91Ó°ÊÓ

By the Journal of Behavioral Decision Making (January 2007),

x is a random variable that takes the number of participants who chose the stated.

57 participants were assigned to a guilty state, i.e.N = 57,

45 are chosen from 57 participants, i.e.r = 45,

10 is the number of samples chosen in 57 participants, i.e., n = 10

x follows a hypergeometric distribution with N = 57, n = 10 and r = 45

a.

The probability of x given by,

$$\begin{gathered} P\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)} \\ \mathrm{Here}, \\ \mathrm{x}=5,\mathrm{N}=57,\mathrm{n}=10\mathrm{and}\mathrm{r}=45 \\ \mathrm{P}\left(\mathrm{x}\right)=\frac{\left(\begin{array}{c}\mathrm{r}\\ \mathrm{x}\end{array}\right)\left(\begin{array}{c}\mathrm{N}-\mathrm{r}\\ \mathrm{n}-\mathrm{x}\end{array}\right)}{\left(\begin{array}{c}\mathrm{N}\\ \mathrm{n}\end{array}\right)} \\ =\frac{\left(\begin{array}{c}45\\ 5\end{array}\right)\left(\begin{array}{c}57-45\\ 10-5\end{array}\right)}{\left(\begin{array}{c}57\\ 10\end{array}\right)} \\ =\frac{\left(\begin{array}{c}45\\ 5\end{array}\right)\left(\begin{array}{c}12\\ 5\end{array}\right)}{\left(\begin{array}{c}57\\ 10\end{array}\right)} \\ =\frac{1221759\mathrm{x}792}{43183019880} \\ =0.0224 \\ \mathrm{P}\left(\mathrm{x}=5\right)=0.0224 \end{gathered}$$

Therefore, the probability is 0.0224

b.

The probability of x given by,

$P\left(x\right)=\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Here,

$$\begin{gathered} \mathrm{x}=8,\mathrm{N}=57,\mathrm{n}=10\mathrm{and}\mathrm{r}=45 \\ \mathrm{P}\left(\mathrm{x}=8\right)=\frac{\left(\begin{array}{c}\mathrm{r}\\ 8\end{array}\right)\left(\begin{array}{c}\mathrm{N}-\mathrm{r}\\ \mathrm{n}-8\end{array}\right)}{\left(\begin{array}{c}\mathrm{N}\\ \mathrm{n}\end{array}\right)} \\ =\frac{\left(\begin{array}{c}45\\ 8\end{array}\right)\left(\begin{array}{c}57-45\\ 10-8\end{array}\right)}{\left(\begin{array}{c}57\\ 10\end{array}\right)} \\ =\frac{\left(\begin{array}{c}45\\ 8\end{array}\right)\left(\begin{array}{c}12\\ 2\end{array}\right)}{\left(\begin{array}{c}57\\ 10\end{array}\right)} \\ =\frac{215553195\mathrm{x}66}{43183019880} \\ =0.3294 \\ \mathrm{P}\left(\mathrm{X}=8\right)=0.3294 \end{gathered}$$

Therefore, the probability is 0.3294

c.

The expected value (mean) of x is given by,

$$\begin{gathered} E\left(x\right)=\underset{x=0}{\overset{n}{∑}}xP\left(X=x\right) \\ =\underset{x=0}{\overset{n}{∑}}x\left\{\frac{\left(\begin{array}{c}r\\ x\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}\right\} \\ =\frac{r}{\left(\begin{array}{c}N\\ n\end{array}\right)}\underset{x=1}{\overset{n}{∑}}\left\{\left(\begin{array}{c}r-1\\ x-1\end{array}\right)\left(\begin{array}{c}N-r\\ n-x\end{array}\right)\right\} \\ =\frac{r}{\left(\begin{array}{c}N\\ n\end{array}\right)}\underset{y=0}{\overset{n}{∑}}\left\{\left(\begin{array}{c}A\\ y\end{array}\right)\left(\begin{array}{c}N-A-1\\ m-y\end{array}\right)\right\}\left[A=r-1,y=x-1,m=n-1\right] \\ =\frac{r}{\left(\begin{array}{c}N\\ n\end{array}\right)}\left(\begin{array}{c}N-1\\ m\end{array}\right) \\ =\frac{r}{\left(\begin{array}{c}N\\ n\end{array}\right)}\left(\begin{array}{c}N-1\\ n-1\end{array}\right) \\ =\frac{r}{\left({\displaystyle \frac{N!}{n!\left(N-n\right)!}}\right)}\left(\frac{(N-1)!}{\left(n-1\right)!\left(N-n\right)!}\right) \\ =\frac{r}{\left(\frac{N(N-1)!}{n\left(n-1\right)!\left(N-n\right)!}\right)}\left(\frac{(N-1)!}{\left(n-1\right)!\left(N-n\right)!}\right) \\ =\frac{nxrx\left(n-1\right)!x\left(N-n\right)!}{Nx\left(N-1\right)!}x\frac{(N-1)!}{\left(n-1\right)!x\left(N-n\right)!} \\ =\frac{nr}{N} \\ E\left(x\right)=\frac{nr}{N} \end{gathered}$$

Therefore, the expected value (mean) of x is $\frac{nr}{N}$