The formula to calculate the mean (expected value) of a probability distribution is:

$\text{Mean} = \sum (x \cdot P(x))$

where $x$ represents the values, and $P(x)$ represents the probabilities.

Given the dataThe formula to calculate the mean (expected value) of a probability distribution is:

$\text{Mean} = \sum (x \cdot P(x))$

where $x$ represents the values, and $P(x)$ represents the probabilities.

Given the

$\text{Mean} = (0 \cdot 0.2679) + (1 \cdot 0.1986) + (2 \cdot 0.0079) + (3 \cdot 0.3987) + (4 \cdot 0.1269)$

Let me calculate this for you.

$\text{Mean} = (0) + (1 \cdot 0.1986) + (2 \cdot 0.0079) + (3 \cdot 0.3987) + (4 \cdot 0.1269)$

$\text{Mean} = 0 + 0.1986 + 0.0158 + 1.1961 + 0.5076$

Now, summing these values:

$\text{Mean} = 0.1986 + 0.0158 + 1.1961 + 0.5076 = 1.9181$

Thus, the mean of the probability distribution, rounded to one decimal, is:

$\boxed{1.9}$

Would you like more details or have any questions?

Related questions:

1. How do you calculate the variance of a probability distribution?
2. What is the formula for standard deviation in probability distributions?
3. How does the mean differ in a normal distribution versus a skewed distribution?
4. What role does the mean play in the law of large numbers?
5. Can probability distributions have more than one mean value?

Tip:

The mean of a probability distribution gives the expected outcome over the long term if you repeat the random process multiple times.