5.1: The possible values of p-hat are 0, 1/3, 2/3, and 1. These correspond to getting 0 persons with lung cancer, 1 with lung cancer, and all 3 with lung cancer.
5.3(a): Here p = .39 and n = 5.
(i) Pr{p-hat = 0.0} = Pr{Y = 0} = 5C0*(0.39)^0(0.61)^5 = 0.084460 = 8.4%
(ii) Pr{p-hat = 0.2} = Pr{Y = 1} = 5C1*(0.39)^1(0.69)^4 = 0.269994 = 27.0%
(iii) Pr{p-hat = 0.4} = Pr{Y = 2} = 5C2*(0.39)^2(0.69)^3 = 0.345238 = 34.5%
(iv) Pr{p-hat = 0.6} = Pr{Y = 3} = 5C3*(0.39)^3(0.69)^2 = 0.220726 = 22.1%
(iii) Pr{p-hat = 0.8} = Pr{Y = 4} = 5C4*(0.39)^4(0.69)^1 = 0.079560 = 8.0%
(iii) Pr{p-hat = 1.0} = Pr{Y = 5} = 5C5*(0.39)^5(0.69)^0 = 0.009022 = 0.1%
5.18: Y = corn plant height ~ N(mu = 145cm, sigma = 22cm)
(a) Pr{135 < Y < 155} = Pr{-0.45 < Z < 0.45} since (155 - 145) / 22 = 0.45
= 0.6736 - 0.3264 = 0.3472 (34.72%)
(b) for n = 16, Pr{135 < Y-bar < 155} = Pr{-1.82 < Z < 1.82} [since
(155-145)/(22/sqrt(16)) = 1.82]
= 0.9656 - 0.0344 = 0.9312 (93.12%)
(c) "percent" is related to "probability", so answer here is same as in (b) (93.12%)
(d) for n = 36, Pr{135 < Y-bar < 155} = Pr{-2.73 < Z < 2.73} [since (155-145)/(22/sqrt(36)) = 2.73]
= 0.9968 - 0.0032 = 0.9936 (99.36%)
5.39:
(a)
Let Y = number of heads.
Pr(Y
= 3) = (120)(5^3)(1-.5)^7 = 0.1172,
Pr(Y
= 4) = (210)(5^4)(1-.5)^6 = 0.2051
Pr(
3 <= Y <= 4) = Pr{Y = 3) + Pr(Y = 4) = 0.1172 + 0.2051 = 0.3223
(b)
normal approximation: mean = 10 * .5 = 5 and SD = sqrt(10 * .5 * .5) = 1.581.
Pr{3
<= Y <= 4} = Pr{2.5 < Y < 4.5} = Pr{(2.5 - 5)/1.581 < Z <
(4.5 - 5)/1.581} = Pr{-1.58 < Z < -.316} = .3745 - .0571 = .3174
5.49: For the normal approximation to the sampling distribution of the sample proportion p-hat, the mean is p = 0.42 and the SD is sqrt[p(1-p)/n] = sqrt[(.42)(.58)/25] = 0.0987. However, it easier to perform a continuity correction by converting to the sample count Y.
For the normal approximation to the sampling distribution of the sample count Y, the mean is np = (25)(0.42) = 10.5 and the SD is sqrt[np(1-p)] = sqrt[(25)(.42)(.58)] = 2.4678.
Pr(p-hat > 0.44) = Pr(Y > (25)(0.44)) = Pr(Y > 11) =(c.c.) Pr(Y > 11.5) = Pr(z > (11.5 – 10.5)/2/4678) = Pr(z > 0.4052). Table 3 gives 0.6590. 1 – 0.6590 = 0.3410.