Answer
a. See Table
b.Suppose we assign the letters A, B, C, D, E and F to the six ages (in years), so that A=15, B=21, C=25, D=28, E=53, F=55.
$6C_{4}$=$\frac{6!}{4!(6-4)!}$ = $\frac{6*5*4*3*2*1}{4*3*2*1*2*1}$ = 15
The 15 possible samples of six ages are:
ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, ADEF, BCDE, BCDF, BCEF, BDEF, CDEF
C. From the table, the sample mean, $x ̅$, we can assume any of the values listed in the table with the corresponding probability.
If sample mean $x ̅$= 30.75:
Population mean = $(55+53+28+25+21+15)/6$ =32.83
Sampling error = $\bar{x}\ -\mu\ $= 32.83-30.75 = 2.08
Work Step by Step
The table is above.
