Here are the solutions to the questions. Q(c) Step 1: Identify the principal amount, interest rate, and time. Assuming the loan amount is the cash price of the camera from Q3.b, which is shs. 30,000. Principal ($P$) = shs. 30,000 Rate ($R$) = 18% p.a. = 0.18 Time ($T$) = 2 years Step 2: Calculate the total amount after 2 years using the compound interest formula $A = P(1 + R)^T$. $$ A = \text{shs. } 30,000 (1 + 0.18)^2 $$ $$ A = \text{shs. } 30,000 (1.18)^2 $$ $$ A = \text{shs. } 30,000 \times 1.3924 $$ $$ A = \text{shs. } 41,772 $$ Step 3: Calculate the total interest paid. $$ \text{Interest} = A - P $$ $$ \text{Interest} = \text{shs. } 41,772 - \text{shs. } 30,000 $$ $$ \text{Interest} = \text{shs. } 11,772 $$ The total interest paid by Kangara is shs. 11,772. $$\boxed{\text{shs. } 11,772}$$ Q4. a) Step 1: Identify the class with the highest frequency. From the table, the highest frequency is 30. Step 2: State the corresponding class interval. The class interval corresponding to the frequency of 30 is 40-44 cm. The modal class is 40-44 cm. $$\boxed{\text{40-44 cm}}$$ Q4. b) i) Step 1: Create a table to calculate midpoints ($x$) and $fx$. | Height (cm) | Frequency (f) | Midpoint (x) | fx | | :---------- | :------------ | :----------- | :--- | | 20-24 | 2 | 22 | 44 | | 25-29 | 15 | 27 | 405 | | 30-34 | 18 | 32 | 576 | | 35-39 | 25 | 37 | 925 | | 40-44 | 30 | 42 | 1260 | | 45-49 | 6 | 47 | 282 | | 50-54 | 3 | 52 | 156 | | 55-59 | 2 | 57 | 114 | | Total | $\sum f = 101$ | | $\sum fx = 3762$ | Step 2: Calculate the mean height using the formula $\bar{x} = \frac{\sum fx}{\sum f}$. $$ \bar{x} = \frac{3762}{101} $$ $$ \bar{x} \approx 37.2475 $$ Rounded to two decimal places, the mean height is 37.25 cm. $$\boxed{37.25 \text{ cm}}$$ Q4. b) ii) Step 1: Calculate the mean height (already done in Q4. b) i)). Mean height = 37.25 cm. Step 2: Determine the median class. Total frequency ($N$) = 101. The position of the median is $\frac{N}{2} = \frac{101}{2} = 50.5$. Create a cumulative frequency (CF) column: | Height (cm) | Frequency (f) | Cumulative Frequency (CF) | | :---------- | :------------ | :------------------------ | | 20-24 | 2 | 2 | | 25-29 | 15 | 17 | | 30-34 | 18 | 35 | | 35-39 | 25 | 60 | <-- Median class (CF > 50.5) | 40-44 | 30 | 90 | | 45-49 | 6 | 96 | | 50-54 | 3 | 99 | | 55-59 | 2 | 101 | The median class is 35-39 cm because its cumulative frequency (60) is the first to exceed 50.5. Step 3: Identify the values for the median formula. Lower boundary of the median class ($L$) = 34.5 cm (since the classes are 30-34 and 35-39, the boundary is $(34+35)/2$). Cumulative frequency before the median class ($CF_b$) = 35. Frequency of the median class ($f_m$) = 25. Class width ($w$) = $39.5 - 34.5 = 5$ cm. Step 4: Calculate the median height using the formula $Median = L + \left(\frac{\frac{N}{2} - CF_b}{f_m}\right) \times w$. $$ Median = 34.5 + \left(\frac{50.5 - 35}{25}\right) \times 5 $$ $$ Median = 34.5 + \left(\frac{15.5}{25}\right) \times 5 $$ $$ Median = 34.5 + 0.62 \times 5 $$ $$ Median = 34.5 + 3.1 $$ $$ Median = 37.6 \text{ cm} $$ Step 5: Calculate the difference between the mean height and the median height. $$ \text{Difference} = |\text{Mean height} - \text{Median height}| $$ $$ \text{Difference} = |37.25 \text{ cm} - 37.6 \text{ cm}| $$ $$ \text{Difference} = |-0.35 \text{ cm}| $$ $$ \text{Difference} = 0.35 \text{ cm} $$ The difference between the mean height and the median height is 0.35 cm. $$\boxed{0.35 \text{ cm}}$$ 3 done, 2 left today. You're making progress.