Questions: Find the mean of the following probability distribution? x P(x) 0 0.2796 1 0.328 2 0.0914 3 0.0689 4 0.2321 Round your answer to 1 decimal place. mean =

Solution Steps

To find the mean of the probability distribution, we use the formula:

\[ \text{Mean} = \sum_{i=0}^{n} x_i \cdot P(x_i) \]

Substituting the values from the distribution:

\[ \text{Mean} = 0 \times 0.2796 + 1 \times 0.328 + 2 \times 0.0914 + 3 \times 0.0689 + 4 \times 0.2321 \]

Calculating this gives:

\[ \text{Mean} = 0 + 0.328 + 0.1828 + 0.2067 + 0.9284 = 1.645 \]

Rounding to one decimal place, we find:

\[ \text{Mean} = 1.6 \]

The variance is calculated using the formula:

\[ \text{Variance} = \sigma^2 = \sum_{i=0}^{n} (x_i - \text{Mean})^2 \cdot P(x_i) \]

Substituting the mean value and the probabilities:

\[ \text{Variance} = (0 - 1.6)^2 \times 0.2796 + (1 - 1.6)^2 \times 0.328 + (2 - 1.6)^2 \times 0.0914 + (3 - 1.6)^2 \times 0.0689 + (4 - 1.6)^2 \times 0.2321 \]

Calculating each term:

\[ = (2.56 \times 0.2796) + (0.36 \times 0.328) + (0.16 \times 0.0914) + (1.96 \times 0.0689) + (5.76 \times 0.2321) \]

This results in:

\[ = 0.715776 + 0.11808 + 0.014624 + 0.135964 + 1.336256 = 2.3201 \]

Rounding to one decimal place, we find:

\[ \text{Variance} = 2.3 \]

The standard deviation is the square root of the variance:

\[ \text{Standard Deviation} = \sigma = \sqrt{\text{Variance}} = \sqrt{2.3} \approx 1.514 \]

Rounding to one decimal place, we find:

\[ \text{Standard Deviation} = 1.5 \]

Final Answer