91Ó°ÊÓ

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:

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

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:

$$\begin{gathered} \mathrm{μ}=∑\left[x×P\left(x\right)\right] \\ =\left(0×0.358\right)+\left(1×0.439\right)+\left(2×0.179\right)+\left(3×0.024\right) \\ =0.869 \\ ≈0.9 \end{gathered}$$

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

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

$\mathrm{σ}=\sqrt{∑\left[x^2×P\left(x\right)\right]-{\mathrm{μ}}^2}$

The calculations are as follows:

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

The standard deviation is given as:

$$\begin{gathered} \mathrm{σ}=\sqrt{∑\left[x^2·P\left(x\right)\right]-{\mathrm{μ}}^2} \\ =\sqrt{1.371-{0.869}^2} \\ =0.7848 \\ ≈0.8 \end{gathered}$$

Thus, the standard deviation of x is 0.8.