Question

look at the table on texts assigned to students.
texts assigned to students

	poetry	prose	total
non - fiction	a	0.9	1.0
total	0.18	0.82	1.0

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 entries in each row should be equal to the row total (which is 1.0 here for both Fiction and Non - Fiction rows). For the Non - Fiction row, we know that the sum of the Poetry and Prose relative frequencies should be 1.0. So, we can use the equation \(a + 0.9=1.0\).

Step2: Solve for \(a\)

To find the value of \(a\), we subtract 0.9 from both sides of the equation \(a + 0.9 = 1.0\). So, \(a=1.0 - 0.9=0.1\)? Wait, no, wait. Wait, we can also 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 texts be \(f\) and Non - Fiction texts be \(n\). The conditional relative frequency for Fiction - Poetry is \(\frac{\text{Fiction - Poetry}}{\text{Fiction Total}} = 0.2\), so Fiction - Poetry \(= 0.2f\). The conditional relative frequency for Non - Fiction - Poetry is \(\frac{\text{Non - Fiction - Poetry}}{\text{Non - Fiction Total}}=a\), so Non - Fiction - Poetry \(=an\). The total Poetry is \(0.2f + an=0.18(f + n)\) (since total is \(f + n\)). Also, from the rows, Fiction Total is \(f\) (so \(f\) is the count for Fiction, and Non - Fiction Total is \(n\)). But maybe easier: in the row for Non - Fiction, the sum of Poetry and Prose is 1.0. Wait, no, the row total is 1.0, so \(a+0.9 = 1.0\) gives \(a = 0.1\)? But wait, let's check the column. The total Poetry is 0.18. The Fiction - Poetry is 0.2 (relative to Fiction row), let's assume the number of Fiction students is \(x\) and Non - Fiction is \(y\). Then the number of Poetry in Fiction is \(0.2x\), in Non - Fiction is \(ay\). The total Poetry is \(0.2x+ay = 0.18(x + y)\). Also, from the rows, Fiction total is \(x\) (so \(x\) is the count, and Non - Fiction total is \(y\)). But also, the Prose column: Fiction - Prose is \(0.8x\), Non - Fiction - Prose is \(0.9y\), total Prose is \(0.8x + 0.9y=0.82(x + y)\). Let's solve \(0.8x+0.9y = 0.82x + 0.82y\). Subtract \(0.8x+0.82y\) from both sides: \(0.08y=0.02x\), so \(x = 4y\). Now, total Poetry: \(0.2x+ay = 0.18(x + y)\). Substitute \(x = 4y\): \(0.2(4y)+ay=0.18(4y + y)\) => \(0.8y+ay = 0.18\times5y\) => \(0.8 + a=0.9\) (divide both sides by \(y\), \(y
eq0\)) => \(a = 0.1\). Wait, but earlier when we did row sum, \(a + 0.9=1.0\) gives \(a = 0.1\). So that's correct.

Wait, no, wait, I made a mistake. Wait, the row for Non - Fiction: the relative frequencies are conditional on the row. So the sum of the two cells in the Non - Fiction row (Poetry and Prose) should be equal to the row total, which is 1.0. So \(a+0.9 = 1.0\), so \(a=1.0 - 0.9 = 0.1\). Wait, but let's check with the column. The total Poetry is 0.18. The Fiction row has 0.2 (Poetry) and Non - Fiction has \(a\) (Poetry). Let the proportion of Fiction in the total be \(p\) and Non - Fiction be \(1 - p\). Then \(0.2p+a(1 - p)=0.18\). Also, from the Prose column: \(0.8p + 0.9(1 - p)=0.82\). Let's solve \(0.8p+0.9 - 0.9p=0.82\) => \(- 0.1p=0.82 - 0.9=-0.08\) => \(p = 0.8\). Then \(1 - p = 0.2\). Now substitute \(p = 0.8\) into the Poetry column equation: \(0.2\times0.8+a\times0.2 = 0.18\) => \(0.16+0.2a = 0.18\) => \(0.2a=0.02\) => \(a = 0.1\). Yes, so that's correct.

Answer:

\(0.1\) (corresponding to the option "0.1")