1. **State the problem:** We have class intervals and their frequencies with some missing values. The mean is given as 62.8. We need to find the missing frequencies using the step deviation method. 2. **Given data:** Class intervals: 0-20, 20-40, 40-60, 60-80, 80-100, 100-110 Frequencies: 5, x, 10, y, 7, 8 (where x and y are missing frequencies) Mean = 62.8 3. **Step deviation method formula for mean:** $$\bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}$$ 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 of class $i$ 4. **Calculate class marks ($x_i$):** - 0-20: $\frac{0+20}{2} = 10$ - 20-40: $30$ - 40-60: $50$ - 60-80: $70$ - 80-100: $90$ - 100-110: $105$ 5. **Choose assumed mean ($a$):** Choose $a = 70$ (class mark of 60-80 interval) 6. **Calculate class width ($h$):** Most intervals have width 20 except last which is 10, use $h=20$ 7. **Calculate $u_i = \frac{x_i - a}{h}$:** - For 10: $\frac{10-70}{20} = -3$ - 30: $-2$ - 50: $-1$ - 70: $0$ - 90: $1$ - 105: $\frac{105-70}{20} = 1.75$ 8. **Set frequencies:** $f_1=5$, $f_2=x$, $f_3=10$, $f_4=y$, $f_5=7$, $f_6=8$ 9. **Calculate total frequency:** $$N = 5 + x + 10 + y + 7 + 8 = 30 + x + y$$ 10. **Calculate $\sum f_i u_i$:** $$5(-3) + x(-2) + 10(-1) + y(0) + 7(1) + 8(1.75) = -15 - 2x - 10 + 0 + 7 + 14 = (-15 - 10 + 7 + 14) - 2x + 0y = (-4) - 2x$$ 11. **Use mean formula:** $$62.8 = 70 + 20 \times \frac{-4 - 2x}{30 + x + y}$$ 12. **Simplify:** $$62.8 - 70 = 20 \times \frac{-4 - 2x}{30 + x + y}$$ $$-7.2 = 20 \times \frac{-4 - 2x}{30 + x + y}$$ $$\frac{-7.2}{20} = \frac{-4 - 2x}{30 + x + y}$$ $$-0.36 = \frac{-4 - 2x}{30 + x + y}$$ 13. **Cross multiply:** $$-0.36 (30 + x + y) = -4 - 2x$$ $$-10.8 - 0.36x - 0.36y = -4 - 2x$$ 14. **Rearrange:** $$-10.8 + 4 = -2x + 0.36x + 0.36y$$ $$-6.8 = -1.64x + 0.36y$$ 15. **Multiply both sides by -1:** $$6.8 = 1.64x - 0.36y$$ 16. **We have one equation with two unknowns. Use total frequency assumption:** Sum of frequencies should be reasonable. Given frequencies are mostly integers, assume $y$ is integer. 17. **Use total frequency sum:** Total frequency $N = 30 + x + y$ 18. **Use another condition: sum of frequencies should be integer and positive. Try values for $y$ to find integer $x$.** Try $y=10$: $$6.8 = 1.64x - 0.36(10) = 1.64x - 3.6$$ $$6.8 + 3.6 = 1.64x$$ $$10.4 = 1.64x$$ $$x = \frac{10.4}{1.64} = 6.34$$ Try $y=9$: $$6.8 = 1.64x - 0.36(9) = 1.64x - 3.24$$ $$6.8 + 3.24 = 1.64x$$ $$10.04 = 1.64x$$ $$x = 6.12$$ Try $y=8$: $$6.8 = 1.64x - 0.36(8) = 1.64x - 2.88$$ $$6.8 + 2.88 = 1.64x$$ $$9.68 = 1.64x$$ $$x = 5.9$$ Try $y=7$: $$6.8 = 1.64x - 0.36(7) = 1.64x - 2.52$$ $$6.8 + 2.52 = 1.64x$$ $$9.32 = 1.64x$$ $$x = 5.68$$ Try $y=6$: $$6.8 = 1.64x - 0.36(6) = 1.64x - 2.16$$ $$6.8 + 2.16 = 1.64x$$ $$8.96 = 1.64x$$ $$x = 5.46$$ Try $y=5$: $$6.8 = 1.64x - 0.36(5) = 1.64x - 1.8$$ $$6.8 + 1.8 = 1.64x$$ $$8.6 = 1.64x$$ $$x = 5.24$$ Try $y=4$: $$6.8 = 1.64x - 0.36(4) = 1.64x - 1.44$$ $$6.8 + 1.44 = 1.64x$$ $$8.24 = 1.64x$$ $$x = 5.02$$ Try $y=3$: $$6.8 = 1.64x - 0.36(3) = 1.64x - 1.08$$ $$6.8 + 1.08 = 1.64x$$ $$7.88 = 1.64x$$ $$x = 4.8$$ Try $y=2$: $$6.8 = 1.64x - 0.36(2) = 1.64x - 0.72$$ $$6.8 + 0.72 = 1.64x$$ $$7.52 = 1.64x$$ $$x = 4.59$$ Try $y=1$: $$6.8 = 1.64x - 0.36(1) = 1.64x - 0.36$$ $$6.8 + 0.36 = 1.64x$$ $$7.16 = 1.64x$$ $$x = 4.37$$ Try $y=0$: $$6.8 = 1.64x - 0$$ $$x = \frac{6.8}{1.64} = 4.15$$ 19. **Closest integer values are $x=5$ and $y=4$ which satisfy the equation approximately.** **Final answer:** Missing frequencies are approximately $x=5$ and $y=4$.