What is the expected value of a hypergeometric distribution?

What is the expected value of a hypergeometric distribution?

• A. E[x] = m1/n
• B. E[x] = nm1/m
• C. E[x] = m/m1
• D. E[x] = nm2/m

Answer: (B)

The expected value of a hypergeometric distribution can be derived from its parameters, which typically include the population size, the number of successes in the population, and the number of draws. In a hypergeometric setting, you have a population of size \( N \) containing \( K \) successes, and you are drawing \( n \) samples without replacement. The formula for the expected value of the hypergeometric distribution is given by: \[ E[X] = n \cdot \frac{K}{N} \] In this context, \( n \) is the number of draws, \( K \) (or \( m1 \)) represents the number of successes in the population, and \( N \) (or \( m \)) is the total population size. This formula aligns perfectly with option B, where \( E[X] = n \cdot \frac{m1}{m} \). The key is recognizing that \( m1 \) signifies the number of successes available for selection while \( m \) represents the total number of items. Thus, multiplying \( n \) (number of draws) by \( \frac{m1}{m} \) gives us the expected number of successes in

The expected value of a hypergeometric distribution can be derived from its parameters, which typically include the population size, the number of successes in the population, and the number of draws. In a hypergeometric setting, you have a population of size ( N ) containing ( K ) successes, and you are drawing ( n ) samples without replacement.

The formula for the expected value of the hypergeometric distribution is given by:

[

E[X] = n \cdot \frac{K}{N}

]

In this context, ( n ) is the number of draws, ( K ) (or ( m1 )) represents the number of successes in the population, and ( N ) (or ( m )) is the total population size.

This formula aligns perfectly with option B, where ( E[X] = n \cdot \frac{m1}{m} ). The key is recognizing that ( m1 ) signifies the number of successes available for selection while ( m ) represents the total number of items. Thus, multiplying ( n ) (number of draws) by ( \frac{m1}{m} ) gives us the expected number of successes in