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

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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