Question

if x takes on the values 0,1,2and3 with probabilities 1/125, 12/125, 48/125 and 64/125 E(x) and E (x²)​

Answer 1

Answer:

The Random Variable X Takes On The Values 0,1,2,3 With Respective Probabilities 1/125, 12/125, 48/125, 64/125 Find E(X), E(X2

Answer 2

$\displaystyle \bf E(x) = 2.4 \: \: and \: \: E( {x}^{2} ) =6.24$

Given :

x takes on the values 0 , 1 , 2 and 3 with probabilities 1/125 , 12/125 , 48/125 and 64/125

To find :

The value of E(x) and E(x²)

Solution :

Step 1 of 2 :

Find the value of E(x)

$\displaystyle \sf E( {x}^{} )$

$\displaystyle \sf = \sum \: {x}^{} p(x)$

$\displaystyle \sf = \bigg( {0}^{} \times \frac{1}{125} \bigg) + \bigg( {1}^{} \times \frac{12}{125} \bigg) + \bigg( {2}^{} \times \frac{48}{125} \bigg) + \bigg( {3}^{} \times \frac{64}{125} \bigg)$

$\displaystyle \sf = 0 + \frac{12}{125} + \frac{96}{125} + \frac{192}{125}$

$\displaystyle \sf = \frac{12 + 96 + 192}{125}$

$\displaystyle \sf = \frac{300}{125}$

$\displaystyle \sf = 2.4$

Step 2 of 2 :

Find the value of E(x²)

$\displaystyle \sf E( {x}^{2} )$

$\displaystyle \sf = \sum \: {x}^{2} p(x)$

$\displaystyle \sf = \bigg( {0}^{2} \times \frac{1}{125} \bigg) + \bigg( {1}^{2} \times \frac{12}{125} \bigg) + \bigg( {2}^{2} \times \frac{48}{125} \bigg) + \bigg( {3}^{2} \times \frac{64}{125} \bigg)$

$\displaystyle \sf = 0 + \frac{12}{125} + \frac{192}{125} + \frac{576}{125}$

$\displaystyle \sf = \frac{12 + 192 + 576}{125}$

$\displaystyle \sf = \frac{780}{125}$

$\displaystyle \sf = 6.24$

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