For the table 10-20: 6, 20-30: 10, 30-40: 9, 40-50: 5 a student took $a=35,h=10$ but wrote $u_i=\frac{a-x_i}{h}$, getting 2, 1, 0, -1. His answer came to 40.67. What is the correct mean?

The definition is $u_i=\frac{x_i-a}{h}$, so with class marks 15, 25, 35, 45 the correct values are -2, -1, 0, 1. Then $\sum f_i{u}_i=-12-10+0+5=-17$ over $\sum f_i=30$, and the mean is $35+10\left(\frac{-17}{30}\right)=35-5.67=29.33$. His reversed sign pushed the correction the wrong way and gave 40.67, which is 11.33 too high. "35" is the assumed mean itself; "5.67" is only the size of the correction.