Math  /  Data & Statistics

QuestionLet ×\times be a discrete random variable with the following PMF. Answer questions 1,2 and 3 Px(x)={1k for x=21k for x=118 for x=018 for x=11k for x=20 otherwise P_{x}(x)=\left\{\begin{array}{ll} \frac{1}{k} & \text { for } x=-2 \\ \frac{1}{k} & \text { for } x=-1 \\ \frac{1}{8} & \text { for } x=0 \\ \frac{1}{8} & \text { for } x=1 \\ \frac{1}{k} & \text { for } x=2 \\ 0 & \text { otherwise } \end{array}\right.
1. (1 point) Find the value of kk A. 0.125 B. 0.25 C. 8 D. 1 E. 4
2. (1 point) Find P(1.5<x<0.5)P(-1.5<x<0.5) A. 0.25 B. 1 C. 0.375 D. 5/325 / 32 E. 0.75
3. (1 point) The E(X)E(X) equals: A. 1 B. 2 C. 0 D. 0.125 E. -0.125

The discrete random variable XX takes the values 1,2 and 3 and has cumulative distribution function F(x)\mathrm{F}(\mathrm{x}) given by \begin{tabular}{|c|c|c|c|} \hline X\mathbf{X} & 1 & 2 & 3 \\ \hline F(x)\mathbf{F}(\mathbf{x}) & 0.4 & 0.4 & 1 \\ \hline \end{tabular}
4. (2 points) The variance of XX equals: A. 1.7 B. 0.96 C. 0.91 D. 0.98 E. 5.8 Page 217

Studdy Solution
Calculate the variance:
Var(X)=E(X2)(E(X))2\text{Var}(X) = E(X^2) - (E(X))^2
Var(X)=5.8(2.2)2\text{Var}(X) = 5.8 - (2.2)^2
Var(X)=5.84.84=0.96\text{Var}(X) = 5.8 - 4.84 = 0.96
The variance is:
0.96 \boxed{0.96}

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