Assignment 2
1． The amount of bleach a machine pours into bottles has a mean of 36 oz. with a
standard deviation of 0.15 oz. Suppose we take a random sample of 36 bottles filled by this machine. What is the probability that the mean of the sample is greater than 35.94 oz?
4.236
15
.03694.35-=-=-=x x
X Z σμ )4.2()4.2(Φ=->Φx The area corresponding to Z=2.4 in Table E.2 is 0.9918
Because the probability that the mean of the sample is greater than 35.94 oz,so the value is 0.9918.
2. Page 186-- Answer questions in 7.47 (b),(c) (b)60.150002
.098.098.099.0)1(60.150002
.098.098.097.0)1(2211=⨯-=--=-=⨯-=--=
n p Z n p Z ππππππ 9452.0)60.1(=Φ
1-2(1-0.9452)=0.8904
Using Table E.2,the area between 97% and 99% with Internet access is 0.8904,only 89.04% of the samples of n=500 would be expected to have sample between 97% and 99% with Internet access. (c))60.1()60.1(60.1500
02.098.098.097.0)1(11>Φ=-<Φ-=⨯-=--=
x x n p Z πππ 9452.0)60.1(=Φ
1-0.9452=0.0548
Using Table E.2,the area less than 97% with Internet access is 0.0548,only
5.48% of the samples of n=500 would be expected to have sample less than 97% with Internet access.
3． Page 216-- Answer questions in 8.54 and Interpret the confidence interval in (a)
and (b).
(a) Use the t-value because of n < 30
n S t n S t 22X X ααμ+≤≤- Absenteeism:days S days X 0.4,7.9==
degrees of freedom ：25-1=24
025.02=α From Table E.3,you can see that 0639.22=αt 65112.17.925
4)0639.2(7.9±=± The construct a 95% confidence interval estimate of the mean number of absences for clerical workers during the year is 351.11049.8≤≤μ.
(b)
confidence interval estimate for the proportion :
n p p Z p n p p Z p )1()1(22-+≤≤--ααπ
48.02512==p 96.12=αZ for 95% confidence 19584.048.02552.048.0)
96.1(48.0±=⨯± The construct a 95% confidence interval estimate of the population proportion of clerical workers absent more than 10 days during the year is 676.0284.0≤≤π.。