To compute the mean and standard deviation for the discrete probability distribution, we will follow these steps:
(a) Finding the Mean (Expected value)
The mean (expected value) of a discrete random variable can be calculated using the formula:
\[ \mu = E(X) = \sum (x_i \cdot p(x_i)) \]
where \(x_i\) are the values of the random variable and \(p(x_i)\) are the corresponding probabilities.
Given the values:
- \( x = -4, -1, 6, 8, 9 \)
- \( p(x) = 0.24, 0.14, 0.27, 0.26, 0.09 \)
Now we will calculate \(E(X)\):
\[ E(X) = (-4 \cdot 0.24) + (-1 \cdot 0.14) + (6 \cdot 0.27) + (8 \cdot 0.26) + (9 \cdot 0.09) \]
Calculating each term:
- \(-4 \cdot 0.24 = -0.96\)
- \(-1 \cdot 0.14 = -0.14\)
- \(6 \cdot 0.27 = 1.62\)
- \(8 \cdot 0.26 = 2.08\)
- \(9 \cdot 0.09 = 0.81\)
Now sum these values:
\[ E(X) = -0.96 - 0.14 + 1.62 + 2.08 + 0.81 = 3.41 \]
Thus, the mean is:
\[ \mu = 3.41 \]
(b) Finding the Standard Deviation
The standard deviation can be calculated using the formula:
\[ \sigma = \sqrt{E(X^2) - (E(X))^2} \]
First, we need to calculate \(E(X^2)\):
\[ E(X^2) = \sum (x_i^2 \cdot p(x_i)) \]
Calculating \(x_i^2\) values:
- \((-4)^2 = 16\)
- \((-1)^2 = 1\)
- \(6^2 = 36\)
- \(8^2 = 64\)
- \(9^2 = 81\)
Now we calculate \(E(X^2)\):
\[ E(X^2) = (16 \cdot 0.24) + (1 \cdot 0.14) + (36 \cdot 0.27) + (64 \cdot 0.26) + (81 \cdot 0.09) \]
Calculating each term:
- \(16 \cdot 0.24 = 3.84\)
- \(1 \cdot 0.14 = 0.14\)
- \(36 \cdot 0.27 = 9.72\)
- \(64 \cdot 0.26 = 16.64\)
- \(81 \cdot 0.09 = 7.29\)
Now sum these values:
\[ E(X^2) = 3.84 + 0.14 + 9.72 + 16.64 + 7.29 = 37.63 \]
Now we can calculate the variance:
\[ \sigma^2 = E(X^2) - (E(X))^2 = 37.63 - (3.41)^2 = 37.63 - 11.6281 = 25.9999 \]
Now, take the square root to find the standard deviation:
\[ \sigma = \sqrt{25.9999} \approx 5.099 \]
So rounded to three decimal places, the standard deviation is:
\[ \sigma \approx 5.099 \]
The final answers are:
- (a) Mean: \(3.410\)
- (b) Standard Deviation: \(5.099\)
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