Question

the chart shows the distribution of nationalities of students at a university. in total, there are 11 nationalities. in the other portion of the pie chart, there is an equal number of students within each nationality. one of the nationalities in the other portion is colombian. what is the probability that a randomly selected student is colombian?

Explanation:

Step1: Calculate the percentage of 'Other'

The sum of all percentages should be \(100\%\).
Let's find the percentage of 'Other' by subtracting the sum of the known percentages from \(100\%\).
The known percentages are: American \(56.0\%\), Canadian \(15.0\%\), Mexican \(5.0\%\), British \(10.5\%\), Australian \(8.3\%\).
The sum of known percentages is \(56.0 + 15.0+5.0 + 10.5+8.3=94.8\%\)
The percentage of 'Other' is \(100 - 94.8 = 5.2\%\)

Step2: Determine the number of nationalities in 'Other'

The total number of nationalities is \(11\). The number of non - 'Other' nationalities is \(5\) (American, Canadian, Mexican, British, Australian). So the number of nationalities in 'Other' is \(11 - 5=6\)

Step3: Calculate the probability for Colombian (since nationalities in 'Other' are equally - distributed)

If the percentage of 'Other' is \(5.2\%\) and there are \(6\) nationalities in 'Other' (with equal number of students in each), then the probability \(P\) that a student is Colombian (one of the 'Other' nationalities) is \(\frac{5.2\%}{6}\)

$$P=\frac{5.2}{6\times100}=\frac{52}{6\times1000}=\frac{13}{1500}\approx0.0087$$

Answer:

\(\frac{13}{1500}\approx0.0087\)