Assume that a discrete random variable has the values of 5, 10, 15, 20, 25, and
ID: 3024283 • Letter: A
Question
Assume that a discrete random variable has the values of 5, 10, 15, 20, 25, and 30. Furthermore, assume that the value 30 has probability of 0.50 associated with it, but each of the other values is equally likely.
What is the probability distribution?
What is the expected value of this variable?
What is the standard deviation of this variable?
What is the probability that this variable is 20 or more?
What is the probability that this variable is more than 20?
What is the probability that such a random variable has the value of 20 or less?
What is the probability that the variable has values between 18 and 28?
What is the interval of values that are within one standard deviation of the mean?
What is the probability that this variable has values within one standard deviation of the mean?
Explanation / Answer
a)
As the probabilities must sum up to 1, then the first 5 values of x must have p = 0.10. Hence,
[ANSWER]
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b)
Consider:
Thus,
E(x) = Expected value = mean = 22.5 [ANSWER]
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c)
As
Var(x) = E(x^2) - E(x)^2 = 81.25
Then
s(x) = sqrt [Var(x)] = 9.013878189 [ANSWER]
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d)
P(20 or more) = P(20) + P(25) + P(30) = 0.1+0.1+0.5 = 0.7 [ANSWER]
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e)
P(more than 20) = P(25) + P(30) = 0.1+0.5 = 0.6 [ANSWER]
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f)
P(20 or less) = P(5) +P(10) + P(15) + P(20) = 0.1+0.1+0.1+0.1 = 0.4 [ANSWER]
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g)
P(between 20 and 28) = P(20) + P(25) = 0.1+0.1 = 0.2 [ANSWER]
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h)
Adding/subtracting one standard deviation from the mean, then it is between
13.48612181 and 31.51387819 [ANSWER]
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i)
P(between 13.49 and 31.51) = P(between 15 and 30) = P(15) + P(20) + P(25) + P(30)
= 0.1+0.1+0.1+0.5
=0.8 [ANSWER]
x P(x) 5 0.1 10 0.1 15 0.1 20 0.1 25 0.1 30 0.5Related Questions
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