Question

look at the table on texts assigned to students.
texts assigned to students
\tpoetry\tprose\ttotal
fiction\t0.2\t0.8\t1.0
non - fiction\ta\t0.9\t1.0
total\t0.18\t0.82\t1.0
which value for a completes the conditional relative frequency table by row?
0.01
0.02
0.1
0.2

Explanation:

Step1: Recall row total property

In a conditional relative frequency table by row, the sum of the relative frequencies in each row should equal 1.0 (since it's a total for the row). For the Non - Fiction row, we know the relative frequency for Prose is 0.9 and the total for the row is 1.0.

Step2: Solve for \(a\)

Using the formula for the sum of relative frequencies in a row: \(a + 0.9=1.0\). To find \(a\), we subtract 0.9 from both sides of the equation. So, \(a = 1.0 - 0.9=0.1\)? Wait, no, wait. Wait, also we can use the column total. The total for the Poetry column is 0.18. The Poetry relative frequency for Fiction is 0.2 and for Non - Fiction is \(a\). Let the number of Fiction books be \(f\) and Non - Fiction be \(n\). The relative frequency of Poetry in Fiction is \(0.2=\frac{\text{Poetry and Fiction}}{\text{Fiction total}}\), and in Non - Fiction is \(a = \frac{\text{Poetry and Non - Fiction}}{\text{Non - Fiction total}}\). The total Poetry is \(0.2\times f+a\times n = 0.18\times(f + n)\). But since the row totals for Fiction and Non - Fiction are 1.0 (so \(f\) and \(n\) are the counts relative to their rows, so we can think of the total number of Fiction and Non - Fiction as some counts, but maybe easier: from the row perspective, for Non - Fiction row, \(a+0.9 = 1.0\) gives \(a = 0.1\)? Wait, but let's check with column. The Poetry column: Fiction has 0.2 (so proportion of Fiction that is Poetry is 0.2) and Non - Fiction has \(a\) (proportion of Non - Fiction that is Poetry is \(a\)). The total Poetry is 0.18. Let’s assume the number of Fiction books is \(x\) and Non - Fiction is \(y\). Then \(0.2x + ay=0.18(x + y)\). But from the row totals, \(x\) (Fiction total) and \(y\) (Non - Fiction total) are such that the row totals are 1.0, so we can consider \(x\) and \(y\) as the number of units in each row. But actually, the conditional relative frequency by row means that for each row, the sum is 1.0. So for Non - Fiction row: \(a+0.9 = 1.0\) => \(a = 0.1\)? Wait, but let's check the column. If \(a = 0.1\), then the Poetry column: 0.2 (Fiction) + 0.1 (Non - Fiction) = 0.3, but the column total is 0.18. Oh, I made a mistake. So we need to use the column total. Let the total number of students (or texts) be \(N\). Let the number of Fiction texts be \(F\) and Non - Fiction be \(NF\). Then the number of Poetry texts is \(0.2F + aNF=0.18(N)\), and \(F + NF=N\). Also, from the row totals, \(F\) (Fiction total) is such that the relative frequency of Fiction is some value, but actually, the table is a conditional relative frequency table by row, so each row sums to 1.0 (relative to the row). So the first row (Fiction) has Poetry: 0.2, Prose: 0.8, total 1.0. Second row (Non - Fiction) has Poetry: \(a\), Prose: 0.9, total 1.0. But the column totals are the relative frequencies across all rows. So the Poetry column total is the sum of the Poetry relative frequencies from each row, weighted by the proportion of each row in the total. Wait, maybe the correct way: in a conditional relative frequency table by row, each cell is (frequency in cell)/(row total). So for the Non - Fiction row, \(a=\frac{\text{Poetry and Non - Fiction}}{\text{Non - Fiction total}}\), and Prose and Non - Fiction is 0.9=\(\frac{\text{Prose and Non - Fiction}}{\text{Non - Fiction total}}\). So \(\text{Poetry and Non - Fiction}+\text{Prose and Non - Fiction}=\text{Non - Fiction total}\), so \(a + 0.9=1.0\), so \(a = 0.1\)? But wait, the column total for Poetry is 0.18. Let's see: if Fiction row has Poetry 0.2 (so if there are \(x\) Fiction texts, 0.2x are Poetry) and Non - Fiction row has \(a\) (so \(ay\) are Poetry, where \(y\) is Non - Fiction texts). The total Poetry is \(0.2x+ay = 0.18(x + y)\). Also, the total Prose is \(0.8x + 0.9y=0.82(x + y)\). Let's solve the Prose equation: \(0.8x+0.9y=0.82x + 0.82y\) => \(0.9y - 0.82y=0.82x - 0.8x\) => \(0.08y = 0.02x\) => \(x = 4y\). Now substitute into the Poetry equation: \(0.2(4y)+ay=0.18(4y + y)\) => \(0.8y+ay = 0.18\times5y\) => \(0.8 + a=0.9\) (dividing both sides by \(y\), \(y eq0\)) => \(a = 0.1\). Wait, so that works. So \(a = 0.1\).

Answer:

0.1