Answer to Question #256790 in Statistics and Probability for jay

Question #256790

C. Solve. Given a population of the scores of BSHS honor students in Mathematics as 1, 3, 4, 6, and 8. Suppose a sample size of 3 have to be drawn from it for the first, second and third rank;



1
Expert's answer
2021-11-09T16:54:06-0500

a.

We have population values "1,3,4,6,8," population size "N=5"

"\\mu=\\dfrac{1+3+4+6+8}{5}=4.4"

b.



"\\sigma^2=\\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2""+(6-4.4)^2)+(8-4.4)^2)=5.84"


c.



"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{5.84}\\approx2.4166"


d. We have population values "1,3,4,6,8" population size "N=5" and sample size "n=3." Thus, the number of possible samples which can be drawn without replacement is




"\\dbinom{N}{n}=\\dbinom{5}{3}=10"


The sampling distribution of the sample means.


"E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=4.4"


The mean of the sampling distribution of the sample means is equal to the the mean of the population.



"E(\\bar{X})=\\mu_{\\bar{X}}=4.4=\\mu"



e.



"Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(\\sum\\bar{X}f(\\bar{X}))^2"





"=\\dfrac{61}{3}-(4.4)^2=\\dfrac{2.92}{3}"





"\\sigma_{\\bar{X}}=\\sqrt{Var(\\bar{X})}=\\sqrt{\\dfrac{2.92}{3}}\\approx0.9866"

Verification:



"Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=\\dfrac{5.84}{3}(\\dfrac{5-3}{5-1})"





"=\\dfrac{2.92}{3}, True"

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