Question: Compute the mean and standard deviation of the random variable with the given discrete probability…

Compute the mean and standard deviation of the random variable with the given discrete probability distribution.

\[ \begin{array}{c|c} x & P(x) \\ \hline -6 & 0.17 \\ -2 & 0.07 \\ 2 & 0.24 \\ 4 & 0.28 \\ 6 & 0.24 \\ \end{array} \]

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Part 1 of 2

(a) Find the mean. Round the answer to three decimal places, if necessary.

The mean is

Part 2 of 2

(b) Find the standard deviation. Round the answer to three decimal places, if necessary.

The standard deviation is

Solution

To compute the mean and standard deviation of the random variable with the given discrete probability distribution, follow these steps: Given: \[ \begin{array}{c|c} x & P(x) \\ \hline -6 & 0.17 \\ -2 & 0.07 \\ 0 & 0.24 \\ 4 & 0.28 \\ 6 & 0.24 \\ \end{array} \] (a) Find the mean. Calculate the mean (\(\mu\)) using the formula: \[ \mu = \sum (x \cdot P(x)) \] \[ \mu = (-6 \cdot 0.17) + (-2 \cdot 0.07) + (0 \cdot 0.24) + (4 \cdot 0.28) + (6 \cdot 0.24) \] \[ \mu = -1.02 - 0.14 + 0 + 1.12 + 1.44 \] \[ \mu = 1.40 \] The mean is 1.40. (b) Find the standard deviation. First, calculate the variance (\(\sigma^2\)) using the formula: \[ \sigma^2 = \sum ((x - \mu)^2 \cdot P(x)) \] Calculate each component: \[ (x - \mu)^2 \cdot P(x) = ((-6 - 1.40)^2 \cdot 0.17), ((-2 - 1.40)^2 \cdot 0.07), ((0 - 1.40)^2 \cdot 0.24), ((4 - 1.40)^2 \cdot 0.28), ((6 - 1.40)^2 \cdot 0.24) \] \[ = (56.56 \cdot 0.17), (11.56 \cdot 0.07), (1.96 \cdot 0.24), (6.76 \cdot 0.28), (21.16 \cdot 0.24) \] \[ = 9.6252, 0.8092, 0.4704, 1.8928, 5.0784 \] \[ \sigma^2 = 9.6252 + 0.8092 + 0.4704 + 1.8928 + 5.0784 \] \[ \sigma^2 = 17.876 \] Then, calculate the standard deviation (\(\sigma\)): \[ \sigma = \sqrt{\sigma^2} = \sqrt{17.876} \] \[ \sigma \approx 4.230 \] The standard deviation is 4.230.