1. **Problem statement:** We have a frequency distribution of blood glucose levels for 100 diabetics with intervals and frequencies given. We need to construct the histogram and cumulative frequency curve, calculate the mean and mode, and find the median approximately both graphically and by interpolation. 2. **Histogram and cumulative frequency curve:** - The histogram bars correspond to the intervals with heights equal to frequencies: 4, 14, 28, 26, 20, 8. - The cumulative frequencies are calculated by summing frequencies up to each class: $$N_1=4, N_2=4+14=18, N_3=18+28=46, N_4=46+26=72, N_5=72+20=92, N_6=92+8=100.$$ - Plotting these cumulative frequencies against the upper class boundaries gives the cumulative frequency curve. 3. **Mean calculation:** - Use the midpoint of each class as representative value: Midpoints: $m_1=\frac{1.15+1.25}{2}=1.20$, $m_2=1.30$, $m_3=1.40$, $m_4=1.50$, $m_5=1.60$, $m_6=1.70$. - Mean formula: $$\bar{x} = \frac{\sum n_i m_i}{\sum n_i} = \frac{4\times1.20 + 14\times1.30 + 28\times1.40 + 26\times1.50 + 20\times1.60 + 8\times1.70}{100}.$$ - Calculate numerator: $$4\times1.20=4.8, 14\times1.30=18.2, 28\times1.40=39.2, 26\times1.50=39, 20\times1.60=32, 8\times1.70=13.6.$$ - Sum numerator: $$4.8 + 18.2 + 39.2 + 39 + 32 + 13.6 = 146.8.$$ - Mean: $$\bar{x} = \frac{146.8}{100} = 1.468.$$ 4. **Mode calculation:** - Mode corresponds to the class with highest frequency, which is $[1.35,1.45[$ with frequency 28. - Using the mode formula for grouped data: $$\text{Mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h,$$ where: - $L=1.35$ (lower boundary of modal class), - $f_1=28$ (frequency of modal class), - $f_0=14$ (frequency before modal class), - $f_2=26$ (frequency after modal class), - $h=0.10$ (class width). - Calculate numerator and denominator: $$f_1 - f_0 = 28 - 14 = 14,$$ $$2f_1 - f_0 - f_2 = 2\times28 - 14 - 26 = 56 - 40 = 16.$$ - Mode: $$1.35 + \frac{14}{16} \times 0.10 = 1.35 + 0.0875 = 1.4375.$$ 5. **Median calculation:** - Total frequency $N=100$, median position is at $\frac{N}{2} = 50$. - From cumulative frequencies, median class is $[1.35,1.45[$ since cumulative frequency before it is 18 and after is 46, next class $[1.45,1.55[$ cumulative frequency is 72 which contains 50. - Using interpolation formula: $$\text{Median} = L + \frac{\frac{N}{2} - F}{f} \times h,$$ where: - $L=1.45$ (lower boundary of median class), - $F=46$ (cumulative frequency before median class), - $f=26$ (frequency of median class), - $h=0.10$ (class width). - Calculate: $$1.45 + \frac{50 - 46}{26} \times 0.10 = 1.45 + \frac{4}{26} \times 0.10 = 1.45 + 0.0154 = 1.4654.$$ **Final answers:** - Mean blood glucose level: $1.468$ g/l - Mode blood glucose level: $1.4375$ g/l - Median blood glucose level (approx.): $1.4654$ g/l