In Exercises 21–25, refer to the accompanying table, which describes the numbers of adults in groups of five who reported sleepwalking (based on data from “Prevalence and Comorbidity of Nocturnal Wandering In the U.S. Adult General Population,” by Ohayon et al., Neurology, Vol. 78, No. 20).

Find the mean and standard deviation for the numbers of sleepwalkers in groups of five.

x	P(x)
0	0.172
1	0.363
2	0.306
3	0.129
4	0.027
5	0.002

The probability distribution for the number of adults in groups who reported sleepwalking is provided.

Let x bethe number of sleepwalkers in the group of five.

The requirements are as follows:

1)The variable x is anumerical random variable.

2)The sum of the probabilities is computed as:

$$\begin{gathered} \sum P\left(x\right)=0.172+0.363+0.306+0.129+0.027+0.002 \\ =0.999 \end{gathered}$$

Therefore,the sum of the probabilities is approximately equal to 1 with a round of error as 0.001.

3) Each value of P(x) is between 0 and 1.

Thus, all the requirements are satisfied.

The mean for a random variable is computed as:

$$\begin{gathered} \mu =\sum \left[x\times P\left(x\right)\right] \\ =\left(0\times 0.172\right)+\left(1\times 0.363\right)+\left(2\times 0.306\right)+...+\left(5\times 0.002\right) \\ =1.48 \\ \approx 1.5 \end{gathered}$$

Thus, the mean number of sleepwalkers is 1.5.

The standard deviation of the random variable x is computed as;

$\sigma =\sqrt{\sum \left[x^2\times P\left(x\right)\right]-{\mu }^2}$

The calculations are as follows;

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0.172\right)+\left(1^2\times 0.363\right)+\left(2^2\times 0.306\right)+...+\left(5^2\times 0.002\right) \\ =3.23 \end{gathered}$$

The standard deviation is given as:

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{3.23-{1.5}^2} \\ =0.98 \\ \approx 1.0 \end{gathered}$$

Thus, the standard deviation of x is 1.0.