
(c) µ = (10)(2/50) = 0.4.
5.100 Suppose n items need to be sampled. P (X \u2265 1) = 1\u2212 (
2
0)(
48
n )
(50n )
= 1\u2212 (50\u2212n)(49\u2212n)
(50)(49)
\u2265 0.9.
The solution is n = 34.
5.101 Define X = number of screens will detect. Then X \u223c b(x; 3, 0.8).
(a) P (X = 0) = (1\u2212 0.8)3 = 0.008.
(b) P (X = 1) = (3)(0.2)2(0.8) = 0.096.
(c) P (X \u2265 2) = P (X = 2) + P (X = 3) = (3)(0.8)2(0.2) + (0.8)3 = 0.896.
5.102 (a) P (X = 0) = (1\u2212 0.8)n \u2264 0.0001 implies that n \u2265 6.
(b) (1\u2212 p)3 \u2264 0.0001 implies p \u2265 0.9536.
5.103 n = 10 and p = 2
50
= 0.04.
P (X \u2265 1) = 1\u2212 P (X = 0) \u2248 1\u2212
(
10
0
)
(0.04)0(1\u2212 0.04)10 = 1\u2212 0.6648 = 0.3351.
The approximation is not that good due to n
N
= 0.2 is too large.
5.104 (a) P =
(22)(
3
0)
(52)
= 0.1.
(b) P =
(21)(
1
1)
(52)
= 0.2.
5.105 n = 200 with p = 0.00001.
(a) P (X \u2265 5) = 1\u2212 P (X \u2264 4) = 1\u2212
4\u2211
x=0
(
200
x
)
(0.00001)x(1\u2212 0.00001)200\u2212x \u2248 0. This
is a rare event. Therefore, the claim does not seem right.
(b) µ = np = (200)(0.00001) = 0.02. Using Poisson approximation,
P (X \u2265 5) = 1\u2212 P (X \u2264 4) \u2248 1\u2212
4\u2211
x=0
e\u22120.02
(0.02)x
x!
= 0.
Chapter 6
Some Continuous Probability
Distributions
6.1 (a) Area=0.9236.
(b) Area=1\u2212 0.1867 = 0.8133.
(c) Area=0.2578\u2212 0.0154 = 0.2424.
(d) Area=0.0823.
(e) Area=1\u2212 0.9750 = 0.0250.
(f) Area=0.9591\u2212 0.3156 = 0.6435.
6.2 (a) The area to the left of z is 1\u2212 0.3622 = 0.6378 which is closer to the tabled value
0.6368 than to 0.6406. Therefore, we choose z = 0.35.
(b) From Table A.3, z = \u22121.21.
(c) The total area to the left of z is 0.5000+0.4838=0.9838. Therefore, from Table
A.3, z = 2.14.
(d) The distribution contains an area of 0.025 to the left of \u2212z and therefore a total
area of 0.025+0.95=0.975 to the left of z. From Table A.3, z = 1.96.
6.3 (a) From Table A.3, k = \u22121.72.
(b) Since P (Z > k) = 0.2946, then P (Z < k) = 0.7054/ From Table A.3, we find
k = 0.54.
(c) The area to the left of z = \u22120.93 is found from Table A.3 to be 0.1762. Therefore,
the total area to the left of k is 0.1762+0.7235=0.8997, and hence k = 1.28.
6.4 (a) z = (17\u2212 30)/6 = \u22122.17. Area=1\u2212 0.0150 = 0.9850.
(b) z = (22\u2212 30)/6 = \u22121.33. Area=0.0918.
(c) z1 = (32\u22123)/6 = 0.33, z2 = (41\u221230)/6 = 1.83. Area = 0.9664\u22120.6293 = 0.3371.
(d) z = 0.84. Therefore, x = 30 + (6)(0.84) = 35.04.
71
72 Chapter 6 Some Continuous Probability Distributions
(e) z1 = \u22121.15, z2 = 1.15. Therefore, x1 = 30 + (6)(\u22121.15) = 23.1 and x2 =
30 + (6)(1.15) = 36.9.
6.5 (a) z = (15\u2212 18)/2.5 = \u22121.2; P (X < 15) = P (Z < \u22121.2) = 0.1151.
(b) z = \u22120.76, k = (2.5)(\u22120.76) + 18 = 16.1.
(c) z = 0.91, k = (2.5)(0.91) + 18 = 20.275.
(d) z1 = (17\u2212 18)/2.5 = \u22120.4, z2 = (21\u2212 18)/2.5 = 1.2;
P (17 < X < 21) = P (\u22120.4 < Z < 1.2) = 0.8849\u2212 0.3446 = 0.5403.
6.6 z1 = [(µ\u2212 3\u3c3)\u2212 µ]/\u3c3 = \u22123, z2 = [(µ+ 3\u3c3)\u2212 µ]/\u3c3 = 3;
P (µ\u2212 3\u3c3 < Z < µ+ 3\u3c3) = P (\u22123 < Z < 3) = 0.9987\u2212 0.0013 = 0.9974.
6.7 (a) z = (32\u2212 40)/6.3 = \u22121.27; P (X > 32) = P (Z > \u22121.27) = 1\u2212 0.1020 = 0.8980.
(b) z = (28\u2212 40)/6.3 = \u22121.90, P (X < 28) = P (Z < \u22121.90) = 0.0287.
(c) z1 = (37\u2212 40)/6.3 = \u22120.48, z2 = (49\u2212 40)/6.3 = 1.43;
So, P (37 < X < 49) = P (\u22120.48 < Z < 1.43) = 0.9236\u2212 0.3156 = 0.6080.
6.8 (a) z = (31.7\u2212 30)/2 = 0.85; P (X > 31.7) = P (Z > 0.85) = 0.1977.
Therefore, 19.77% of the loaves are longer than 31.7 centimeters.
(b) z1 = (29.3\u2212 30)/2 = \u22120.35, z2 = (33.5\u2212 30)/2 = 1.75;
P (29.3 < X < 33.5) = P (\u22120.35 < Z < 1.75) = 0.9599\u2212 0.3632 = 0.5967.
Therefore, 59.67% of the loaves are between 29.3 and 33.5 centimeters in length.
(c) z = (25.5\u2212 30)/2 = \u22122.25; P (X < 25.5) = P (Z < \u22122.25) = 0.0122.
Therefore, 1.22% of the loaves are shorter than 25.5 centimeters in length.
6.9 (a) z = (224 \u2212 200)/15 = 1.6. Fraction of the cups containing more than 224 mil-
limeters is P (Z > 1.6) = 0.0548.
(b) z1 = (191\u2212 200)/15 = \u22120.6, Z2 = (209\u2212 200)/15 = 0.6;
P (191 < X < 209) = P (\u22120.6 < Z < 0.6) = 0.7257\u2212 0.2743 = 0.4514.
(c) z = (230 \u2212 200)/15 = 2.0; P (X > 230) = P (Z > 2.0) = 0.0228. Therefore,
(1000)(0.0228) = 22.8 or approximately 23 cups will overflow.
(d) z = \u22120.67, x = (15)(\u22120.67) + 200 = 189.95 millimeters.
6.10 (a) z = (10.075\u2212 10.000)/0.03 = 2.5; P (X > 10.075) = P (Z > 2.5) = 0.0062.
Therefore, 0.62% of the rings have inside diameters exceeding 10.075 cm.
(b) z1 = (9.97\u2212 10)/0.03 = \u22121.0, z2 = (10.03\u2212 10)/0.03 = 1.0;
P (9.97 < X < 10.03) = P (\u22121.0 < Z < 1.0) = 0.8413\u2212 0.1587 = 0.6826.
(c) z = \u22121.04, x = 10 + (0.03)(\u22121.04) = 9.969 cm.
6.11 (a) z = (30\u2212 24)/3.8 = 1.58; P (X > 30) = P (Z > 1.58) = 0.0571.
(b) z = (15 \u2212 24)/3.8 = \u22122.37; P (X > 15) = P (Z > \u22122.37) = 0.9911. He is late
99.11% of the time.
Solutions for Exercises in Chapter 6 73
(c) z = (25\u2212 24)/3.8 = 0.26; P (X > 25) = P (Z > 0.26) = 0.3974.
(d) z = 1.04, x = (3.8)(1.04) + 24 = 27.952 minutes.
(e) Using the binomial distribution with p = 0.0571, we get
b(2; 3, 0.0571) =
(
3
2
)
(0.0571)2(0.9429) = 0.0092.
6.12 µ = 99.61 and \u3c3 = 0.08.
(a) P (99.5 < X < 99.7) = P (\u22121.375 < Z < 1.125) = 0.8697\u2212 0.08455 = 0.7852.
(b) P (Z > 1.645) = 0.05; x = (1.645)(0.08) + 99.61 = 99.74.
6.13 z = \u22121.88, x = (2)(\u22121.88) + 10 = 6.24 years.
6.14 (a) z = (159.75 \u2212 174.5)/6.9 = \u22122.14; P (X < 159.75) = P (Z < \u22122.14) = 0.0162.
Therefore, (1000)(0.0162) = 16 students.
(b) z1 = (171.25\u2212 174.5)/6.9 = \u22120.47, z2 = (182.25\u2212 174.5)/6.9 = 1.12.
P (171.25 < X < 182.25) = P (\u22120.47 < Z < 1.12) = 0.8686\u2212 0.3192 = 0.5494.
Therefore, (1000)(0.5494) = 549 students.
(c) z1 = (174.75\u2212 174.5)/6.9 = 0.04, z2 = (175.25\u2212 174.5)/6.9 = 0.11.
P (174.75 < X < 175.25) = P (0.04 < Z < 0.11) = 0.5438\u2212 0.5160 = 0.0278.
Therefore, (1000)(0.0278)=28 students.
(d) z = (187.75\u2212 174.5)/6.9 = 1.92; P (X > 187.75) = P (Z > 1.92) = 0.0274.
Therefore, (1000)(0.0274) = 27 students.
6.15 µ = \$15.90 and \u3c3 = \$1.50.
(a) 51%, since P (13.75 < X < 16.22) = P
(
13.745\u221215.9
1.5
< Z < 16.225\u221215.9
1.5
)
= P (\u22121.437 < Z < 0.217) = 0.5871\u2212 0.0749 = 0.5122.
(b) \$18.36, since P (Z > 1.645) = 0.05; x = (1.645)(1.50) + 15.90 + 0.005 = 18.37.
6.16 (a) z = (9.55 \u2212 8)/0.9 = 1.72. Fraction of poodles weighing over 9.5 kilograms =
P (X > 9.55) = P (Z > 1.72) = 0.0427.
(b) z = (8.65\u2212 8)/0.9 = 0.72. Fraction of poodles weighing at most 8.6 kilograms =
P (X < 8.65) = P (Z < 0.72) = 0.7642.
(c) z1 = (7.25\u2212 8)/0.9 = \u22120.83 and z2 = (9.15\u2212 8)/0.9 = 1.28.
Fraction of poodles weighing between 7.3 and 9.1 kilograms inclusive
= P (7.25 < X < 9.15) = P (\u22120.83 < Z < 1.28) = 0.8997\u2212 0.2033 = 0.6964.
6.17 (a) z = (10, 175\u2212 10, 000)/100 = 1.75. Proportion of components exceeding 10.150
kilograms in tensile strength= P (X > 10, 175) = P (Z > 1.75) = 0.0401.
(b) z1 = (9, 775 \u2212 10, 000)/100 = \u22122.25```