Question

the random variable w has a geometric distribution with p = 0.25. approximately how far do the values of w typically vary, on average, from the mean of the distribution?
a 1.73
b 2.00
c 3.00
d 3.46
e 4.00

Explanation:

Step1: Recall Geometric Distribution Stats

For a geometric distribution with success probability \( p \), the mean \( \mu = \frac{1 - p}{p} \) (wait, no: correct formula for mean of geometric (number of trials until first success) is \( \mu = \frac{1}{p} \), and standard deviation \( \sigma = \frac{\sqrt{1 - p}}{p} \). Wait, let's confirm: The geometric distribution (modeling number of trials \( X \) until first success) has \( E[X] = \frac{1}{p} \), \( \text{Var}(X) = \frac{1 - p}{p^2} \), so standard deviation \( \sigma = \frac{\sqrt{1 - p}}{p} \).

Step2: Identify the Measure

The question asks "how far do the values of \( W \) typically vary, on average, from the mean" – this is the standard deviation.

Step3: Plug in \( p = 0.25 \)

First, calculate \( 1 - p = 1 - 0.25 = 0.75 \). Then, \( \sqrt{1 - p} = \sqrt{0.75} \approx 0.8660 \). Then, divide by \( p = 0.25 \): \( \sigma = \frac{0.8660}{0.25} \approx 3.464 \approx 3.46 \). Wait, but wait: Wait, maybe I mixed up the geometric distribution definition. Wait, some definitions use geometric as number of failures before first success, with mean \( \frac{1 - p}{p} \). Wait, let's check the problem statement: "geometric distribution with \( p = 0.25 \)". Let's confirm the formula for standard deviation.

Wait, let's re-express: If \( W \) is the number of trials until first success (geometric distribution, parameter \( p \)), then:

Mean \( \mu = \frac{1}{p} \)

Variance \( \text{Var}(W) = \frac{1 - p}{p^2} \)

Standard deviation \( \sigma = \sqrt{\frac{1 - p}{p^2}} = \frac{\sqrt{1 - p}}{p} \)

So with \( p = 0.25 \):

\( \sqrt{1 - 0.25} = \sqrt{0.75} \approx 0.8660 \)

\( \frac{0.8660}{0.25} = 3.464 \approx 3.46 \), which is option D.

Wait, but let's check again. Wait, maybe the problem is using the other geometric definition (number of failures before first success, \( Y \), where \( Y = X - 1 \), \( X \) is trials until success). Then \( E[Y] = \frac{1 - p}{p} \), \( \text{Var}(Y) = \frac{1 - p}{p^2} \), same variance as \( X \). So standard deviation is same. So regardless, the standard deviation formula is \( \frac{\sqrt{1 - p}}{p} \).

So plugging \( p = 0.25 \):

\( \sqrt{1 - 0.25} = \sqrt{0.75} \approx 0.866 \)

\( 0.866 / 0.25 = 3.464 \approx 3.46 \), which matches option D.

Wait, but let's check the answer options. Option D is 3.46. So that's the standard deviation, which is the measure of typical variation from the mean.

Answer:

D. 3.46