Math  /  Data & Statistics

QuestionCollege students are randomly selected and arranged in groups of three. The random variable xx 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. \begin{tabular}{c|c} \hline x\mathbf{x} & P(x)\mathbf{P ( x )} \\ \hline 0 & 0.102 \\ \hline 1 & 0.356 \\ \hline 2 & 0.396 \\ \hline 3 & 0.146 \\ \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 numerical values of the random variable xx are not associated with probabilities. C. No, the random variable xx is categorical instead of numerical. D. No, the sum of all the probabilities is not equal to 1 . E. No, not every probability is between 0 and 1 inclusive.

Studdy Solution

STEP 1

What is this asking? We're checking if this table is a legit probability distribution, and if so, we need to find its mean and standard deviation! Watch out! A probability distribution needs to sum up to **one** and have probabilities between **zero** and **one**!

STEP 2

1. Check if it's a probability distribution
2. Calculate the mean
3. Calculate the standard deviation

STEP 3

Looking at the table, all the P(x)P(x) values are between 0\textbf{0} and 1\textbf{1}, awesome!

STEP 4

Let's add up all the probabilities: 0.102+0.356+0.396+0.146=1.0000.102 + 0.356 + 0.396 + 0.146 = \textbf{1.000}.
Perfect! It adds up to 1\textbf{1}!

STEP 5

The number of students taking online courses, xx, is definitely a number.
So, we're good here!

STEP 6

Since all the conditions are met, this is a valid probability distribution!

STEP 7

The mean, also called the expected value, is calculated as: μ=xP(x)\mu = \sum x \cdot P(x).
Basically, we multiply each value of xx by its probability and add them all up!

STEP 8

μ=(00.102)+(10.356)+(20.396)+(30.146)\mu = (0 \cdot 0.102) + (1 \cdot 0.356) + (2 \cdot 0.396) + (3 \cdot 0.146) μ=0+0.356+0.792+0.438\mu = 0 + 0.356 + 0.792 + 0.438 μ=1.586\mu = \textbf{1.586} So, the mean is 1.586\textbf{1.586}!
On average, about 1.586\textbf{1.586} students in a group of three take online courses.

STEP 9

The standard deviation is calculated using this formula: σ=[(xμ)2P(x)]\sigma = \sqrt{\sum[(x - \mu)^2 \cdot P(x)]}.
It's a bit more complex, but we can handle it!

STEP 10

First, we find (xμ)2(x - \mu)^2 for each xx: (01.586)2=2.515396(0 - 1.586)^2 = 2.515396 (11.586)2=0.344596(1 - 1.586)^2 = 0.344596(21.586)2=0.171396(2 - 1.586)^2 = 0.171396(31.586)2=1.983796(3 - 1.586)^2 = 1.983796

STEP 11

Now, multiply each squared difference by its corresponding probability: 2.5153960.102=0.2566107922.515396 \cdot 0.102 = 0.256610792 0.3445960.356=0.1227537760.344596 \cdot 0.356 = 0.1227537760.1713960.396=0.0678044160.171396 \cdot 0.396 = 0.0678044161.9837960.146=0.2896933561.983796 \cdot 0.146 = 0.289693356

STEP 12

Add up the results from the previous sub-step: 0.256610792+0.122753776+0.067804416+0.289693356=0.736862340.256610792 + 0.122753776 + 0.067804416 + 0.289693356 = \textbf{0.73686234}

STEP 13

Finally, take the square root of the sum: σ=0.736862340.858\sigma = \sqrt{0.73686234} \approx \textbf{0.858}

STEP 14

The table *does* show a probability distribution.
The correct answer is A.
The mean is 1.586\textbf{1.586} and the standard deviation is approximately 0.858\textbf{0.858}.

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