It would help if you typed the questions with your solutions. It is a lot of trouble to switch between the browser and image file viewer windows.
In would also help if each of the questions and solutions were in seperate thresds.
CB
Dear
please find the attached file of questions.
please check my working .
i will be very thankful to you.
Ans 2 b)
there are 5C3 (because order does not matter)ways of having a group of 3 from this finite population. So there are 10 groups and you have to find the mean of each of the 10 groups.
6 6 9->7
6 6 15->9
6 6 18->10
6 9 15->10
6 9 15->10
6 15 18->13
6 15 18->13
6 9 18->11
6 9 18->11
9 15 18->14
Then it is asking you to find the average of the averages that you have found.
(7+9+10+10+10+13+13+11+11+14)/10=10.8
The variance is
[(7-10.8)2+(9-10.8)2+(10-10.8)2+(10-10.8)2+(10-10.8)2+ (13-10.8)2+(13-10.8)2+(11-10.8)2+(11-10.8)2+(14-10.8)2]/(10-1)=4.4
Ans 3 c)
estimator with min. standard error that is variance of T is better estimator.
variance of T1=V(X1+X2+ X3 /3)
= V( X1 /3) +V(X2/3) + V(X3 /3)
= 1/9 V(X1) + 1/9 V(X2)+ 1/9V(X3) = 1/3 σ2
SIMILARLY FIND v(T2)
v(T2)=3/8σ2
as variance of T1 is less,it is better estimator
Ans 1 .b)
If all 5 are selected without replacement, there is only one way in which all 5 have tomatoes, C(5,5) = 1. Total number of ways of selecting 5 cans from 10 = C(10,5) = 180
Hence, required probability of selecting all tomatoes = 1/180 ≈ 0. 005556
[IMG]file:///C:/DOCUME%7E1/KHYBER%7E1/LOCALS%7E1/Temp/msohtml1/01/clip_image001.gif[/IMG]
The total number of ways of selecting 3 or more tomatoes is:
C(5,3)*C(5,2) + C(5,4)*C(5,1) + C(5,5) = 126
Total number of ways of selecting 5 cans from 10 = C(10,5) = 180
Hence, required probability of selecting 3 or more tomatoes = 126/180 = 0.7