A) Does the data represent a valid probability distribution?

To determine whether the given data forms a valid probability distribution, the following two conditions must be met:

1. The probabilities $P(x)$ must all be between 0 and 1, inclusive.

   • The given values of $P(x)$ are:
     $P(0) = 0.033, P(1) = 0.153, P(2) = 0.314, P(3) = 0.314, P(4) = 0.153, P(5) = 0.033$.
     All of these probabilities are between 0 and 1, so this condition is satisfied.

2. The sum of all probabilities must equal 1.

   • Let's check the sum of the probabilities: $0.033 + 0.153 + 0.314 + 0.314 + 0.153 + 0.033 = 1.000$ Since the sum equals 1, this condition is also satisfied.

Since both conditions are met, yes, the given data represents a probability distribution.

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B) Find the mean of the random variable $x$.

The mean (or expected value $\mu$) of a discrete random variable is given by the formula: $\mu = \sum [x \cdot P(x)]$

Let's calculate the mean:

$\mu = (0 \cdot 0.033) + (1 \cdot 0.153) + (2 \cdot 0.314) + (3 \cdot 0.314) + (4 \cdot 0.153) + (5 \cdot 0.033)$ $\mu = 0 + 0.153 + 0.628 + 0.942 + 0.612 + 0.165 = 2.5$

Thus, the mean of the random variable $x$ is 2.5.

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C) Find the standard deviation of the random variable $x$.

The standard deviation $\sigma$ is the square root of the variance. The variance $\sigma^2$ is given by: $\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2$

First, calculate $\sum [x^2 \cdot P(x)]$: $\sum [x^2 \cdot P(x)] = (0^2 \cdot 0.033) + (1^2 \cdot 0.153) + (2^2 \cdot 0.314) + (3^2 \cdot 0.314) + (4^2 \cdot 0.153) + (5^2 \cdot 0.033)$ $= 0 + 0.153 + 1.256 + 2.826 + 2.448 + 0.825 = 7.508$

Now, calculate the variance: $\sigma^2 = 7.508 - (2.5)^2 = 7.508 - 6.25 = 1.258$

Finally, calculate the standard deviation: $\sigma = \sqrt{1.258} \approx 1.122$

Thus, the standard deviation is approximately 1.122.

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Summary:

A) The data represents a valid probability distribution.
B) The mean is $2.5$.
C) The standard deviation is approximately $1.122$.

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Do you want details on any of the steps or have additional questions?

Here are 5 related questions to expand on this:

1. How does the concept of expected value relate to real-life decision-making scenarios?
2. What other methods can be used to validate a probability distribution?
3. Can we use similar methods to calculate expected values for continuous random variables?
4. How does changing probabilities affect the mean and standard deviation of a distribution?
5. What role does variance play in understanding the spread of a distribution?

Tip: Variance measures how spread out the values of a random variable are from the mean. The higher the variance, the more spread out the values.