a census was conducted that determined the land area and populations of towns in a certain region. the…

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 = \n\npop. > 20,000 pop. < 20,000 total\n< 20 sq. mi. 3 29 32\n> 20 sq. mi. 12 11 23\ntotal 15 40 55\n\npop. > 20,000 pop. < 20,000 total\n< 20 sq. mi. 0.2 0.73 a\n> 20 sq. mi. 0.8 0.28 b\ntotal 1.0 1.0 1.0

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 = \n\npop. > 20,000 pop. < 20,000 total\n< 20 sq. mi. 3 29 32\n> 20 sq. mi. 12 11 23\ntotal 15 40 55\n\npop. > 20,000 pop. < 20,000 total\n< 20 sq. mi. 0.2 0.73 a\n> 20 sq. mi. 0.8 0.28 b\ntotal 1.0 1.0 1.0

Answer

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)