Question 1

We need to calculate the mean and mode of the given data.

Step 1 — Calculate the Mean

Let's find the class marks for each interval. We will use the assumed mean method. Let the assumed mean (a) be 30. We calculate dᵢ = xᵢ - a and fᵢdᵢ.

| Age (in years) | Number of patients (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ - 30 | fᵢdᵢ | | :--- | :---: | :---: | :---: | :---: | | 5 - 15 | 6 | 10 | -20 | -120 | | 15 - 25 | 11 | 20 | -10 | -110 | | 25 - 35 | 21 | 30 | 0 | 0 | | 35 - 45 | 23 | 40 | 10 | 230 | | 45 - 55 | 14 | 50 | 20 | 280 | | 55 - 65 | 5 | 60 | 30 | 150 | | Total | 80 | | | 430 |

The sum of frequencies (∑fᵢ) is 80. The sum of fᵢdᵢ (∑fᵢdᵢ) is 430. The mean formula is:

$\text{Mean} (\bar{x}) = a + \frac{\sum f_i d_i}{\sum f_i}$

$= 30 + \frac{430}{80}$

$= 30 + 5.375$

$\boxed{35.38 \text{ years}}$

Step 2 — Calculate the Mode

Let's find the modal class first. The maximum frequency is 23. This frequency belongs to the class 35 - 45. So, the modal class is 35 - 45. The lower limit (l) is 35. The class size (h) is 10. The frequency of the modal class (f₁) is 23. The frequency of the preceding class (f₀) is 21. The frequency of the succeeding class (f₂) is 14. The mode formula is:

$\text{Mode} = l + \left[ \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h$

$= 35 + \left[ \frac{23 - 21}{2 \times 23 - 21 - 14} \right] \times 10$

$= 35 + \left[ \frac{2}{46 - 21 - 14} \right] \times 10$

$= 35 + \left[ \frac{2}{11} \right] \times 10$

$= 35 + \frac{20}{11}$

$= 35 + 1.81$

$\boxed{36.81 \text{ years}}$

Step 3 — Compare and Interpret

The mean age of patients is 35.38 years. This is the average age of patients. The modal age of patients is 36.8 years. This is the most frequent age of patients. Both values are close. The most common age is slightly higher than the average age.

Answer

(i) The mean of the data is 35.38 years. (ii) The mode of the data is 36.8 years. (iii) The mean represents the average age of patients. The mode represents the most frequent age of patients. Both values are close. The most common age is slightly higher than the average age.