flity Distributions
Question 5, 5.1.12
Part 2 of 3
4 correct
Points: 0 of 1
Close
Groups of adults are randomly selected and arranged in groups of three. The random variable $x$ is the number in the group who say that they would feel comfortable in a self-driving vehicle. 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.
\begin{tabular}{c|c}
\hline $\mathbf{x}$ & $\mathbf{P}(\mathbf{x})$ \\
\hline 0 & 0.352 \\
\hline 1 & 0.432 \\
\hline 2 & 0.188 \\
\hline 3 & 0.028 \\
\hline
\end{tabular}
Does the table show a probability distribution? Select all that apply.
A. Yes, the table shows a probability distribution.
B. No, the random variable x's number values are not associated with probabilities.
C. No, not every probability is between 0 and 1 inclusive.
D. No, the random variable $x$ is categorical instead of numerical.
E. No, the sum of all the probabilities is not equal to 1 .
Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. $\mu=\square$ adult(s) (Round to one decimal place as needed.)
B. The table does not show a probability distribution.

Step 1 ：First, we need to check if the given table represents a probability distribution. A probability distribution must satisfy two conditions: (1) all probabilities must be between 0 and 1 inclusive, and (2) the sum of all probabilities must equal 1.

Step 2 ：Looking at the table, we can see that all probabilities are between 0 and 1 inclusive, so the first condition is satisfied.

Step 3 ：To check the second condition, we sum up all the probabilities: $0.352 + 0.432 + 0.188 + 0.028 = 1$. So, the second condition is also satisfied.

Step 4 ：Therefore, the table does represent a probability distribution.

Step 5 ：Next, we need to find the mean of the random variable $x$. The mean, or expected value, of a discrete random variable is given by the sum of the product of each outcome and its probability. In mathematical terms, if $x$ is a random variable and $P(x)$ is the probability of $x$, then the mean $\mu$ is given by $\mu = \sum xP(x)$.

Step 6 ：Substituting the given values into the formula, we get $\mu = 0*0.352 + 1*0.432 + 2*0.188 + 3*0.028$.

Step 7 ：Calculating the above expression, we get $\mu = 0 + 0.432 + 0.376 + 0.084 = 0.892$.

Step 8 ：So, the mean of the random variable $x$ is $0.892$.

Step 9 ：Finally, we need to find the standard deviation of the random variable $x$. The standard deviation is a measure of the amount of variation or dispersion of a set of values. The standard deviation $\sigma$ of a random variable $x$ with mean $\mu$ is given by $\sigma = \sqrt{\sum (x - \mu)^2 P(x)}$.

Step 10 ：Substituting the given values into the formula, we get $\sigma = \sqrt{(0 - 0.892)^2*0.352 + (1 - 0.892)^2*0.432 + (2 - 0.892)^2*0.188 + (3 - 0.892)^2*0.028}$.

Step 11 ：Calculating the above expression, we get $\sigma = \sqrt{0.281 + 0.043 + 0.042 + 0.066} = \sqrt{0.432}$.

Step 12 ：So, the standard deviation of the random variable $x$ is $\sqrt{0.432}$, or approximately $0.657$ when rounded to three decimal places.