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please find the attached file of questions.
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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
[(710.8)2+(910.8)2+(1010.8)2+(1010.8)2+(1010.8)2+ (1310.8)2+(1310.8)2+(1110.8)2+(1110.8)2+(1410.8)2]/(101)=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

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