Question

The probability distribution in the table displays the probability of a certain golf score at theend of a local amateur golf tournament. Find the variance, o?, of the probability distribution.

Answer 1

C) 1.0

Explanation:

Given μ = -0.6

μ = ∑x·p(x)

Using this equation:

σ² = ∑x²·p(x)-μ²

∑x²·p(x) = (-3)² · 0.03 + (-2)² · 0.14 + (-1)² · 0.38 + 0² · 0.3 + 1² · 0.15

∑x²·p(x) = 1.36

σ² = 1.36 - (-0.6)² = 1

Answer: C) 1

Answer 2

The variance formula is:

`\begin{gathered} Variance=\sum_^(x^2)\text{ - }(\sum_^x)^2 \\ \sum_^(x^2)\text{ = }\sum_^x^2p(x)=0.12+0.56+0.38+0+0.15=1.21 \\ \sum_^x=\sum_^xp(x)=\text{ -0.09 - 0.28 -0.38+0+0.15= -0.6} \\ \\ Var\imaginaryI ance=\sum_^(x^2)\text{ -\lparen}\sum_^x)^2 \\ Variance=1.21\text{ - \lparen-0.6\rparen}^2 \\ Variance=1.21\text{ - }0.36 \\ Variance=0.8 \end{gathered}`