Answer to Question #316469 in Statistics and Probability for Valeng

Question

1. A six pumpkins are displayed in a store a population consists of 6 values (19,14,15,9,10. and 17). Suppose that random sample size 2 are taken form this population

a. How many samples are possible? List them and compute the mean of each sample

b. Construct the sampling distribution of the sample means

c. Construct the histogram of the sampling distribution of the sample means

Answer 1

$a:\\The\,\,number\,\,of\,\,samples\,\,is\,\,C_{6}^{2}=15\\\left( 9,10 \right) ,\bar{x}=9.5\\\left( 9,14 \right) ,\bar{x}=11.5\\\left( 9,15 \right) ,\bar{x}=12\\\left( 9,17 \right) ,\bar{x}=13\\\left( 9,19 \right) ,\bar{x}=14\\\left( 10,14 \right) ,\bar{x}=12\\\left( 10,15 \right) ,\bar{x}=12.5\\\left( 10,17 \right) ,\bar{x}=13.5\\\left( 10,19 \right) ,\bar{x}=14.5\\\left( 14,15 \right) ,\bar{x}=14.5\\\left( 14,17 \right) ,\bar{x}=15.5\\\left( 14,19 \right) ,\bar{x}=16.5\\\left( 15,17 \right) ,\bar{x}=16\\\left( 15,19 \right) ,\bar{x}=17\\\left( 17,19 \right) ,\bar{x}=18\\$

$P\left( \bar{x}=9.5 \right) =\frac{1}{15}\\P\left( \bar{x}=11.5 \right) =\frac{1}{15}\\P\left( \bar{x}=12 \right) =\frac{2}{15}\\P\left( \bar{x}=12.5 \right) =\frac{1}{15}\\P\left( \bar{x}=13 \right) =\frac{1}{15}\\P\left( \bar{x}=13.5 \right) =\frac{1}{15}\\P\left( \bar{x}=14 \right) =\frac{1}{15}\\P\left( \bar{x}=14.5 \right) =\frac{2}{15}\\P\left( \bar{x}=15.5 \right) =\frac{1}{15}\\P\left( \bar{x}=16 \right) =\frac{1}{15}\\P\left( \bar{x}=16.5 \right) =\frac{1}{15}\\P\left( \bar{x}=17 \right) =\frac{1}{15}\\P\left( \bar{x}=18 \right) =\frac{1}{15}$