Question

college students are randomly selected and arranged in groups of three. the random variable x is the number in the group who say that they take one or more online courses. 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.
c. no, the numerical values of the random variable x are not associated with probabilities.
d. no, not every probability is between 0 and 1 inclusive.
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. μ = 1.6 students (round to one decimal place as needed.)
b. the table does not show a probability distribution.
find the standard deviation of the random variable x. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. σ = students (round to one decimal place as needed.)
b. the table does not show a probability distribution.

Explanation:

Step1: Check probability - distribution requirements

We need to check if $\sum_{x}P(x)=1$ and $0\leq P(x)\leq1$ for all $x$. Here, $P(0) = 0.103$, $P(1)=0.351$, $P(2)=0.399$, $P(3)=0.147$.
\[

$$\begin{align*} \sum_{x = 0}^{3}P(x)&=0.103 + 0.351+0.399 + 0.147\\ &=1 \end{align*}$$

\]
and $0\leq0.103\leq1$, $0\leq0.351\leq1$, $0\leq0.399\leq1$, $0\leq0.147\leq1$. So, it is a probability - distribution.

Step2: Calculate the mean $\mu$

The formula for the mean of a discrete probability distribution is $\mu=\sum_{x}x\cdot P(x)$.
\[

$$\begin{align*} \mu&=0\times0.103+1\times0.351 + 2\times0.399+3\times0.147\\ &=0 + 0.351+0.798 + 0.441\\ &=1.6 \end{align*}$$

\]

Step3: Calculate the variance $\sigma^{2}$

The formula for the variance is $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
\[

$$\begin{align*} (0 - 1.6)^{2}\times0.103&=( - 1.6)^{2}\times0.103=2.56\times0.103 = 0.26368\\ (1 - 1.6)^{2}\times0.351&=( - 0.6)^{2}\times0.351=0.36\times0.351 = 0.12636\\ (2 - 1.6)^{2}\times0.399&=(0.4)^{2}\times0.399 = 0.16\times0.399=0.06384\\ (3 - 1.6)^{2}\times0.147&=(1.4)^{2}\times0.147=1.96\times0.147 = 0.28812 \end{align*}$$

\]
\[

$$\begin{align*} \sigma^{2}&=0.26368+0.12636 + 0.06384+0.28812\\ &=0.742 \end{align*}$$

\]

Step4: Calculate the standard - deviation $\sigma$

The standard - deviation $\sigma=\sqrt{\sigma^{2}}$. So, $\sigma=\sqrt{0.742}\approx0.9$.

Answer:

The mean $\mu = 1.6$ students and the standard deviation $\sigma\approx0.9$ students.