Chapter 14, Problem 14.34E

a.

To determine

To find: The probability that he or she would choose the pair of socks in the center position.

To find: The distribution of X, and the number of subjects among the 100 who would choose the pair of socks in the center position.

Answer to Problem 14.34E

The probability that he or she would choose a pair of socks in the center position is $\frac{1}{5}$.

The distribution of X, and the number of subjects among the 100 who would choose the pair of socks in the center position are binomial.

Explanation of Solution

Given info:

There are five identical pairs of white socks which were vertically attached to a blue background for display. Also, there were 100 participants and they were asked to select the preferred pair of socks.

Calculation:

There is only one center position out of five positions. The probability of selecting a pair of socks in each position is,

$\begin{array}{c}\text{Probability}=\frac{\text{Number of center position}}{\text{Total number of positions}}\\ =\frac{1}{5}\end{array}$.

Thus, the probability that he or she would choose a pair of socks in the center position is $\frac{1}{5}$.

Binomial distribution:

The random variable x is defined as the number of success observations follow binomial distribution if the distribution has fixed number of observations (n) in which all the observations are independent and have two possible outcomes (success and failure). Also, for each observation, p represents the probability of success.

The probability using binomial distribution is given by,

$P\left(X=x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Where, n is the number of trials, x is the number of successes among n trials, p is the probability of success and q is the probability of failure.

The variable X represents the number of subjects among the 100 who would choose the pair of socks in the center position. Here the n represents the participants. Also, there are two possible outcomes (selecting a pair of socks in the center position and not selecting a pair of socks in the center position) with probability of success (p) which is 0.20.

Since, there are  two possible outcomes with fixed number of observations, it can be stated that the distributions of participants selecting a pair of socks follow binomial distribution with $n=100$ with probability $p=0.20$.

Thus, X follows the binomial distribution with $n=100,p=0.20$.

b.

To determine

To find: The mean and standard deviation of X.

Answer to Problem 14.34E

The mean of X is 20.

The standard deviation of X is 4.

Explanation of Solution

Calculation:

Mean:

The mean is calculated by using the following formula:

$\mu =np$

Substitute n as 100 and p as 0.20

$\begin{array}{c}\mu =\left(100\right)\left(0.20\right)\\ =20\end{array}$

Thus, the mean of X is 20.

The standard deviation of X is:

The standard deviation is calculated by using the following formula:

$\sigma =\sqrt{np\left(1-p\right)}$

Substitute n as 100, p as 0.20 and q as $0.80\left(=1-0.20\right)$

$\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{\left(100\right)\left(0.20\right)\left(0.80\right)}\\ =\sqrt{16}\\ =4\end{array}$

Thus, the standard deviation of X is 4.

c.

To determine

To find: The probability that 34 or more subjects chose the pair of socks in center using normal approximation.

To find: The exact binomial probability and compare the results.

Answer to Problem 14.34E

The approximate probability that 34 or more subjects chose pair of socks in center is 0.

The exact probability is larger than the approximate probability by 0.9996.

The exact probability of choosing a pair of socks in the center is 0.9996.

Explanation of Solution

Given info:

In an experiment, a pair of socks in the center was chosen by the 34 subjects.

Calculation:

Normal approximation:

Rule of thumb:

• If the sample size n is large, the binomial distribution can be approximated to normal $N\left(np,\sqrt{np\left(1-p\right)}\right)$.
• Normal approximation can be used when the sample size n is large, in such a way that $np\ge 10\text{and}n\left(1-p\right)\ge 10$

Here, $n=100,p=0.2$.

$\begin{array}{c}np=\left(100\right)\left(0.20\right)\\ =20\\ \ge 10\end{array}$

$\begin{array}{c}n\left(1-p\right)=\left(100\right)\left(1-0.2\right)\\ =100\times 0.8\\ =80\\ \ge 10\end{array}$

Thus, the two conditions are satisfied when the sample size n is large, so that the binomial distribution can be safely approximated to normal distribution.

The approximate probability obtaining 34 or more subjects choosing pair of socks in the center:

Mean:

The mean is calculated by using the following formula:

$\mu =np$

Substitute n as 100 and p as 0.20

$\begin{array}{c}\mu =\left(100\right)\left(0.20\right)\\ =20\end{array}$

Thus, the value of mean is 20.

Standard deviation:

The standard deviation is calculated by using the following formula:

$\sigma =\sqrt{np\left(1-p\right)}$

Substitute n as 100, p as 0.20 and q as $0.80\left(=1-0.20\right)$

$\begin{array}{c}\sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{\left(100\right)\left(0.20\right)\left(0.80\right)}\\ =\sqrt{16}\\ =4\end{array}$

Thus, the value of standard deviation is 2.3324.

That is, $P\left(X\ge 34\right)$

$\begin{array}{c}P\left(X\ge 34\right)=P\left(\frac{X-20}{4}\ge \frac{34-20}{4}\right)\\ =P\left(z\ge \frac{14}{4}\right)\\ =P\left(z\ge 7\right)\\ =1-P\left(z<7\right)\end{array}$

From the TABLE A: Standard Normal cumulative proportions, and the area to the left of $z<7$ are 1.

$\begin{array}{c}P\left(X\ge 34\right)=1-P\left(z<7\right)\\ =1-1\\ =0\end{array}$

Thus, the approximate probability that 34 or more subjects chose pair of socks in center is 0.

The exact probability that 34 or more subjects chose pair of socks in center:

$\begin{array}{c}P\left(X\ge 34\right)=1-P\left(X<34\right)\\ =1-P\left(X\le 33\right)\end{array}$

Software procedure:

Step-by-step procedure to obtain the ‘Binomial probability’ using the MINITAB software:

• Choose Calc > Probability Distributions > Binomial Distribution.
• Choose Cumulative Probability.
• Enter Number of trials as 100 and Event probability as 0.2.
• In Input constant, enter 33.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(X\le 33\right)=0.9993$

Therefore,

$\begin{array}{c}P\left(X\ge 34\right)=1-P\left(X<34\right)\\ =1-P\left(X\le 33\right)\\ =1-0.9993\\ =0.0007\end{array}$

Thus, the exact probability that 34 or more subjects choosing a pair of socks in the center is 0.0007.

Justification:

Here, approximate probability that 34 or more subjects chose pair of socks in center is 0 and exact probability that 34 or more subjects chose pair of socks in center is 0.0007.

Therefore, the exact probability and approximate probability are approximately equal with small difference 0.0007.

d.

To determine

To explain: Whether this experiment supports the “center stage effect” or not.

Answer to Problem 14.34E

No, this experiment does not support the “center stage effect”.

Explanation of Solution

From part (c), the probability that 34 or more subjects chose pair of socks in center is very low.

Thus, this experiment does not support the “center stage effect”