Answer:
(b) ![\frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D)
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.
![\begin{equation*} \begin{matrix}Proportion of males & Probability \\0 & 9/16 \\1 & 6/16 \\2 & 1/16 \end{matrix} \end{equation*}](https://tex.z-dn.net/?f=%20%20%5Cbegin%7Bequation%2A%7D%20%5Cbegin%7Bmatrix%7DProportion%20of%20males%20%26%20Probability%20%5C%5C0%20%26%209%2F16%20%5C%5C1%20%26%206%2F16%20%5C%5C2%20%26%201%2F16%20%5Cend%7Bmatrix%7D%20%5Cend%7Bequation%2A%7D%20)
b. The mean of the sampling distribution
![\mu = (\frac{9}{16} X0 ) +(\frac{6}{16} X 1)+(\frac{1}{16} X 2) = \frac{8}{16} = \frac{1}{2}](https://tex.z-dn.net/?f=%20%5Cmu%20%3D%20%28%5Cfrac%7B9%7D%7B16%7D%20X0%20%29%20%2B%28%5Cfrac%7B6%7D%7B16%7D%20X%201%29%2B%28%5Cfrac%7B1%7D%7B16%7D%20X%202%29%20%3D%20%5Cfrac%7B8%7D%7B16%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20)
c. No; the proportion of males is
while the mean is
.
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.