How is the variance of a hypergeometric distribution calculated?

Study for the Society of Actuaries Exam P. Immerse in flashcards and multiple-choice questions, each with hints and explanations. Gear up for your exam success!

Multiple Choice

How is the variance of a hypergeometric distribution calculated?

The variance of a hypergeometric distribution is calculated using the formula that accounts for the finite population and the sampling without replacement characteristic of this distribution. In this context, "n" represents the number of draws (sample size), "m1" is the number of successes in the population, "m" is the total population size, and "m2" is the number of failures, which can be derived as m2 = m - m1. The correct formula for the variance of a hypergeometric distribution reflects the adjustment needed when sampling without replacement. It is given by: \[ \text{var}(X) = n \left(\frac{m_1}{m}\right) \left(\frac{m_2}{m}\right) \left(\frac{m - n}{m - 1}\right) \] This formula incorporates three main components: 1. \(\frac{m_1}{m}\) represents the proportion of successes in the population, 2. \(\frac{m_2}{m}\) represents the proportion of failures in the population, 3. \(\frac{m - n}{m - 1}\) adjusts for the lack of replacement when drawing from a finite

The variance of a hypergeometric distribution is calculated using the formula that accounts for the finite population and the sampling without replacement characteristic of this distribution. In this context, "n" represents the number of draws (sample size), "m1" is the number of successes in the population, "m" is the total population size, and "m2" is the number of failures, which can be derived as m2 = m - m1.

The correct formula for the variance of a hypergeometric distribution reflects the adjustment needed when sampling without replacement. It is given by:

[

\text{var}(X) = n \left(\frac{m_1}{m}\right) \left(\frac{m_2}{m}\right) \left(\frac{m - n}{m - 1}\right)

]

This formula incorporates three main components:

  1. (\frac{m_1}{m}) represents the proportion of successes in the population,

  2. (\frac{m_2}{m}) represents the proportion of failures in the population,

  3. (\frac{m - n}{m - 1}) adjusts for the lack of replacement when drawing from a finite

More practice questions