A population consists of the numbers 2, 4, 8, 10 and 5. Let us list all the possible samples of size 3 from this population and construct the sampling distribution of the sample mean.
A population consists of the numbers 2, 4, 8, 10 and 5
Sampling distribution of size 3 can be given as
P(Xˉ=4.66)=15P(\bar X=4.66) = \dfrac{1}{5}P(Xˉ=4.66)=51
P(Xˉ=6.66)=110P(\bar X=6.66) = \dfrac{1}{10}P(Xˉ=6.66)=101
P(Xˉ=5.66)=110P(\bar X=5.66) = \dfrac{1}{10}P(Xˉ=5.66)=101
P(Xˉ=5)=15P(\bar X=5) = \dfrac{1}{5}P(Xˉ=5)=51
P(Xˉ=7.33)=110P(\bar X=7.33) = \dfrac{1}{10}P(Xˉ=7.33)=101
P(Xˉ=6.33)=110P(\bar X=6.33) = \dfrac{1}{10}P(Xˉ=6.33)=101
P(Xˉ=3.66)=110P(\bar X=3.66) = \dfrac{1}{10}P(Xˉ=3.66)=101
P(Xˉ=7.66)=110P(\bar X=7.66) = \dfrac{1}{10}P(Xˉ=7.66)=101
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Dear Dexter Prades, a solution of the question has already been published.
Please ma'am / sir , I need the answer! The deadline for this is may 29 please! I'm begging!