Question 1

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

Step 1 — Calculate the Mean

We first find the class mark ($x_i$) for each interval. We use the formula $x_i = (\text{Upper limit} + \text{Lower limit}) / 2$. Let's choose 135 as the assumed mean ($a$). The class size ($h$) is 20. We calculate $d_i = x_i - a$ and $u_i = d_i / h$. Then we find the product $f_i u_i$.

| Monthly consumption (in units) | Number of consumers ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - 135$ | $u_i = d_i / 20$ | $f_i u_i$ | | :--- | :--- | :--- | :--- | :--- | :--- | | 65 - 85 | 4 | 75 | -60 | -3 | -12 | | 85 - 105 | 5 | 95 | -40 | -2 | -10 | | 105 - 125 | 13 | 115 | -20 | -1 | -13 | | 125 - 145 | 20 | 135 | 0 | 0 | 0 | | 145 - 165 | 14 | 155 | 20 | 1 | 14 | | 165 - 185 | 8 | 175 | 40 | 2 | 16 | | 185 - 205 | 4 | 195 | 60 | 3 | 12 | | Total | 68 | | | | 7 |

The sum of frequencies ($\sum f_i$) is 68. The sum of $f_i u_i$ ($\sum f_i u_i$) is 7. We use the step deviation method to find the mean.

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

$= 135 + \left(\frac{7}{68}\right) \times 20$

$= 135 + \frac{140}{68}$

$= 135 + 2.0588...$

$\boxed{137.058 \text{ units}}$

Step 2 — Calculate the Mode

We identify the modal class, which has the highest frequency. The highest frequency is 20, corresponding to the class 125 - 145. The lower limit ($l$) of the modal class is 125. The class size ($h$) is 20. The frequency ($f_1$) of the modal class is 20. The frequency ($f_0$) of the class preceding the modal class is 13. The frequency ($f_2$) of the class succeeding the modal class is 14.

We use the formula for the mode.

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

$= 125 + \left[\frac{20 - 13}{2 \times 20 - 13 - 14}\right] \times 20$

$= 125 + \left[\frac{7}{40 - 27}\right] \times 20$

$= 125 + \left[\frac{7}{13}\right] \times 20$

$= 125 + \frac{140}{13}$

$= 125 + 10.769...$

$\boxed{135.76 \text{ units}}$

Step 3 — Calculate the Median

We need to calculate the cumulative frequency.

| Monthly consumption (in units) | Number of consumers ($f_i$) | Cumulative frequency (cf) | | :--- | :--- | :--- | | 65 - 85 | 4 | 4 | | 85 - 105 | 5 | 9 | | 105 - 125 | 13 | 22 | | 125 - 145 | 20 | 42 | | 145 - 165 | 14 | 56 | | 165 - 185 | 8 | 64 | | 185 - 205 | 4 | 68 |

The total number of consumers ($n$) is 68. We find $n/2$, which is $68/2 = \textbf{34}$. The cumulative frequency just greater than 34 is 42. This corresponds to the median class 125 - 145. The lower limit ($l$) of the median class is 125. The class size ($h$) is 20. The frequency ($f$) of the median class is 20. The cumulative frequency ($cf$) of the class preceding the median class is 22.

We use the formula for the median.

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

$= 125 + \left[\frac{34 - 22}{20}\right] \times 20$

$= 125 + \left[\frac{12}{20}\right] \times 20$

$= 125 + 12$

$\boxed{137 \text{ units}}$

Step 4 — Compare the Measures

We have found the mean, mode, and median. The mean is approximately 137.058 units. The mode is approximately 135.76 units. The median is 137 units. The three measures of central tendency are approximately the same.

Answer

(i) The median of the data is 137 units. (ii) The mean of the data is approximately 137.058 units. (iii) The mode of the data is approximately 135.76 units. (iv) The mean, median, and mode are approximately the same.