Answer to Question #317969 in Statistics and Probability for Alicehyejin

Question #317969

Suppose the population consists of the scores of 6 students in a certain examination, as follows: 9, 11, 13, 15, 17, and 19 . Byusing the sampling distribution, estimate the population mean using a random variable of the size n=2

1
Expert's answer
2022-03-28T06:26:50-0400

The  samples:(9,11),xˉ=10(9,13),xˉ=11(9,15),xˉ=12(9,17),xˉ=13(9,19),xˉ=14(11,13),xˉ=12(11,15),xˉ=13(11,17),xˉ=14(11,19),xˉ=15(13,15),xˉ=14(13,17),xˉ=15(13,19),xˉ=16(15,17),xˉ=16(15,19),xˉ=17(17,19),xˉ=18P(xˉ=10)=115P(xˉ=11)=115P(xˉ=12)=215P(xˉ=13)=215P(xˉ=14)=315=15P(xˉ=15)=215P(xˉ=16)=215P(xˉ=17)=115P(xˉ=18)=115Exˉ=10115+11115+12215+13215+1415+15215+16215+17115+18115=14The\,\,samples:\\\left( 9,11 \right) ,\bar{x}=10\\\left( 9,13 \right) ,\bar{x}=11\\\left( 9,15 \right) ,\bar{x}=12\\\left( 9,17 \right) ,\bar{x}=13\\\left( 9,19 \right) ,\bar{x}=14\\\left( 11,13 \right) ,\bar{x}=12\\\left( 11,15 \right) ,\bar{x}=13\\\left( 11,17 \right) ,\bar{x}=14\\\left( 11,19 \right) ,\bar{x}=15\\\left( 13,15 \right) ,\bar{x}=14\\\left( 13,17 \right) ,\bar{x}=15\\\left( 13,19 \right) ,\bar{x}=16\\\left( 15,17 \right) ,\bar{x}=16\\\left( 15,19 \right) ,\bar{x}=17\\\left( 17,19 \right) ,\bar{x}=18\\P\left( \bar{x}=10 \right) =\frac{1}{15}\\P\left( \bar{x}=11 \right) =\frac{1}{15}\\P\left( \bar{x}=12 \right) =\frac{2}{15}\\P\left( \bar{x}=13 \right) =\frac{2}{15}\\P\left( \bar{x}=14 \right) =\frac{3}{15}=\frac{1}{5}\\P\left( \bar{x}=15 \right) =\frac{2}{15}\\P\left( \bar{x}=16 \right) =\frac{2}{15}\\P\left( \bar{x}=17 \right) =\frac{1}{15}\\P\left( \bar{x}=18 \right) =\frac{1}{15}\\E\bar{x}=10\cdot \frac{1}{15}+11\cdot \frac{1}{15}+12\cdot \frac{2}{15}+13\cdot \frac{2}{15}+14\cdot \frac{1}{5}+15\cdot \frac{2}{15}+16\cdot \frac{2}{15}+17\cdot \frac{1}{15}+18\cdot \frac{1}{15}=14


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
New on Blog