X is a hypergeometric random variable The population size is

X is a hypergeometric random variable. The population size is 90 and there are 20 successes in the population. The sample size is 29. What is the expectation of X, rounded to two decimal places?

X is a continuous random variable with cumulative distribution function F(x). If F(0) = 0.058, F(2) = 0.441, and F(5) = 0.592, what is the probability that 2 < X < 5? Calculate your result rounded to three decimal places.

X is a negative binomial random variable with the probability of success on each trial = 0.74 and the required number of successes = 9. Calculate the expectation of X, rounded to two decimal places.

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Solution

1)

X is a hypergeometric random variable. The population size is 90 and there are 20 successes in the population. The sample size is 29. What is the expectation of X, rounded to two decimal places?

Note that

E(x) = n k / N

where

n = sample size = 29
k = number of successes in the population = 20
N = population size = 90

Thus,

E(x) = 29*20/90 = 6.44 [answer]

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2)

F(2<x<5) = F(5) - F(2) = 0.592 - 0.441 = 0.151 [answer]

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3)

Here,

E(x) = k / p

where

k = number of successes
p = probability of a success
x = the number of trials

Hence,

E(x) = 9 / 0.74 = 12.16 [answer]