Sampling Distribution of mean

(a) α = 1 - the sum of all the other numbers
= 0.2
(b) f(100) = 0.02 + 0.06 + 0.02 = 0.1
f(110) = 0.09 + 0.15 + 0.06 = 0.3
f(120) = 0.09 + 0.2 + 0.11 = 0.4
f(130) = 0.01 + 0.05 + 0.04 = 0.1
f(140) = 0.02 + 0.03 + 0.05 = 0.1
(c) Expected weight = 100(0.1) + 110(0.3) + 120(0.4) + 130(0.1) + 140(0.1)
= 118 lb
(d) Variance is 100^2(0.1) + 110^2(0.3) + 120^2(0.4) + 130^2(0.1) + 140^2(0.1) - 118^2
= 116
Standard deviation = √116 lb
(e) The sample mean for 36 students is 118 lb
The standard deviation of the sample mean = √116 / √36 = 1.795 lb
Probability of total weight is at most 4350 lb
= Probability of mean weight is at most 4350/36 = 120.833 lb
= Pr(X < 120.833)
= Pr[Z < (120.833 - 118)/1.795]
= Pr(Z < 1.578)
= 0.9428

2009-11-29 01:10:04 補充：
Yes, I am assuming normal distribution following central limit theorem. I think 36 is sufficiently large to apply this assumption to get approximate number.