The frequency distribution below summarizes the home sale prices in the city of Summerhill for the month of June. Determine the width of each class. Round to the nearest whole number as needed.

[tex]\[
\begin{tabular}{r|c}
Sale Price (in thousand [tex]$\$[/tex]$) & Frequency \\
\hline
[tex]$80.0-110.9$[/tex] & 2 \\
[tex]$111.0-141.9$[/tex] & 5 \\
[tex]$142.0-172.9$[/tex] & 7 \\
[tex]$173.0-203.9$[/tex] & 10 \\
[tex]$204.0-234.9$[/tex] & 3 \\
[tex]$235.0-265.9$[/tex] & 1 \\
\end{tabular}
\][/tex]

A. 61
B. 31
C. 30
D. 28



Answer :

To determine the width of each class for the frequency distribution given, follow these steps:

1. Identify the Class Boundaries:
Look at any class interval from the table. The boundaries are the values at the start and the end of the interval. Let's use the first interval, [tex]$80.0-110.9$[/tex].

2. Calculate the Class Width:
Subtract the lower boundary from the upper boundary.
[tex]$[tex]$ \text{Upper boundary} - \text{Lower boundary} = 110.9 - 80.0 $[/tex]$[/tex]

3. Compute the Difference:
[tex]$[tex]$ 110.9 - 80.0 = 30.9 $[/tex]$[/tex]

4. Round to the Nearest Whole Number:
Round [tex]$30.9$[/tex] to the nearest whole number, which gives:
[tex]$[tex]$ 31 $[/tex]$[/tex]

So, the width of each class, rounded to the nearest whole number, is \( 31 \).

Therefore, the correct answer is:
[tex]\[ \boxed{31} \][/tex]