# check my working

#### samileo31

[FONT=&quot]Dear
[/FONT]
[FONT=&quot]please find the attached file of questions.
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[FONT=&quot]i will be very thankful to you.
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[FONT=&quot]
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[FONT=&quot]Ans 2 b)[/FONT]
[FONT=&quot]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[/FONT]
[FONT=&quot]The variance is[/FONT]
[FONT=&quot][(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[/FONT]

[FONT=&quot]Ans 3 c)[/FONT]
[FONT=&quot]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)[/FONT]
[FONT=&quot]= 1/9 V(X1) + 1/9 V(X2)+ 1/9V(X3) = 1/3 σ2[/FONT]
[FONT=&quot]SIMILARLY FIND v(T2)[/FONT]
[FONT=&quot]v(T2)=3/8σ2[/FONT]
[FONT=&quot]as variance of T1 is less,it is better estimator[/FONT]

[FONT=&quot]Ans 1 .b)[/FONT]
[FONT=&quot]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 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[/FONT]
[FONT=&quot]Total number of ways of selecting 5 cans from 10 = C(10,5) = 180[/FONT]
[FONT=&quot]Hence, required probability of selecting 3 or more tomatoes = 126/180 = 0.7[/FONT]
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#### CaptainBlack

MHF Hall of Fame
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

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