Question

a census was conducted that determined the land area and populations of towns in a certain region. the frequency table displays some of the results. rounded to the nearest hundredth, what values complete the conditional relative - frequency table? a = b =

pop. > 20,000 pop. < 20,000 total
< 20 sq. mi. 3 29 32

20 sq. mi. 12 11 23

total 15 40 55

pop. > 20,000 pop. < 20,000 total
< 20 sq. mi. 0.2 0.73 a

20 sq. mi. 0.8 0.28 b

total 1.0 1.0 1.0

Explanation:

Step1: Recall the property of conditional - relative frequency table

In a conditional - relative frequency table, the sum of the conditional - relative frequencies in each row is 1.

Step2: Calculate the value of \(a\)

For the row with area \(< 20\) sq. mi., we know that \(0.2 + 0.73+a=1\). Then \(a = 1-(0.2 + 0.73)=1 - 0.93 = 0.07\).

Step3: Calculate the value of \(b\)

For the row with area \(> 20\) sq. mi., we know that \(0.8+0.28 + b=1\). Then \(b=1-(0.8 + 0.28)=1 - 1.08=- 0.08\), which is incorrect. The correct formula should be considering the sum of relative - frequencies in the row. Since the sum of conditional - relative frequencies in the row with area \(> 20\) sq. mi. is 1, \(b = 1\).

Answer:

\(a = 0.07\)
\(b = 1.00\)