Question

a recent report indicated that 22 percent of the households in a certain community speak a language other than english at home. a reporter will randomly select a household from the community until the first household that speaks a language other than english at home is selected. let random variable ( q ) represent the number of attempts needed until the first household that speaks a language other than english at home is selected. the random variable ( q ) has a geometric distribution with ( p = 0.22 ). which of the following is closest to the variance of the random variable?

a ( 0.0484 )
b ( 3.5454 )
c ( 4.0144 )
d ( 4.5455 )
e ( 16.1157 )

Explanation:

Step1: Recall Geometric Distribution Variance Formula

The variance of a geometric distribution (for the number of trials until the first success) is given by $\frac{1 - p}{p^2}$, where $p$ is the probability of success on a single trial.

Step2: Identify the Value of \( p \)

Here, \( p = 0.22 \), so \( 1 - p = 1 - 0.22 = 0.78 \).

Step3: Substitute into the Formula

Calculate the variance: $\frac{0.78}{(0.22)^2}$. First, compute \( (0.22)^2 = 0.0484 \). Then, divide \( 0.78 \) by \( 0.0484 \): $\frac{0.78}{0.0484} \approx 16.1157$? Wait, no, wait. Wait, maybe I mixed up the geometric distribution parameterization. Wait, actually, there are two parameterizations of the geometric distribution: one where \( Q \) is the number of trials until the first success (including the success), and the other where it's the number of failures before the first success. Wait, the formula for the variance when \( Q \) is the number of trials until first success (so \( P(Q = k) = (1 - p)^{k - 1}p \) for \( k = 1, 2, \dots \)) is \( \frac{1 - p}{p^2} \)? Wait, no, let's check again. Wait, the mean of the geometric distribution (number of trials until first success) is \( \frac{1}{p} \), and the variance is \( \frac{1 - p}{p^2} \)? Wait, no, actually, no: the variance of the geometric distribution (for the number of trials until first success) is \( \frac{1 - p}{p^2} \)? Wait, let's recalculate. Wait, \( p = 0.22 \), so \( 1 - p = 0.78 \), \( p^2 = 0.0484 \). Then \( 0.78 / 0.0484 \approx 16.1157 \)? But that's option E. Wait, but maybe I made a mistake. Wait, no, wait, maybe the problem is using the other parameterization: the number of failures before the first success. Wait, no, the problem says "the number of attempts needed until the first household that speaks a language other than English at home is selected". So attempts include the successful one. So \( Q \) is the number of trials until first success, so the variance should be \( \frac{1 - p}{p^2} \). Wait, but let's check the options. Wait, option E is 16.1157, which is 0.78 / 0.0484 ≈ 16.1157. But wait, maybe I messed up the formula. Wait, no, let's check the formula again. The variance of a geometric distribution (for the number of trials until the first success) is \( \frac{1 - p}{p^2} \). So with \( p = 0.22 \), that's \( (1 - 0.22) / (0.22)^2 = 0.78 / 0.0484 ≈ 16.1157 \), which is option E. Wait, but that contradicts my initial thought. Wait, maybe I confused the variance formula. Wait, let's confirm: for the geometric distribution where \( X \) is the number of trials until the first success, the probability mass function is \( P(X = k) = (1 - p)^{k - 1}p \) for \( k = 1, 2, \dots \). The mean is \( \frac{1}{p} \), and the variance is \( \frac{1 - p}{p^2} \). So with \( p = 0.22 \), variance is \( (1 - 0.22) / (0.22)^2 = 0.78 / 0.0484 ≈ 16.1157 \), which is option E. Wait, but let's check the options again. The options are A: 0.0484 (which is \( p^2 \)), B: 3.5454 (which is \( 1/p - 1 \)? Wait, 1/0.22 ≈ 4.5455, minus 1 is 3.5455, which is option B. Wait, maybe the problem is using the geometric distribution for the number of failures before the first success. Let's check that. If \( Y \) is the number of failures before the first success, then \( P(Y = k) = (1 - p)^k p \) for \( k = 0, 1, 2, \dots \). The mean of \( Y \) is \( \frac{1 - p}{p} \), and the variance is also \( \frac{1 - p}{p^2} \)? Wait, no, the variance of \( Y \) (number of failures before first success) is \( \frac{1 - p}{p^2} \), same as the variance of \( X \) (number of trials until first success)? Wait, no, that can't be. Wait, no, actually, the variance of \( X \) (trials until first success) is \( \frac{1 - p}{p^2} \), and the variance of \( Y \) (failures before first success) is also \( \frac{1 - p}{p^2} \)? Wait, no, let's calculate the variance of \( X \). Let \( X \) be the number of trials until first success. Then \( E[X] = \frac{1}{p} \), \( E[X^2] = \frac{2 - p}{p^2} \), so variance \( Var(X) = E[X^2] - (E[X])^2 = \frac{2 - p}{p^2} - \frac{1}{p^2} = \frac{1 - p}{p^2} \). So that's correct. So if \( X \) is the number of trials until first success, variance is \( \frac{1 - p}{p^2} \). So with \( p = 0.22 \), that's \( 0.78 / 0.0484 ≈ 16.1157 \), which is option E. So the correct answer should be E.

Answer:

E. 16.1157