How do you compute the variance of the probability distribution in the table provided? OutcomeProbability500.5510.2520.1530.2?

Var(X)=1.4&nbsp;Explanation:&nbsp;Let X be the random variable that represents a potential result:x50515253P(X=x)0.50.20.10.2&nbsp;First we quickly check that&nbsp;∑P(x)=1&nbsp;which is indeed the case.&nbsp;The, we calculate&nbsp;x2,xP(x),and&nbsp;x2P(x):&nbsp;x50515253P(X=x)0.50.20.10.2xP(X=x)2510.25.210.6x22500260127042809x2P(X=x)1250520.2270.4561.8&nbsp;So then the Expectation is calculated using:&nbsp;E(X)=∑xP(x)&nbsp;=25+10.2+5.2+10.6&nbsp;=51&nbsp;Next prior to calculating the Variance we calculate&nbsp;E(X2):&nbsp;E(X2)=∑x2P(x)&nbsp;=1250+520.2+270.4+561.8&nbsp;=2602.4&nbsp;After that, we can determine the variance:Var(X)=E(X2)−E2(X)&nbsp;=2602.4−(51)2&nbsp;=2602.4−2601&nbsp;=1.4&nbsp;We can also calculate the Standard Deviation (if required); as&nbsp;σ2=Var(X)RIghtarrowσ=1.18&nbsp;(3sf)

Expert answers to "How do you compute the variance of the..."

Aryan Phillips

Answered question

2022-02-12

How do you compute the variance of the probability distribution in the table provided?

\begin{array}{cc} Outcome & Probability \\ 50 & 0.5 \\ 51 & 0.2 \\ 52 & 0.1 \\ 53 & 0.2 ? \end{array}

Answer & Explanation

Hannah Escobar

Beginner2022-02-13Added 9 answers

$V a r (X) = 1.4$
Explanation:
Let X be the random variable that represents a potential result:
$\begin{array}{cc} x & 50 & 51 & 52 & 53 \\ P (X = x) & 0.5 & 0.2 & 0.1 & 0.2 \end{array}$
First we quickly check that $\sum P (x) = 1$ which is indeed the case.
The, we calculate $x^{2}, x P (x), and x^{2} P (x)$ :
$\begin{array}{cc} x & 50 & 51 & 52 & 53 \\ P (X = x) & 0.5 & 0.2 & 0.1 & 0.2 \\ x P (X = x) & 25 & 10.2 & 5.2 & 10.6 \\ x^{2} & 2500 & 2601 & 2704 & 2809 \\ x^{2} P (X = x) & 1250 & 520.2 & 270.4 & 561.8 \end{array}$
So then the Expectation is calculated using:
$E (X) = \sum x P (x)$
$= 25 + 10.2 + 5.2 + 10.6$
$= 51$
Next prior to calculating the Variance we calculate $E (X^{2})$ :
$E (X^{2}) = \sum x^{2} P (x)$
$= 1250 + 520.2 + 270.4 + 561.8$
$= 2602.4$
After that, we can determine the variance:
$V a r (X) = E (X^{2}) - E^{2} (X)$
$= 2602.4 - {(51)}^{2}$
$= 2602.4 - 2601$
$= 1.4$
We can also calculate the Standard Deviation (if required); as
$σ^{2} = V a r (X) R I g h t a r r o w σ = 1.18$ (3sf)

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