The table shows the age, in years, of employees in a company.

\begin{tabular}{|c|c|}
\hline Age [tex]$(a)$[/tex] in years & Frequency \\
\hline [tex]$18 \leqslant a\ \textless \ 20$[/tex] & 3 \\
\hline [tex]$20 \leqslant a\ \textless \ 22$[/tex] & 2 \\
\hline [tex]$22 \leqslant a\ \textless \ 24$[/tex] & 7 \\
\hline [tex]$24 \leqslant a\ \textless \ 26$[/tex] & 8 \\
\hline [tex]$26 \leqslant a$[/tex] & 0 \\
\hline
\end{tabular}

a) Find the modal class interval.

[tex]$\square$[/tex] < [tex]$a\ \textless \ $[/tex] [tex]$\square$[/tex]

b) Work out an estimate of the mean age of these employees. Optional working:



Answer :

Let's break down the problem as follows:

### Part (a) Find the modal class interval

The modal class interval is the class interval with the highest frequency.

Given the frequencies:

- [tex]\(3\)[/tex] for [tex]\(18 \leq a < 20\)[/tex]
- [tex]\(2\)[/tex] for [tex]\(20 \leq a < 22\)[/tex]
- [tex]\(7\)[/tex] for [tex]\(22 \leq a < 24\)[/tex]
- [tex]\(8\)[/tex] for [tex]\(24 \leq a < 26\)[/tex]
- [tex]\(0\)[/tex] for [tex]\(26 \leq a\)[/tex]

The highest frequency is [tex]\(8\)[/tex], which corresponds to the interval [tex]\(24 \leq a < 26\)[/tex].

Therefore, the modal class interval is:

[tex]\[ 24 < a < 26 \][/tex]

### Part (b) Work out an estimate of the mean age of these employees

An estimate of the mean age can be found using the midpoints of each age interval and their corresponding frequencies.

#### Step-by-step solution:

1. Identify the midpoints of each age interval:

- For [tex]\(18 \leq a < 20\)[/tex], the midpoint is [tex]\( \frac{18 + 20}{2} = 19\)[/tex]
- For [tex]\(20 \leq a < 22\)[/tex], the midpoint is [tex]\( \frac{20 + 22}{2} = 21\)[/tex]
- For [tex]\(22 \leq a < 24\)[/tex], the midpoint is [tex]\( \frac{22 + 24}{2} = 23\)[/tex]
- For [tex]\(24 \leq a < 26\)[/tex], the midpoint is [tex]\( \frac{24 + 26}{2} = 25\)[/tex]
- For [tex]\(26 \leq a\)[/tex], the upper bound is not defined, thus calculations involving this class will be disregarded as their frequency is zero.

2. Multiply each midpoint by its corresponding frequency to get a value we denote as the product of the midpoint and frequency:

- [tex]\(19 \times 3 = 57\)[/tex]
- [tex]\(21 \times 2 = 42\)[/tex]
- [tex]\(23 \times 7 = 161\)[/tex]
- [tex]\(25 \times 8 = 200\)[/tex]
- Since the last interval has a frequency of 0, it doesn't contribute to the total.

3. Sum the products obtained in step 2:

[tex]\[ 57 + 42 + 161 + 200 = 460 \][/tex]

4. Calculate the total number of employees by summing the frequencies:

[tex]\[ 3 + 2 + 7 + 8 + 0 = 20 \][/tex]

5. Compute the mean age by dividing the sum of the products by the total number of employees:

[tex]\[ \text{Mean Age} = \frac{460}{20} \approx 23 \][/tex]

Optional working:
- Sum of midpoints × frequencies = [tex]\(460\)[/tex]
- Total number of employees = [tex]\(20\)[/tex]
- Mean age = [tex]\(\frac{460}{20} \approx 23\)[/tex]

Thus, the estimated mean age of these employees is approximately [tex]\( \boxed{23} \)[/tex] years.