Answer:
[tex]\dfrac{2}{3}[/tex]
Step-by-step explanation:
Given that:
The number of observations is:
6, 10, 12
If we are to use a simple random sampling without replacement, then we will have:
(6,10) (6,12) (10,12)
Here;
the sample size n = 2
The population size N = 3
For (6,10) ; The sample mean = [tex]\dfrac{6+10}{2}[/tex]
= [tex]\dfrac{16}{2}[/tex]
= 8
For (6,12) ; The sample mean = [tex]\dfrac{6+12}{2}[/tex]
= [tex]\dfrac{18}{2}[/tex]
= 9
For (10, 12) ; The sample mean = [tex]\dfrac{10+12}{2}[/tex]
= [tex]\dfrac{22}{2}[/tex]
= 11
The probability distribution of sample mean(x) is:
X 8 9 11
P(X=x) [tex]\dfrac{1}{3}[/tex] [tex]\dfrac{1}{3}[/tex] [tex]\dfrac{1}{3}[/tex]
Thus, the probability that the sample mean is larger than 8 is:
P(X> 8) = P(X = 9) + P(X + 11)
P(X> 8) = [tex]\dfrac{1}{3}+\dfrac{1}{3}[/tex]
P(X > 8) = [tex]\dfrac{1+1}{3}[/tex]
P(X> 8) = [tex]\dfrac{2}{3}[/tex]
