91Ó°ÊÓ

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

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

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

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

The standard deviation of the random variable x is computed as

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

The calculations are as follows.

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

The standard deviation is given as

$$\begin{gathered} \mathrm{σ}=\sqrt{∑\left[x^2·P\left(x\right)\right]-{\mathrm{μ}}^2} \\ =\sqrt{3.46-{1.648}^2} \\ =0.8626 \\ ≈0.9 \end{gathered}$$

Thus, the standard deviation of x is 0.9.