college students are randomly selected and arranged in groups of three. the random variable x is the number…

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.\n| x | p(x) |\n|----|----| \n| 0 | 0.101 |\n| 1 | 0.353 |\n| 2 | 0.404 |\n| 3 | 0.142 |\nc. no, the numerical values of the random variable x are not associated with probabilities.\nd. no, the random variable x is categorical instead of numerical.\ne. no, not every probability is between 0 and 1 inclusive.\nfind the mean of the random variable x. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. μ = 1.6 students (round to one decimal place as needed.)\nb. the table does not show a probability distribution.\nfind 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.\na. σ = students (round to one decimal place as needed.)\nb. the table does not show a probability distribution.

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.\n| x | p(x) |\n|----|----| \n| 0 | 0.101 |\n| 1 | 0.353 |\n| 2 | 0.404 |\n| 3 | 0.142 |\nc. no, the numerical values of the random variable x are not associated with probabilities.\nd. no, the random variable x is categorical instead of numerical.\ne. no, not every probability is between 0 and 1 inclusive.\nfind the mean of the random variable x. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. μ = 1.6 students (round to one decimal place as needed.)\nb. the table does not show a probability distribution.\nfind 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.\na. σ = students (round to one decimal place as needed.)\nb. the table does not show a probability distribution.

Answer

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