SL 4.2—Presentation of data

The total number of households is $n=3+12+8+2=25$.

Relative frequency $=\frac{f}{n}$ where $f$ is the frequency of the value of interest, here $f=8$ for 2 cars.

$\frac{8}{25}=0.32$.

So the relative frequency of households owning 2 cars is $0.32$.

$0.08=\frac{2}{25}$ uses the frequency for 3 cars instead of 2 cars.

$0.48=\frac{12}{25}$ uses the frequency for 1 car instead of 2 cars.

$2.00$ incorrectly divides $n$ by $f$ instead of $f$ by $n$.

Only $0.32$ correctly divides the frequency of 2 cars (8) by the total (25).

IB class intervals for continuous data must be written as inequalities with no gaps and no overlaps, e.g. $a\le x<b$.

With $10\le x<20$, a value of exactly 20 falls into the next interval $20\le x<30$, not both.

Every value of $x$ belongs to exactly one interval under this scheme.

Only option B satisfies this requirement.

Gap notation like $10-19$ then $20-29$ creates ambiguity about where a value like 19.5 belongs.

Strict inequalities $10<x<20$ leave the boundary value $x=20$ belonging to no interval.

Using $\le$ on both ends means a value like $x=20$ belongs to two intervals simultaneously, which is invalid.

Only the half-open inequality notation in B assigns every value to exactly one class.