For the table 10-20: 7, 20-30: 12, 30-40: 14, 40-50: 9, 50-60: 8, take the assumed mean to be a = 15 (the FIRST class mark, not the middle one) with h = 10. Find the mean.

With $a=15$ the values $u_i=\frac{x_i-15}{10}$ are 0, 1, 2, 3, 4, so $\sum f_i{u}_i=0+12+28+27+32=99$ and $\sum f_i=50$. Mean $=15+10\left(\frac{99}{50}\right)=15+19.8=34.8$, the same value the direct method gives $(\frac{1740}{50})$. Any class mark may serve as a; the middle one is only a convenience. "16.98" omits the factor h; "19.8" is the correction alone; "54.8" adds the correction to the middle class mark 35 out of habit.