a. To verify P(X) is a probability function we have to show that ∑ X P ( X ) = 1 \sum_{X} P(X) = 1 ∑ X P ( X ) = 1 .
∑ x P ( x ) = P ( 18 ) + P ( 19 ) + P ( 20 ) + P ( 21 ) + P ( 22 ) + P ( 23 ) = 0.15 + 0.10 + 0.32 + 0.05 + 0.13 + 0.25 = 1 \begin{aligned}
\sum_{x} P(x) &= P(18) + P(19) + P(20) + P(21)+P(22)+P(23)\\
&= 0.15+0.10+0.32+0.05+0.13+0.25\\
&=1
\end{aligned} x ∑ P ( x ) = P ( 18 ) + P ( 19 ) + P ( 20 ) + P ( 21 ) + P ( 22 ) + P ( 23 ) = 0.15 + 0.10 + 0.32 + 0.05 + 0.13 + 0.25 = 1
Hence P(X) is a probability function.
e. Variance ( X ) = E ( X 2 ) − ( E ( X ) ) 2 \text{Variance}(X) = E(X^2) - (E(X))^2 Variance ( X ) = E ( X 2 ) − ( E ( X ) ) 2 .
E ( X 2 ) = ∑ x X 2 P ( X ) = 1 8 2 ∗ 0.15 + 1 9 2 ∗ 0.10 + 2 0 2 ∗ 0.32 + 2 1 2 ∗ 0.05 + 2 2 2 ∗ 0.13 + 2 3 2 ∗ 0.25 = 429.92 \begin{aligned}
E(X^2) &= \sum_{x} X^2 P(X)\\
& = 18^2*0.15+19^2*0.10+20^2*0.32~+~\\
& ~~~~~~~~~~~~~21^2*0.05+22^2*0.13+23^2*0.25\\
&=429.92
\end{aligned} E ( X 2 ) = x ∑ X 2 P ( X ) = 1 8 2 ∗ 0.15 + 1 9 2 ∗ 0.10 + 2 0 2 ∗ 0.32 + 2 1 2 ∗ 0.05 + 2 2 2 ∗ 0.13 + 2 3 2 ∗ 0.25 = 429.92
E ( X ) = ∑ x X P ( X ) = 18 ∗ 0.15 + 19 ∗ 0.10 + 20 ∗ 0.32 + 21 ∗ 0.05 + 22 ∗ 0.13 + 23 ∗ 0.25 = 20.66 \begin{aligned}
E(X) &= \sum_{x} X P(X)\\
& = 18*0.15+19*0.10+20*0.32~+~\\
& ~~~~~~~~~~~~~21*0.05+22*0.13+23*0.25\\
&=20.66
\end{aligned} E ( X ) = x ∑ XP ( X ) = 18 ∗ 0.15 + 19 ∗ 0.10 + 20 ∗ 0.32 + 21 ∗ 0.05 + 22 ∗ 0.13 + 23 ∗ 0.25 = 20.66
∴ Variance ( X ) = 429.92 − 20.6 6 2 = 429.92 − 426.8356 = 3.0844 \therefore \text{Variance}(X) = 429.92-20.66^2 = 429.92-426.8356 = 3.0844 ∴ Variance ( X ) = 429.92 − 20.6 6 2 = 429.92 − 426.8356 = 3.0844
f. Standard deviation ( X ) = Var ( X ) = 3.0844 = 1.7562 \text{Standard deviation}(X) = \sqrt{\text{Var}(X)} = \sqrt{3.0844} = 1.7562 Standard deviation ( X ) = Var ( X ) = 3.0844 = 1.7562
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