1. **State the problem:** We have class intervals 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 with frequencies 5, $x$, 10, missing, 7, 8 respectively. The mean is given as 62.8. We need to find the missing frequencies using the step deviation method. 2. **Step deviation method formula:** $$\bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}$$ where: - $a$ = assumed mean - $h$ = class width - $f_i$ = frequency of class $i$ - $u_i = \frac{x_i - a}{h}$, where $x_i$ is the class mark 3. **Calculate class marks ($x_i$):** - 0-20: $10$ - 20-40: $30$ - 40-60: $50$ - 60-80: $70$ - 80-100: $90$ - 100-120: $110$ 4. **Choose assumed mean ($a$):** Choose $a=70$ (middle class mark for 60-80) 5. **Class width ($h$):** $h=20$ 6. **Calculate $u_i = \frac{x_i - a}{h}$:** - For 10: $\frac{10-70}{20} = -3$ - For 30: $\frac{30-70}{20} = -2$ - For 50: $\frac{50-70}{20} = -1$ - For 70: $\frac{70-70}{20} = 0$ - For 90: $\frac{90-70}{20} = 1$ - For 110: $\frac{110-70}{20} = 2$ 7. **Assign frequencies:** - $f_1=5$ - $f_2=x$ - $f_3=10$ - $f_4=y$ (missing frequency) - $f_5=7$ - $f_6=8$ 8. **Calculate $\sum f_i u_i$ and $\sum f_i$:** $$\sum f_i u_i = 5(-3) + x(-2) + 10(-1) + y(0) + 7(1) + 8(2) = -15 - 2x - 10 + 0 + 7 + 16 = (-15 - 10 + 7 + 16) - 2x = (-2) - 2x$$ $$\sum f_i = 5 + x + 10 + y + 7 + 8 = 30 + x + y$$ 9. **Use mean formula:** $$62.8 = 70 + 20 \times \frac{-2 - 2x}{30 + x + y}$$ 10. **Simplify:** $$62.8 - 70 = 20 \times \frac{-2 - 2x}{30 + x + y}$$ $$-7.2 = 20 \times \frac{-2 - 2x}{30 + x + y}$$ $$\frac{-7.2}{20} = \frac{-2 - 2x}{30 + x + y}$$ $$-0.36 = \frac{-2 - 2x}{30 + x + y}$$ 11. **Cross multiply:** $$-0.36 (30 + x + y) = -2 - 2x$$ $$-10.8 - 0.36x - 0.36y = -2 - 2x$$ 12. **Rearrange:** $$-10.8 + 2 - 0.36x + 2x - 0.36y = 0$$ $$-8.8 + 1.64x - 0.36y = 0$$ 13. **Express $y$ in terms of $x$:** $$1.64x - 0.36y = 8.8$$ $$-0.36y = 8.8 - 1.64x$$ $$y = \frac{1.64x - 8.8}{0.36}$$ 14. **Use total frequency if known or additional info to find $x$ and $y$.** Since no total frequency is given, assume $y$ is an integer and frequencies are positive. Try $x=6$: $$y = \frac{1.64(6) - 8.8}{0.36} = \frac{9.84 - 8.8}{0.36} = \frac{1.04}{0.36} \approx 2.89$$ Try $x=5$: $$y = \frac{8.2 - 8.8}{0.36} = \frac{-0.6}{0.36} = -1.67$$ (not possible) Try $x=7$: $$y = \frac{11.48 - 8.8}{0.36} = \frac{2.68}{0.36} \approx 7.44$$ 15. **Choose $x=6$, $y=3$ (approximate integers) as missing frequencies.** **Final answer:** $$x = 6, \quad y = 3$$