To determine whether the table represents a valid probability distribution, we must verify if it satisfies the following key requirements:

1. The sum of all probabilities must equal 1.
2. Each probability must be between 0 and 1 inclusive.
3. The random variable $x$ must represent numerical values (not categorical).

Let’s check each requirement with the provided dataTo determine whether the table represents a valid probability distribution, we must verify if it satisfies the following key requirements:

1. The sum of all probabilities must equal 1.
2. Each probability must be between 0 and 1 inclusive.
3. The random variable $x$ must represent numerical values (not categorical).

Let’s check each requirement with the provided

$x$	$P(x)$
0	0.104
1	0.353
2	0.398
3	0.145

Step 1: Check if the probabilities are between 0 and 1

Each of the probabilities provided is within the range [0, 1]:

• $P(0) = 0.104$
• $P(1) = 0.353$
• $P(2) = 0.398$
• $P(3) = 0.145$

Thus, the probabilities satisfy this condition.

Step 2: Check if the sum of the probabilities equals 1

The sum of the probabilities is: $0.104 + 0.353 + 0.398 + 0.145 = 1.000$ The sum is exactly 1, which also satisfies this requirement.

Step 3: Check if $x$ represents numerical values

The random variable $x$ represents the number of students in the group who take online courses, which is a numerical value.

Conclusion

Since all the requirements for a probability distribution are satisfied, the table does represent a valid probability distribution. Therefore, Option A is correct.

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Part 2: Finding the Mean and Standard Deviation

Now, let's calculate the mean $\mu$ and standard deviation $\sigma$.

Mean $\mu$:

The mean of a probability distribution is calculated as: $\mu = \sum x \cdot P(x)$ Using the values in the table: $\mu = (0 \cdot 0.104) + (1 \cdot 0.353) + (2 \cdot 0.398) + (3 \cdot 0.145)$ $\mu = 0 + 0.353 + 0.796 + 0.435 = 1.584$

Standard Deviation $\sigma$:

The standard deviation is given by: $\sigma = \sqrt{\sum (x - \mu)^2 \cdot P(x)}$ First, calculate $(x - \mu)^2 \cdot P(x)$ for each $x$:

$(0 - 1.584)^2 \cdot 0.104 = 1.583^2 \cdot 0.104 = 2.505 \cdot 0.104 = 0.2605$ $(1 - 1.584)^2 \cdot 0.353 = (-0.584)^2 \cdot 0.353 = 0.341 \cdot 0.353 = 0.1203$ $(2 - 1.584)^2 \cdot 0.398 = 0.416^2 \cdot 0.398 = 0.173 \cdot 0.398 = 0.0688$ $(3 - 1.584)^2 \cdot 0.145 = 1.416^2 \cdot 0.145 = 2.006 \cdot 0.145 = 0.291$

Now sum the values and take the square root: $\sigma = \sqrt{0.2605 + 0.1203 + 0.0688 + 0.291} = \sqrt{0.7406} \approx 0.8609$

Final Answer:

• The mean is $\mu \approx 1.584$.
• The standard deviation is $\sigma \approx 0.8609$.

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Do you want more details or have any questions?
Here are 5 related questions:

1. How is the mean of a probability distribution used in practical scenarios?
2. How do you interpret standard deviation in a probability distribution?
3. What is the difference between a probability mass function and a probability density function?
4. How would you calculate the variance of this distribution?
5. What happens if the sum of the probabilities is less than or greater than 1?

Tip: Always check that probabilities sum to 1 before calculating the mean and standard deviation of a distribution.