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weeeeeb [17]
3 years ago
5

Consider a game in which a player rolls a number cube to determine the number of points earned. If a player rolls a number great

er than or equal to 4, the number of the roll is added to the total points. Any other roll is deducted from the player’s total. What is the expected value of the points earned on a single roll in this game?
Mathematics
1 answer:
Vanyuwa [196]3 years ago
3 0
E_{roll}= \sum_{i=1}^{n} (x_i)(p_i)
E_{roll}= ((-1)(\frac{1}{6}))+((-2)(\frac{1}{6}))+((-3)(\frac{1}{6}))+((4)(\frac{1}{6}))+((5)(\frac{1}{6}))+((6)(\frac{1}{6}))
E_{roll}= (\frac{-1}{6})+(\frac{-1}{3})+(\frac{-1}{2})+(\frac{2}{3})+(\frac{5}{6})+(1)((6)(\frac{1}{6}))
E_{roll}= \frac{3}{2}


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Question options :

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Step-by-step explanation:

Given the following :

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According to the empirical rule ;

Assuming a normal distribution with x being random variables ;

About 68% of x-values lie between -1 to 1 standard deviation of the mean. With about 95% of the x values lying between - 2 and +2 standard deviation of mean. With 99.7% falling between - 3 to 3 standard deviations from the mean.

Using the empirical rule :

95% will fall between + or - 2 standard deviation of the mean.

Lower limit = - 2(3) = - 6

Upper limit = 2(3) = 6

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0.3y+ z y ​ 0, point, 3, y, plus, start fraction, y, divided by, z, end fraction when y=10y=10y, equals, 10 and z=5z=5z, equals,
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