1. **Problem Statement:** (a)(i) Find the mean of the monthly allowances of 7 students: 1200, 1100, 800, 600, 550, 900, 750. (a)(ii) Calculate the first (Q1) and third quartiles (Q3) of the data. (b)(i) Identify the variable of interest and its type from the frequency distribution. (b)(ii) Construct a "less than" ogive and estimate the median and percentage of students with allowance more than 950. (b)(iii) Calculate the mean and standard deviation from the grouped data. --- 2. **Formulas and Rules:** - Mean for raw data: $$\text{Mean} = \frac{\sum x_i}{n}$$ where $x_i$ are data points and $n$ is number of points. - Quartiles: Sort data, Q1 is median of lower half, Q3 is median of upper half. - For grouped data mean: $$\bar{x} = \frac{\sum f_i m_i}{\sum f_i}$$ where $f_i$ is frequency and $m_i$ is class midpoint. - Standard deviation: $$s = \sqrt{\frac{\sum f_i (m_i - \bar{x})^2}{\sum f_i - 1}}$$ - Median from ogive: value at cumulative frequency $\frac{n}{2}$. - Percentage more than a value: $$\frac{\text{number above}}{\text{total}} \times 100\%$$ --- 3. **(a)(i) Mean Calculation:** Data: 1200, 1100, 800, 600, 550, 900, 750 $$\sum x_i = 1200 + 1100 + 800 + 600 + 550 + 900 + 750 = 5900$$ Number of data points $n=7$ $$\text{Mean} = \frac{5900}{7} = 842.86$$ Interpretation: On average, the students receive about 842.86 per month. --- 4. **(a)(ii) Quartiles Calculation:** Sort data: 550, 600, 750, 800, 900, 1100, 1200 Median (Q2) is 800 (middle value). Lower half: 550, 600, 750 Upper half: 900, 1100, 1200 Q1 = median of lower half = 600 Q3 = median of upper half = 1100 --- 5. **(b)(i) Variable and Type:** Variable: Monthly allowance Type: Quantitative continuous variable (amount of money) --- 6. **(b)(ii) Less than ogive and estimates:** Class intervals and frequencies: 500-<600:8, 600-<700:10, 700-<800:17, 800-<900:20, 900-1000:24, 1000-1100:13, 1100-<1200:8 Cumulative frequencies (less than): <600:8 <700:8+10=18 <800:18+17=35 <900:35+20=55 <1000:55+24=79 <1100:79+13=92 <1200:92+8=100 Median position: $\frac{100}{2}=50$ Median class: 800-<900 (cumulative freq before 55, after 35) Median estimate: $$L=800, F=35, f=20, n=100$$ $$\text{Median} = L + \frac{\frac{n}{2} - F}{f} \times \text{class width} = 800 + \frac{50-35}{20} \times 100 = 800 + 0.75 \times 100 = 875$$ Percentage with allowance more than 950: Classes above 950: 950-1000 (part of 900-1000), 1000-1100, 1100-1200 Cumulative frequency less than 950: Between 900-1000 class, 950 is halfway, frequency 24 Assuming uniform distribution, frequency below 950 in this class: $$24 \times \frac{950-900}{100} = 24 \times 0.5 = 12$$ Total less than 950: $$55 + 12 = 67$$ More than 950: $$100 - 67 = 33$$ Percentage: $$\frac{33}{100} \times 100\% = 33\%$$ --- 7. **(b)(iii) Mean and Standard Deviation:** Class midpoints $m_i$: 550, 650, 750, 850, 950, 1050, 1150 Frequencies $f_i$: 8, 10, 17, 20, 24, 13, 8 Calculate mean: $$\sum f_i m_i = 8\times550 + 10\times650 + 17\times750 + 20\times850 + 24\times950 + 13\times1050 + 8\times1150$$ $$= 4400 + 6500 + 12750 + 17000 + 22800 + 13650 + 9200 = 86300$$ $$\bar{x} = \frac{86300}{100} = 863$$ Calculate variance: $$\sum f_i (m_i - \bar{x})^2 = 8(550-863)^2 + 10(650-863)^2 + 17(750-863)^2 + 20(850-863)^2 + 24(950-863)^2 + 13(1050-863)^2 + 8(1150-863)^2$$ Calculate each term: $$8 \times 97489 = 779912$$ $$10 \times 45369 = 453690$$ $$17 \times 12769 = 217073$$ $$20 \times 169 = 3380$$ $$24 \times 7569 = 181656$$ $$13 \times 34969 = 454597$$ $$8 \times 82849 = 662792$$ Sum: $$779912 + 453690 + 217073 + 3380 + 181656 + 454597 + 662792 = 2757100$$ Variance: $$s^2 = \frac{2757100}{100 - 1} = \frac{2757100}{99} = 27849.49$$ Standard deviation: $$s = \sqrt{27849.49} = 166.87$$ --- **Final answers:** (a)(i) Mean = 842.86 (a)(ii) Q1 = 600, Q3 = 1100 (b)(i) Variable: Monthly allowance, Type: Quantitative continuous (b)(ii) Median = 875, Percentage more than 950 = 33% (b)(iii) Mean = 863, Standard deviation = 166.87