In Exercises 7–14, determine whether a probability

distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

A sociologist randomly selects single adults for different groups of three, and the random variable xis the number in the group who say that the most fun way to flirt is in person(based on a Microsoft Instant Messaging survey).

x	P(x)
0	0.091
1	0.334
2	0.408
3	0.166

The probability distribution for the group thatsays that the most fun way to flirt is in person is provided.

The variable x is the number in the group that says that the most fun way to flirt is in person.

The requirements are as follows.

1)The variable x is anumerical random variable.

2)The sum of the probabilities is computed as

$$\begin{gathered} \sum P\left(x\right)=0.091+0.334+0.408+0.166 \\ =0.999 \end{gathered}$$

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

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

Thus, the distribution is a valid probability distribution as all the requirements are satisfied.

The mean for the random variable is computed as

$$\begin{gathered} \mu =\sum \left[x\times P\left(x\right)\right] \\ =\left(0\times 0.091\right)+\left(1\times 0.334\right)+\left(2\times 0.408\right)+\left(3\times 0.166\right) \\ =1.648 \\ \approx 1.6 \end{gathered}$$Thus, the mean value of the random variable x is 1.6.

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.

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0.091\right)+\left(1^2\times 0.334\right)+\left(2^2\times 0.408\right)+\left(3^2\times 0.166\right) \\ =3.46 \end{gathered}$$

The standard deviation is given as

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{3.46-{1.648}^2} \\ =0.8626 \\ \approx 0.9 \end{gathered}$$

Thus, the standard deviation of x is 0.9.