Question

proportion of population
15/64
6/64
1/64
skin pigmentation
which conclusion is supported by the graph?
more than half of the population has the central skin pigmentation.
the lightest and the darkest skin pigmentations are equally common in the population.
the central skin pigmentation makes up more of the population than the rest of the skin pigmentation groups combined.
fewer people have skin pigmentation lighter than the central peak than people who have pigmentation darker
save and exit

Explanation:

Step1: Analyze Option 1

Sum the proportions of non - central groups: Let's assume the central proportion is \(p_c\), and non - central proportions are \(p_1,p_2,\cdots\). From the graph, the non - central proportions are \(\frac{1}{64}+\frac{6}{64}+\frac{15}{64}+\frac{15}{64}+\frac{6}{64}+\frac{1}{64}=\frac{1 + 6+15 + 15+6 + 1}{64}=\frac{44}{64}\). The central proportion: Let's say the total is 64 (from denominators). The central bar's proportion: If we assume the central is the tallest bar. Let's calculate the sum of non - central: as above \(\frac{44}{64}\), so central is \(1-\frac{44}{64}=\frac{20}{64}\)? Wait, no, maybe I misread. Wait the y - axis has 1/64, 6/64, 15/64. Wait the bars: the first (lightest) is 1/64, then 6/64, 15/64, then the central (tallest) which is more than 15/64? Wait no, the graph has a normal curve over the bars. Wait maybe the bars are: lightest (1/64), next (6/64), next (15/64), central (let's say \(x\)), then next (15/64), next (6/64), darkest (1/64). So sum of all: \(1/64+6/64 + 15/64+x+15/64+6/64+1/64=1\). So \(x=1-(1 + 6+15+15+6 + 1)/64=1 - 44/64=20/64\). Now, "More than half of the population has the central skin pigmentation": \(20/64\approx0.3125<0.5\), so this is wrong.

Step2: Analyze Option 2

The lightest skin pigmentation proportion is \(1/64\), the darkest is \(1/64\). So they are equally common. This seems correct. Let's check other options.

Step3: Analyze Option 3

Central proportion is \(20/64\), sum of others is \(44/64\). \(20/64<44/64\), so central is less than others combined. So this is wrong.

Step4: Analyze Option 4

Proportion of lighter than central: \(1/64+6/64 + 15/64=\frac{22}{64}\). Proportion of darker than central: \(15/64+6/64+1/64=\frac{22}{64}\). So they are equal, so "Fewer people have skin pigmentation lighter than the central peak than people who have pigmentation darker" is wrong.

Answer:

B. The lightest and the darkest skin pigmentations are equally common in the population. (Wait, the option labels: the second option is "The lightest and the darkest skin pigmentations are equally common in the population.")