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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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LekaFEV [45]

Answer:

The margin of error is approximately 0.3

Step-by-step explanation:

The following information has been provided;

The sample size, n =225 students

The sample mean number of hours spent studying per week = 20.6

The standard deviation = 2.7

The question requires us to determine the margin of error that would be associated with a 90% confidence level. In constructing confidence intervals of the population mean, the margin of error is defined as;

The product of the associated z-score and the standard error of the sample mean. The standard error of the sample mean is calculated as;

\frac{sigma}{\sqrt{n} }

where sigma is the standard deviation and n the sample size. The z-score associated with a 90% confidence level, from the given table, is 1.645.

The margin of error is thus;

1.645*\frac{2.7}{\sqrt{225}}=0.2961

Therefore, the margin of error is approximately 0.3

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Step-by-step explanation:

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