[FONT="]Dear
[/FONT]
[FONT="]please find the attached file of questions.
[/FONT]
[FONT="]please check my working .[/FONT]
[FONT="]i will be very thankful to you.
[/FONT]
[FONT="]
[/FONT]
[FONT="]Ans 2 b)[/FONT]
[FONT="]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="]The variance is[/FONT]
[FONT="][(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[/FONT]
[FONT="]Ans 3 c)[/FONT]
[FONT="]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="]= 1/9 V(X1) + 1/9 V(X2)+ 1/9V(X3) = 1/3 σ2[/FONT]
[FONT="]SIMILARLY FIND v(T2)[/FONT]
[FONT="]v(T2)=3/8σ2[/FONT]
[FONT="]as variance of T1 is less,it is better estimator[/FONT]
[FONT="]Ans 1 .b)[/FONT]
[FONT="]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="]Total number of ways of selecting 5 cans from 10 = C(10,5) = 180[/FONT]
[FONT="]Hence, required probability of selecting 3 or more tomatoes = 126/180 = 0.7[/FONT]
[FONT="] [/FONT]
[FONT="][/FONT]
[/FONT]
[FONT="]please find the attached file of questions.
[/FONT]
[FONT="]please check my working .[/FONT]
[FONT="]i will be very thankful to you.
[/FONT]
[FONT="]
[/FONT]
[FONT="]Ans 2 b)[/FONT]
[FONT="]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="]The variance is[/FONT]
[FONT="][(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[/FONT]
[FONT="]Ans 3 c)[/FONT]
[FONT="]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="]= 1/9 V(X1) + 1/9 V(X2)+ 1/9V(X3) = 1/3 σ2[/FONT]
[FONT="]SIMILARLY FIND v(T2)[/FONT]
[FONT="]v(T2)=3/8σ2[/FONT]
[FONT="]as variance of T1 is less,it is better estimator[/FONT]
[FONT="]Ans 1 .b)[/FONT]
[FONT="]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="]Total number of ways of selecting 5 cans from 10 = C(10,5) = 180[/FONT]
[FONT="]Hence, required probability of selecting 3 or more tomatoes = 126/180 = 0.7[/FONT]
[FONT="] [/FONT]
[FONT="][/FONT]
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