Standard Deviation of a Probability Distribution

Consider two dice – one we will call the “fair die” and the other one will be called the “loaded die”. The fair die is the familiar one where each possible number (1 through 6) has the same chance of being rolled. The loaded die is designed in a special way that 3’s or 4’s are relatively likely to occur, and the remaining numbers (1, 2, 5, and 6) are unlikely to occur. The table below gives the probabilities of the possible rolls for both dice.

Fair Die		Loaded Die
Roll	Probability	Roll	Probability
1	1/6	1	1/12
2	1/6	2	1/12
3	1/6	3	1/3
4	1/6	4	1/3
5	1/6	5	1/12
6	1/6	6	1/12

How can we distinguish the fair and loaded dice? An obvious way is to roll each a number of times and see if we can distinguish the patterns of rolls that we get. We first rolled the fair die 20 times with the results

3, 3, 5, 6, 6, 1, 2, 1, 4, 3, 2, 5, 6, 4, 2, 5, 6, 1, 2, 3 (mean 3.5)

We then rolled the loaded die 20 times with the results

3, 2, 1, 4, 4, 1, 4, 3, 3, 3, 1, 3, 3, 5, 3, 3, 3, 6, 3, 4 (mean 3.1)

Below we have displayed back to back stemplots of the rolls from the two dice.

P6.clip3.png

What do we see? For the fair die, the rolls appear to be evenly spread out among the six possible numbers. In contrast, the rolls for the loaded die tend to concentrate on the values and 3 and 4, and the remaining numbers were less likely to occur.

Can we compute a summary number to contrast the probability distributions for the fair and loaded dice?

We have already discussed one summary number for a random variable, the mean m. This number represents the average outcome for the random variable when one performs the experiment many times.

Let’s compute the mean for the two probability distributions. For the fair die, the mean is given by

$\mu$ (Fair Die) = (1) (1/6) + (2) (1/6) + (3) (1/6) + (4) (1/6) + (5) (1/6) + (6) (1/6) = 3.5,

and for the loaded die the mean is given by

$\mu$ (Loaded Die) = (1) (1/12) + (2) (1/12) + (3) (1/3) + (4) (1/3) + (5) (1/12) + (6) (1/12) = 3.5.

The means of the two probability distributions are the same – this means that we’ll tend to get the same average roll when we roll the fair die and roll the loaded die many times.

But we know from our rolling data that the two probability distributions are different. For the loaded die, we’re more likely to roll 3’s or 4’s. In other words, for the loaded die, it is more likely to roll a number close to the mean value $\mu$ = 3.5.

The standard deviation of a random variable X, denoted by the Greek letter $\sigma$ , measures how close the random variable is to the mean $\mu$ . It is called a standard deviation since it represents an “average” (or standard) distance (or deviation) from the mean $\mu$ .

To find the standard deviation $σ$ for a random variable, we

(Compute deviations.) For each value of the random variable X, we compute the difference (or deviation) of X from the mean value $\mu$ .
(Square deviations.) We then square each of the differences that we computed in step 1.
(Compute the average squared deviation.) We find the average squared deviation, by multiplying each squared deviation by the corresponding probability, and summing the products.
(Take square root.) The standard deviation $\sigma$ is the square root of the average squared deviation.

We illustrate in the following table the computation of the standard deviation for the roll of the fair die and for the roll of the loaded die.

Computation of the standard deviation $\sigma$ for the Fair Die.

Roll	$R - μ$	$(R - μ)^{2}$	Probability	$(R - μ)^{2}$ x Probability
1	1 – 3.5 = -2.5	$LaTeX: \left(-2.5\right)^2=6.25$	1/6	6.25 x (1/6)
2	2 – 3.5 = -1.5	$LaTeX: \left(-1.5\right)^2=2.25$	1/6	2.25 x (1/6)
3	3 – 3.5 = -0.5	$LaTeX: \left(–0.5\right)^2=0.25$	1/6	0.25 x (1/6)
4	4 – 3.5 = 0.5	$LaTeX: \left(0.5\right)^2=0.25$	1/6	0.25 x (1/6)
5	5 – 3.5 = 1.5	$LaTeX: \left(1.5\right)^2=2.25$	1/6	2.25 x (1/6)
6	6 – 3.5 = 2.5	$LaTeX: \left(2.5\right)^2=6.25$	1/6	6.25 x (1/6)
SUM				2.917
				$\sqrt{2.917}$ = 1.71

Computation of the standard deviation $σ$ for the Loaded Die.

Roll	$R - μ$	$(R - μ)^{2}$	Probability	$(R - μ)^{2}$ x Probability
1	1 – 3.5 = -2.5	$LaTeX: \left(-2.5\right)^2=6.25$	1/12	6.25 x (1/12)
2	2 – 3.5 = -1.5	$LaTeX: \left(-1.5\right)^2=2.25$	1/12	2.25 x (1/12)
3	3 – 3.5 = -0.5	$LaTeX: \left(–0.5\right)^2=0.25$	1/3	0.25 x (1/3)
4	4 – 3.5 = 0.5	$LaTeX: \left(0.5\right)^2=0.25$	1/3	0.25 x (1/3)
5	5 – 3.5 = 1.5	$LaTeX: \left(1.5\right)^2=2.25$	1/12	2.25 x (1/12)
6	6 – 3.5 = 2.5	$LaTeX: \left(2.5\right)^2=6.25$	1/12	6.25 x (1/12)
SUM				$\sqrt{1.583}$ = 1.26

We see from our calculations that

$\sigma$ (Fair Die) = 1.71, $\sigma$ (Loaded Die) = 1.26

What does this mean? Since the loaded die roll has a smaller standard deviation, this means that the roll of the loaded die tends to be closer to the mean (3.5) than for the fair die. When we roll the loaded die many times, we will notice a smaller spread or variation in the rolls than when we roll the fair die many times.

Practice: the Standard Deviation

1. Suppose you spin two spinners of the type shown below where each spinner is equally likely to land 1, 2, or 3. Find the probability distribution for the sum S of the two spins.

P6.cliip4.png

2. Find the mean and standard deviation of S.

3. Suppose you spin the two different spinners shown below. Find the probability distribution for the sum T of the two spins.

P6.clip5.png

4. Graph the probability distributions for S and T. By looking at the graphs, how do the two distributions differ?

5. Compute the means and standard deviations for S and T and place these values next to the graphs you drew in part 4. Do S and T differ with respect to their average value? Do the two random variables differ with respect to their spread?

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