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.

Groups of adults are randomly selected and arranged in groups of three. The random variable xis the number in the group who say that they would feel comfortable in a self driving vehicle (based on a TE Connectivity survey).

x	P(x)
0	0.358
1	0.439
2	0.179
3	0.024

The probability distribution for the group who say they would feel comfortable in a self-driving vehicle is provided.

The variable x is the number in the group who say that they would feel comfortable in a self-driving vehicle.

The requirements are as follows:

1) The variable x is anumerical random variable.

2) The sum of the probabilities is computed as:

$$\begin{gathered} \sum P\left(x\right)=0.358+0.439+0.179+0.024 \\ =1 \end{gathered}$$

Therefore,the sum of the probabilities is equals to 1.

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:

$$\begin{gathered} \mu =\sum \left[x\times P\left(x\right)\right] \\ =\left(0\times 0.358\right)+\left(1\times 0.439\right)+\left(2\times 0.179\right)+\left(3\times 0.024\right) \\ =0.869 \\ \approx 0.9 \end{gathered}$$

Thus, the mean value of random variable x is 0.9.

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:

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0.358\right)+\left(1^2\times 0.439\right)+\left(2^2\times 0.179\right)+\left(3^2\times 0.024\right) \\ =1.371 \end{gathered}$$

The standard deviation is given as:

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{1.371-{0.869}^2} \\ =0.7848 \\ \approx 0.8 \end{gathered}$$

Thus, the standard deviation of x is 0.8.