The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
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$$\begin{array} { | l | l | l | l | l | l | l | l | l | l | }
\hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \
\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \
\hline
\end{array}$$ What is the standard deviation of this distribution?

Question

The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. Let's call X the number of color-blind individuals in a random sample of 8 Caucasian American males. The table gives the probability distribution for X.
?
?
$$\begin{array} { | l | l | l | l | l | l | l | l | l | l | } 
\hline X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \
\hline P ( X ) & 0.513 & 0.357 & 0.109 & 0.019 & 0.002 & 0.000 & 0.000 & 0.000 & 0.000 \
\hline
\end{array}$$ What is the standard deviation of this distribution?

Quizplus · Accepted Answer

The formula for standard deviation of a probability distribution is:
$$\sigma = \sqrt{\sum(x-\mu)^2P(x)}$$
where $\mu$ is the mean of the distribution.
In this case, the probability distribution is given and we can calculate the mean as:
$$\mu = \sum(x \cdot P(x)) = (0 \cdot 0.513) + (1 \cdot 0.357) + (2 \cdot 0.109) + (3 \cdot 0.019) + (4 \cdot 0.002) + (5 \cdot 0) + (6 \cdot 0) + (7 \cdot 0) + (8 \cdot 0) = 1.016$$
Now we can plug in the values and calculate the standard deviation:
$$\sigma = \sqrt{(0-1.016)^2 \cdot 0.513 + (1-1.016)^2 \cdot 0.357 + (2-1.016)^2 \cdot 0.109 + (3-1.016)^2 \cdot 0.019 + (4-1.016)^2 \cdot 0.002} \approx 0.766$$
Therefore, the answer is B, 0.766.

The Frequency of Color Blindness (Dyschromatopsia) in the Caucasian American