Answer to Question #342948 in Statistics and Probability for Yoongi

Question #342948

Consider all possible samples of size 2 that can be drawn with replacement from the population 1, 5, and 8. Compute the following:


1. Population mean


2. Population variance


3. Population standard deviation


4. Illustrate the probability histogram of the sampling distribution of the means

1
Expert's answer
2022-05-23T09:26:14-0400

We have population values 1,5,8 population size N=3 and sample size n=2.

1.Population mean

"\\mu=\\dfrac{1+5+8}{3}=\\dfrac{14}{3}"


2. Population variance 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{\\dfrac{121}{9}+\\dfrac{1}{9}+\\dfrac{100}{9}}{3}=\\dfrac{74}{9}"



3. Population standard deviation


"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{\\dfrac{74}{9}}=\\dfrac{\\sqrt{74}}{3}\\approx2.867442"

4. The number of possible samples which can be drawn with replacement is "N^n=3^2=9."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 1,1 & 2\/2 \\\\\n \\hdashline\n 2 & 1,5 & 6\/2 \\\\\n \\hdashline\n 3 & 1,8 & 9\/2 \\\\\n \\hdashline\n 4 & 5,1 & 6\/2 \\\\\n \\hdashline\n 5 & 5,5 & 10\/2 \\\\\n \\hdashline\n 6 & 5,8 & 13\/2 \\\\\n \\hdashline\n 7 & 8,1 & 9\/2 \\\\\n \\hdashline\n 8 & 8,5 & 13\/2 \\\\\n \\hdashline\n 9 & 8,8 & 16\/2 \\\\\n \\hdashline \n\\end{array}"




"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f(\\bar{X}) &\\bar{X} f(\\bar{X}) &\\bar{X}^2 f(\\bar{X})\\\\ \\hline\n 2\/2 & 1\/9 & 2\/18 & 4\/36\\\\\n \\hdashline\n 6\/2 & 2\/9 & 12\/18 & 72\/36\\\\\n \\hdashline\n 9\/2 & 2\/9 & 18\/18 & 162\/36\\\\\n \\hdashline\n 10\/2 & 1\/9 & 10\/18 & 100\/36\\\\\n \\hdashline\n 13\/2 & 2\/9 & 26\/18 & 338\/36\\\\\n \\hdashline\n 16\/2 & 1\/9 & 16\/18 & 256\/36\\\\\n \\hdashline\n\\end{array}"


Mean of sampling distribution 

"\\mu_{\\bar{X}}=E(\\bar{X})=\\sum\\bar{X}_if(\\bar{X}_i)=\\dfrac{14}{3}=\\mu"


The variance of sampling distribution 

"Var(\\bar{X})=\\sigma^2_{\\bar{X}}=\\sum\\bar{X}_i^2f(\\bar{X}_i)-\\big[\\sum\\bar{X}_if(\\bar{X}_i)\\big]^2""=\\dfrac{932}{36}-(\\dfrac{14}{3})^2=\\dfrac{37}{9}= \\dfrac{\\sigma^2}{n}"

The standard error of the mean

"\\sigma_{\\bar{X}}=\\sqrt{\\dfrac{37}{9}}\\approx2.0276"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
APPROVED BY CLIENTS