Answer:
(b) [TeX] \frac{1}{2}[/TeX]
Step-by-step explanation:
1 male named Bart (b)
3 females named Charlene(c), Diana(d), and Erin(e).
Since there is replacement, the possible samples are:
bb, bc, be, bd, cb, cc, cd, ce, db, dc, dd, de, eb, ec, ed, and ee.
Total Number of pairs = 16
Event of picking 2 males:bb
Event of 1 male:bc,cb,bd,db,be,eb
Event of picking 0 males:
cc, cd, ce, dc, dd, de, ec, ed, and ee.
[TeX] \begin{equation*} \begin{matrix}Proportion of males & Probability \\0 & 9/16 \\1 & 6/16 \\2 & 1/16 \end{matrix} \end{equation*} [/TeX]
b. The mean of the sampling distribution
[TeX] \mu = (\frac{9}{16} X0 ) +(\frac{6}{16} X 1)+(\frac{1}{16} X 2) = \frac{8}{16} = \frac{1}{2} [/TeX]
c. No; the proportion of males is [TeX] \frac{1}{4}[/TeX] while the mean is [TeX] \frac{1}{2}[/TeX].
B. No, the sample mean is not equal to the population proportion of males. These values are not always equal, because proportion is an unbiased estimator.