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Math  /  Data & Statistics

QuestionQuestion \#2 (pg. 183) Compute the mean and variance of the following discrete probability distribution. \begin{tabular}{|cc|} \hline X\mathbf{X} & P(x)\mathbf{P}(\mathbf{x}) \\ \hline 2 & .5 \\ 8 & .3 \\ 10 & .2 \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. The random variable X X takes on values 2,8, 2, 8, and 10 10 .
2. The probabilities associated with each value of X X are P(2)=0.5 P(2) = 0.5 , P(8)=0.3 P(8) = 0.3 , and P(10)=0.2 P(10) = 0.2 .
3. The mean of a discrete probability distribution is calculated using the formula: \mu = \sum (x \cdot P(x)) \]
4. The variance of a discrete probability distribution is calculated using the formula: \sigma^2 = \sum ((x - \mu)^2 \cdot P(x)) \]

STEP 2

1. Compute the mean of the distribution.
2. Compute the variance of the distribution.

STEP 3

Calculate the mean (μ\mu):
μ=(2⋅0.5)+(8⋅0.3)+(10⋅0.2)\mu = (2 \cdot 0.5) + (8 \cdot 0.3) + (10 \cdot 0.2)

STEP 4

Perform the calculations:
μ=1+2.4+2=5.4\mu = 1 + 2.4 + 2 = 5.4

STEP 5

Calculate the variance (σ2\sigma^2):
σ2=((2−5.4)2⋅0.5)+((8−5.4)2⋅0.3)+((10−5.4)2⋅0.2)\sigma^2 = ((2 - 5.4)^2 \cdot 0.5) + ((8 - 5.4)^2 \cdot 0.3) + ((10 - 5.4)^2 \cdot 0.2)

STEP 6

Perform the calculations:
σ2=((−3.4)2⋅0.5)+((2.6)2⋅0.3)+((4.6)2⋅0.2)\sigma^2 = ((-3.4)^2 \cdot 0.5) + ((2.6)^2 \cdot 0.3) + ((4.6)^2 \cdot 0.2)
σ2=(11.56⋅0.5)+(6.76⋅0.3)+(21.16⋅0.2)\sigma^2 = (11.56 \cdot 0.5) + (6.76 \cdot 0.3) + (21.16 \cdot 0.2)
σ2=5.78+2.028+4.232=12.04\sigma^2 = 5.78 + 2.028 + 4.232 = 12.04
The mean is:
μ=5.4 \mu = 5.4
The variance is:
σ2=12.04 \sigma^2 = 12.04

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