A rowing team consists of four rowers who weigh 152, 156, 160, and 164 pounds. Find all possible random samples with replacement of size two and compute the sample mean for each one. Use them to find the probability distribution, the mean, and the standard deviation of the sample mean X−−.X-.

 A rowing team consists of four rowers who weigh 152, 156, 160, and 164 pounds. Find all possible random samples with replacement of size two and compute the sample mean for each one. Use them to find the probability distribution, the mean, and the standard deviation of the sample mean 

X−−.$\stackrel{-}{\mathit{X}}.$

Solution

The following table shows all possible samples with replacement of size two, along with the mean of each:

Sample	Mean		Sample	Mean		Sample	Mean		Sample	Mean
152, 152	152		156, 152	154		160, 152	156		164, 152	158
152, 156	154		156, 156	156		160, 156	158		164, 156	160
152, 160	156		156, 160	158		160, 160	160		164, 160	162
152, 164	158		156, 164	160		160, 164	162		164, 164	164

The table shows that there are seven possible values of the sample mean X−−.$\stackrel{-}{\mathit{X}}.$ The value x−−=152$\stackrel{-}{\mathit{x}}=152$ happens only one way (the rower weighing 152 pounds must be selected both times), as does the value x−−=164$\stackrel{-}{\mathit{x}}=164$, but the other values happen more than one way, hence are more likely to be observed than 152 and 164 are. Since the 16 samples are equally likely, we obtain the probability distribution of the sample mean just by counting:

x−−P(x−−)152116154216156316158416160316162216164116$$\begin{array}{c}\stackrel{}{\mathit{<br>}}\end{array}$$$$\begin{array}{cccccccc}\stackrel{-}{\mathit{x}}& 152& 154& 156& 158& 160& 162& 164\\ P\left(\stackrel{-}{\mathit{x}}\right)& \frac{1}{16}& \frac{2}{16}& \frac{3}{16}& \frac{4}{16}& \frac{3}{16}& \frac{2}{16}& \frac{1}{16}\end{array}$$

Now we apply the formulas from Section 4.2.2 "The Mean and Standard Deviation of a Discrete Random Variable" in Chapter 4 "Discrete Random Variables" for the mean and standard deviation of a discrete random variable to X−−.$\stackrel{-}{\mathit{X}}.$ For μX−−${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ we obtain.

μX−−=Σx−− P(x−−)=152(116)+154(216)+156(316)+158(416)+160(316)+162(216)+164(116)=158$$\begin{array}{ll}\hfill {\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}& ={\displaystyle \mathrm{\Sigma }\stackrel{-}{\mathit{x}}P\left(\stackrel{-}{\mathit{x}}\right)}\hfill \\ \hfill & =152\left(\frac{1}{16}\right)+154\left(\frac{2}{16}\right)+156\left(\frac{3}{16}\right)+158\left(\frac{4}{16}\right)+160\left(\frac{3}{16}\right)+162\left(\frac{2}{16}\right)+164\left(\frac{1}{16}\right)\hfill \\ \hfill & =158\hfill \end{array}$$

For σX−−${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}$ we first compute Σx−−2P(x−−)${\displaystyle \mathrm{\Sigma }{\stackrel{-}{\mathit{x}}}^{2}P\left(\stackrel{-}{\mathit{x}}\right)}$:

1522(116)+1542(216)+1562(316)+1582(416)+1602(316)+1622(216)+1642(116)$${152}^2\left(\frac{1}{16}\right)+{154}^2\left(\frac{2}{16}\right)+{156}^2\left(\frac{3}{16}\right)+{158}^2\left(\frac{4}{16}\right)+{160}^2\left(\frac{3}{16}\right)+{162}^2\left(\frac{2}{16}\right)+{164}^2\left(\frac{1}{16}\right)$$

which is 24,974, so that

σX−−=Σx−−2P(x−−)−μ2x−−−−−−−−−−−−−−√=24,974−1582−−−−−−−−−−−√=10−−√