Solution to HW # 5
Ex 5.2 (a): = 0.5*0.333+5*0.167+7*.375+11*.125 = 5
(0.55)2(0.333) + (55)2(0.167) + (75)2(0.375) + (115)2(0.125) = 12.74 The median can be located anywhere between 5 and 7.
(b) = (0.55)3(0.333) + (55)3(0.167) + (75)3(0.375) + (11
From Text book 1. Exercise. 6.3 (a), (b) (Use JMP to generate samples)
Solution. In general, the agreement between the theoretical distributions and the simulated samples will improve with an increase in the sample size. However, the random variability wi
HW Chp#7 7.1. (a) For N=40000, we may ignore the effect of the finite population correction factor., since it is close enough to 1(Remember, as a matter of convenience, the correction factor is usually omitted if N>20n). First, population variance is 4*=
HW # 7 8.2. We have, n = 40 p = 14/40 = .35 alpha = 0.5 To find the confidence interval, use the formula for finding confidence interval on proportions (8.10)
which gives a Confidence Interval (0.2021, .4978) Alternate and preferred method: Use JMP. The g
HW # 8
From Chapter 9 Ex. 9.11 Here, we have to perform a hypothesis test about a single proportion. The null hypothesis is H0: p = 0.5 The alternate hypothesis is HA: p>0.5. Since the sample size is 50% of the population i.e. 500, it is sufficiently larg
ME335Engineering StatisticsFall 2008 HW # 3 Solution
4.3 (a)
x
x
x
f ( x)dx x 1,
x ke
0
x

x 4
dx x 1 , 4k x 1 , k=1/4
x 1 1 4 e dx = e  2  e 2 =0.4712 4
(b)Pcfw_2<X<8=
x
2 x
8
f ( x)dx =
8 2
x 7  1 4 f (c) Pcfw_X>7= ( x )dx = e dx = e 4 = 0.173