Groups of people aged 15–65 are randomly selected and arranged in groups of six. The random variable xis the number in the group who say that their family and / or partner contribute most to their happiness (based on a Coca-Cola survey). The accompanying table lists

the values of xalong with their corresponding probabilities. Does the table describe a probability distribution? If so, find the mean and standard deviation.

x	P(x)
0	0+
1	0.003
2	0.025
3	0.111
4	0.279
5	0.373
6	0.208

The variable x is the number of members in a group of six who say that their family and/or partner are the highest contributors to their happiness (based on a Coca-Cola survey).

The requirements for a probability distribution table are as follows.

1)The variable x represents the count of members who favor the particular decision in the group of 6. Thus, it is a numerical variable.

2)The sum of the probabilities is computed as

$$\begin{gathered} \sum P\left(x\right)=0+0.003+0.025+...+0.373+0.208 \\ =0.999 \end{gathered}$$

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

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

Therefore, the provided table describes a probability distribution.

The mean for the random variable is computed as

$$\begin{gathered} \mu =\sum \left[x\times P\left(x\right)\right] \\ =\left(0\times 0\right)+\left(1\times 0.003\right)+\left(2\times 0.025\right)+...+\left(6\times 0.208\right) \\ =4.615 \\ \approx 4.6 \end{gathered}$$

Thus, the mean is 4.6.

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 calculation isas follows.

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0\right)+\left(1^2\times 0.003\right)+\left(2^2\times 0.025\right)+...+\left(6^2\times 0.208\right) \\ =22.379 \end{gathered}$$

The standard deviation is given as

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{22.379-{4.6}^2} \\ =1.104 \\ \approx 1.0 \end{gathered}$$

Thus, the standard deviation is 1.0.