07-15-2010, 09:18 PM
Idea solution Please don't Copy
b) A finite population consists of values 6, 6, 9, 15 and 18. Calculate the sample means for all possible random samples of size n=3, that can be drawn from this population without replacement. Make the sampling distribution of sample mean and find the mean and variance of this distribution.
Solution :
there are 5C3 (because order does not matter)ways of having a ***** 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
b) Ten vegetables cans, all of the same size, have lost their labels. It is known that 5 contain tomatoes and 5 contain corns. If 5 are selected at random, what is the probability that all contain tomatoes? What is the probability that 3 or more contain tomatoes?
Solution:
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
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
b) A finite population consists of values 6, 6, 9, 15 and 18. Calculate the sample means for all possible random samples of size n=3, that can be drawn from this population without replacement. Make the sampling distribution of sample mean and find the mean and variance of this distribution.
Solution :
there are 5C3 (because order does not matter)ways of having a ***** 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
b) Ten vegetables cans, all of the same size, have lost their labels. It is known that 5 contain tomatoes and 5 contain corns. If 5 are selected at random, what is the probability that all contain tomatoes? What is the probability that 3 or more contain tomatoes?
Solution:
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
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