Question 6

Let's find the median, mean, and mode for the given data.

Step 1 — Find the Median

First, we make a cumulative frequency table. We find the total number of surnames. Then we find N/2 to locate the median class.

| Number of letters | Frequency (fᵢ) | Cumulative frequency (cf) | | :--- | :--- | :--- | | 1 - 4 | 6 | 6 | | 4 - 7 | 30 | 36 | | 7 - 10 | 40 | 76 | | 10 - 13 | 16 | 92 | | 13 - 16 | 4 | 96 | | 16 - 19 | 4 | 100 | | Total (N) | 100 | |

The total number of surnames is 100. So, N/2 is 50. The cumulative frequency just above 50 is 76. This corresponds to the class interval 7 - 10. This is our median class.

The lower limit (l) of the median class is 7. The cumulative frequency (cf) before the median class is 36. The frequency (f) of the median class is 40. The class size (h) is 3.

We use the median formula.

$\text{Median} = l + \left[ \frac{N/2 - cf}{f} \right] \times h$

$= 7 + \left[ \frac{50 - 36}{40} \right] \times 3$

$= 7 + \left[ \frac{14}{40} \right] \times 3$

$= 7 + 0.35 \times 3$

$= 7 + 1.05$

$\boxed{8.05}$

Step 2 — Find the Mean

Let's calculate the class marks ($x_i$). We will use the step deviation method. Let the assumed mean (a) be 11.5. The class size (h) is 3.

| Number of letters | fᵢ | xᵢ | dᵢ = xᵢ - 11.5 | uᵢ = dᵢ / 3 | fᵢuᵢ | | :--- | :--- | :--- | :--- | :--- | :--- | | 1 - 4 | 6 | 2.5 | -9 | -3 | -18 | | 4 - 7 | 30 | 5.5 | -6 | -2 | -60 | | 7 - 10 | 40 | 8.5 | -3 | -1 | -40 | | 10 - 13 | 16 | 11.5 | 0 | 0 | 0 | | 13 - 16 | 4 | 14.5 | 3 | 1 | 4 | | 16 - 19 | 4 | 17.5 | 6 | 2 | 8 | | Total | 100 | | | | -106 |

The sum of frequencies ($\sum f_i$) is 100. The sum of $f_i u_i$ ($\sum f_i u_i$) is -106.

We use the mean formula.

$\text{Mean}(\bar{x}) = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$

$= 11.5 + \left( \frac{-106}{100} \right) \times 3$

$= 11.5 + (-1.06) \times 3$

$= 11.5 - 3.18$

$\boxed{8.32}$

Step 3 — Find the Mode

We find the class with the highest frequency. The highest frequency is 40. This belongs to the class interval 7 - 10. This is our modal class.

The lower limit (l) of the modal class is 7. The class size (h) is 3. The frequency ($f_1$) of the modal class is 40. The frequency ($f_0$) before the modal class is 30. The frequency ($f_2$) after the modal class is 16.

We use the mode formula.

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

$= 7 + \left[ \frac{40 - 30}{2 \times 40 - 30 - 16} \right] \times 3$

$= 7 + \left[ \frac{10}{80 - 30 - 16} \right] \times 3$

$= 7 + \left[ \frac{10}{34} \right] \times 3$

$= 7 + 0.2941 \times 3$

$= 7 + 0.8823$

$\boxed{7.88}$

Answer

(i) The median number of letters in the surnames is 8.05. (ii) The mean number of letters in the surnames is 8.32. (iii) The modal size of the surnames is 7.88.