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.

x	p(x)
1	0.353
2	0.404
3	0.142

c. no, the numerical values of the random variable x are not associated with probabilities.
d. no, the random variable x is categorical instead of numerical.
e. no, not every probability is between 0 and 1 inclusive.
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

All probabilities \(P(x)\) are between \(0\) and \(1\) inclusive (\(0.101\), \(0.353\), \(0.404\), \(0.142\)). Also, \(\sum_{x = 0}^{3}P(x)=0.101 + 0.353+0.404 + 0.142=1\). So, it is a probability - distribution.

Step2: Calculate the mean \(\mu\)

The formula for the mean of a discrete probability distribution is \(\mu=\sum_{x}x\cdot P(x)\). So, \(\mu=0\times0.101 + 1\times0.353+2\times0.404 + 3\times0.142=0 + 0.353+0.808+0.426 = 1.587\approx1.6\).

Step3: Calculate the variance \(\sigma^{2}\)

The formula for the variance is \(\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)\).
\((0 - 1.6)^{2}\times0.101+(1 - 1.6)^{2}\times0.353+(2 - 1.6)^{2}\times0.404+(3 - 1.6)^{2}\times0.142\)
\(=( - 1.6)^{2}\times0.101+( - 0.6)^{2}\times0.353+(0.4)^{2}\times0.404+(1.4)^{2}\times0.142\)
\(=2.56\times0.101 + 0.36\times0.353+0.16\times0.404+1.96\times0.142\)
\(=0.25856+0.12708+0.06464+0.27832 = 0.7286\).

Step4: Calculate the standard - deviation \(\sigma\)

The standard - deviation \(\sigma=\sqrt{\sigma^{2}}\), so \(\sigma=\sqrt{0.7286}\approx0.9\).

Answer:

For the mean: A. \(\mu = 1.6\) students
For the standard deviation: A. \(\sigma=0.9\) students