1. **Stating the problem:** We are given grouped frequency distributions for Thika road service time (x) and Mombasa road service time (y). We need to calculate various summary statistics including mean, coefficient of variation, mode, median, geometric mean, quartiles, and Pearson’s 1st coefficient of skewness. 2. **Grouping data:** Thika road (x): Class intervals and frequencies: 10–19:18, 20–29:12, 30–39:7, 40–49:5, 50–59:8, 60–69:10, Total=60 Mombasa road (y): Class intervals and frequencies: 10–19:10, 20–29:8, 30–39:6, 40–49:6, 50–59:14, 60–69:6, Total=50 3. **Calculate midpoints (class marks):** For each class interval $[a,b]$, midpoint $m = \frac{a+b}{2}$ Thika midpoints: 14.5, 24.5, 34.5, 44.5, 54.5, 64.5 Mombasa midpoints: same as above because class intervals are same 4. **Calculate arithmetic mean:** $$\bar{x} = \frac{\sum f_i m_i}{\sum f_i}$$ Thika: $\sum f_i m_i = 18(14.5)+12(24.5)+7(34.5)+5(44.5)+8(54.5)+10(64.5)$ $=261+294+241.5+222.5+436+645=2100$ $$\bar{x} = \frac{2100}{60} = 35$$ Mombasa: $\sum f_i m_i=10(14.5)+8(24.5)+6(34.5)+6(44.5)+14(54.5)+6(64.5)$ $=145+196+207+267+763+387=1965$ $$\bar{y} = \frac{1965}{50}=39.3$$ 5. **Calculate standard deviation:** Need $\sum f_i m_i^2$ first. Thika: $\sum f_i m_i^2 = 18(14.5^2)+ 12(24.5^2)+ 7(34.5^2)+5(44.5^2)+8(54.5^2)+10(64.5^2)$ $=18(210.25)+12(600.25)+7(1190.25)+5(1980.25)+8(2970.25)+10(4160.25)$ $=3784.5+7203+8331.75+9901.25+23762+41602.5=94585$ Variance for Thika: $$s_x^2 = \frac{\sum f_i m_i^2}{\sum f_i} - \bar{x}^2 = \frac{94585}{60} - 35^2 = 1576.42 - 1225 = 351.42$$ $$s_x = \sqrt{351.42} \approx 18.75$$ Mombasa: $\sum f_i m_i^2= 10(210.25)+8(600.25)+6(1190.25)+6(1980.25)+14(2970.25)+6(4160.25)$ $=2102.5+4802+7141.5+11881.5+41583.5+24961.5=92372.5$ Variance for Mombasa: $$s_y^2 = \frac{92372.5}{50} - 39.3^2 = 1847.45 - 1544.49 = 302.96$$ $$s_y = \sqrt{302.96} \approx 17.41$$ 6. **Coefficient of variation (CV):** $$CV = \frac{s}{\bar{x}} \times 100\%$$ Thika: $$CV_x = \frac{18.75}{35} \times 100 = 53.57\%$$ Mombasa: $$CV_y = \frac{17.41}{39.3} \times 100 = 44.29\%$$ 7. **Mode:** Modal class is the class with highest frequency. Thika modal class = 10–19 (18 customers) Mode for grouped data: $$Mode = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$$ Where: - $L=10$ lower class boundary of modal class - $f_1=18$ frequency of modal class - $f_0=12$ freq. of class before modal (20–29) - $f_2=7$ freq. after modal (30–39) - $h=10$ class width $$Mode_x = 10 + \frac{(18-12)}{(2\times 18 - 12 -7)} \times 10 = 10 + \frac{6}{17} \times 10 = 10 + 3.53 = 13.53$$ Mombasa modal class = 50–59 (14 customers) $$L=50, f_1=14, f_0=6 (40–49), f_2=6 (60–69), h=10$$ $$Mode_y = 50 + \frac{(14-6)}{(2\times 14 -6 -6)} \times 10 = 50 + \frac{8}{16} \times 10 = 50 + 5 = 55$$ 8. **Median:** Median position for Thika: $\frac{60}{2}=30^{th}$ value Cumulative frequency table (Thika): 10–19:18, 20–29:30 (18+12), 30–39:37, 40–49:42, 50–59:50, 60–69:60 Median class = 20–29 Median formula: $$Median = L + \frac{\frac{N}{2} - F}{f} \times h$$ Where: - $L=20$ lower boundary - $N=60$ - $F=18$ cumulative frequency before median class - $f=12$ frequency of median class - $h=10$ $$Median_x=20 + \frac{30 - 18}{12} \times 10 = 20 + \frac{12}{12} \times 10 = 20 + 10 = 30$$ Mombasa median: $\frac{50}{2} = 25^{th}$ value Cumulative freq: 10,18,24,30,44,50 Median class = 50–59 $$Median_y=50 + \frac{25 - 30}{14} \times 10=50 + \frac{-5}{14} \times 10=50 - 3.57=46.43$$ 9. **Geometric mean:** $$GM=\exp\left(\frac{\sum f_i \ln m_i}{\sum f_i}\right)$$ Thika: $$\sum f_i \ln m_i = 18\ln 14.5 + 12\ln 24.5 + 7\ln 34.5 + 5\ln 44.5 + 8\ln 54.5 + 10\ln 64.5$$ Approximate values: $18(2.674)$ + $12(3.199)$ + $7(3.542)$ + $5(3.794)$ + $8(3.999)$ + $10(4.167)$ $=48.13 + 38.39 + 24.79 + 18.97 + 31.99 + 41.67 = 203.94$ $$GM_x= \exp\left(\frac{203.94}{60}\right) = \exp(3.399) = 29.97$$ Mombasa: $10\ln 14.5 + 8\ln 24.5 + 6\ln 34.5 + 6\ln 44.5 + 14\ln 54.5 + 6\ln 64.5$ $=10(2.674)+8(3.199)+6(3.542)+6(3.794)+14(3.999)+6(4.167)$ $=26.74+25.59+21.25+22.76+55.99+25.00=177.33$ $$GM_y=\exp\left(\frac{177.33}{50}\right)=\exp(3.547)=34.69$$ 10. **Quartiles:** Using cumulative frequencies for quartiles: For Thika (N=60): 1st quartile $Q_1 = \frac{60}{4} = 15^{th}$ value, median at 30, 3rd quartile $Q_3 = 45^{th}$ value. Classes: - $Q_1$ is in 10–19 (freq=18) - $Q_3$ is in 50–59 (cum freq up to 50) $$Q_1 = 10 + \frac{15-0}{18} \times 10 = 10 + 8.33 = 18.33$$ $$Q_3 = 50 + \frac{45-42}{8} \times 10 = 50 + 3.75 = 53.75$$ For Mombasa (N=50): $Q_1 = 12.5^{th}$ value; cumulative freq shows 10–19:10, 20–29:18, so $Q_1$ in 20–29. $Q_3 = 37.5^{th}$ value; cumulative freq 40–49:30, 50–59:44, so $Q_3$ in 50–59 class. $$Q_1 = 20 + \frac{12.5 - 10}{8} \times 10 = 20 + 3.13 = 23.13$$ $$Q_3 = 50 + \frac{37.5 - 30}{14} \times 10 = 50 + 5.36 = 55.36$$ 11. **Pearson’s 1st coefficient of skewness (SK1):** $$SK1 = \frac{\bar{x} - Mode}{s}$$ Thika: $$SK1_x = \frac{35 - 13.53}{18.75} = \frac{21.47}{18.75} = 1.15$$ Mombasa: $$SK1_y = \frac{39.3 - 55}{17.41} = \frac{-15.7}{17.41} = -0.90$$ **Interpretation:** Positive skew for Thika (right skewed) and negative skew for Mombasa (left skewed). --- **Final answers:** - Thika mean = 35, CV = 53.57%, Mode = 13.53, Median =30, GM=29.97, Q1=18.33, Q3=53.75, SK1=1.15 - Mombasa mean = 39.3, CV=44.29%, Mode=55, Median=46.43, GM=34.69, Q1=23.13, Q3=55.36, SK1=-0.90