Question

Guilt in decision making. The Journal of Behavioral Decision Making (January 2007) published a study of how guilty feelings impact on-the-job decisions. In one experiment, 57 participants were assigned to a guilty state through a reading/writing task. Immediately after the task, the participants were presented with a decision problem where the stated option had predominantly negative features (e.g., spending money on repairing a very old car). Of these 57 participants, 45 chose the stated option. Suppose 10 of the 57 guilty-state participants are selected at random. Define x as the number in the sample of 10 who chose the stated option.

a. Find P ( X = 5 ).

b. Find P ( X = 8 ).

c. What is the expected value (mean) of x?

Short answer

$$\begin{gathered} \mathrm{a}.\mathrm{P}\left(\mathrm{X}=5\right)=0.0224 \\ \mathrm{b}.\mathrm{P}\left(\mathrm{X}=8\right)=0.3294 \\ \mathrm{c}.\mathrm{The}\mathrm{expected}\mathrm{value}\left(\mathrm{mean}\right)\mathrm{of}\mathrm{x}\mathrm{is}\frac{\mathrm{nr}}{\mathrm{N}} \end{gathered}$$

Step-by-step solution

Given information

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

Finding the probability

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

Finding the probability

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

Finding the expected value (mean)

c.

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

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

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