Answer to Question #317983 in Statistics and Probability for Alicehyejin

Question

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

Answer 1

$The\,\,samples:\\\left( 9,11 \right) ,\bar{x}=10\\\left( 9,13 \right) ,\bar{x}=11\\\left( 9,15 \right) ,\bar{x}=12\\\left( 9,17 \right) ,\bar{x}=13\\\left( 9,19 \right) ,\bar{x}=14\\\left( 11,13 \right) ,\bar{x}=12\\\left( 11,15 \right) ,\bar{x}=13\\\left( 11,17 \right) ,\bar{x}=14\\\left( 11,19 \right) ,\bar{x}=15\\\left( 13,15 \right) ,\bar{x}=14\\\left( 13,17 \right) ,\bar{x}=15\\\left( 13,19 \right) ,\bar{x}=16\\\left( 15,17 \right) ,\bar{x}=16\\\left( 15,19 \right) ,\bar{x}=17\\\left( 17,19 \right) ,\bar{x}=18\\P\left( \bar{x}=10 \right) =\frac{1}{15}\\P\left( \bar{x}=11 \right) =\frac{1}{15}\\P\left( \bar{x}=12 \right) =\frac{2}{15}\\P\left( \bar{x}=13 \right) =\frac{2}{15}\\P\left( \bar{x}=14 \right) =\frac{3}{15}=\frac{1}{5}\\P\left( \bar{x}=15 \right) =\frac{2}{15}\\P\left( \bar{x}=16 \right) =\frac{2}{15}\\P\left( \bar{x}=17 \right) =\frac{1}{15}\\P\left( \bar{x}=18 \right) =\frac{1}{15}\\E\bar{x}=10\cdot \frac{1}{15}+11\cdot \frac{1}{15}+12\cdot \frac{2}{15}+13\cdot \frac{2}{15}+14\cdot \frac{1}{5}+15\cdot \frac{2}{15}+16\cdot \frac{2}{15}+17\cdot \frac{1}{15}+18\cdot \frac{1}{15}=14$