Concept:If x1, x2, x3, ¦xn are the number of observations (Or class marks of classes) with respective frequencies f1, f2, f3, ¦ fn, thenThe sum of observations = f1x1 + f2x2 + f3x3 +¦.+ fnxnThe total number of observations = f1+ f2 +¦..+ fnTherefore, the mean of the data, x? = (f1x1+ f2x2 + f3x3 + ¦.+ fnxn)/( f1+f2+¦ + fn).The mean of the data, x? = \(\displaystyle ?\frac{f_ix_i}{N} \)Where, xi is the class marks of classes Class mark = (lower limit + upper limit)/2Calculation:Total frequency = 120? 17 + p + q + 32 + p - 3q + 19 = 120? 2p - 2q = 52? p - q = 26 ------(1)We know mean x? = \(\displaystyle ?\frac{f_ix_i}{N} \),? Class mark of 1st class intreval (x1) = \(\displaystyle \frac{x_l + x_u}{2}\) = \(\displaystyle \frac{0 + 20}{2}\) = 10Similarly, we get? x2 = \(\displaystyle \frac{20+40}{2}\) = 30? x3 = \(\displaystyle \frac{40 + 60}{2}\) = 50? x4 = \(\displaystyle \frac{60 + 80}{2}\) = 70? x5 = \(\displaystyle \frac{80 + 100}{2}\) = 90? x? = \(\displaystyle ?\frac{f_ix_i}{N} \)? x? = \(\displaystyle \frac{f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5}{N}\)? 50 = \(\displaystyle \frac{17×10+(p+q)×30+32×50+(p-3q)×70+19×90}{120}\)? 6000 = 170 +30p + 30q + 1600 + 70p - 210 q + 1710? 6000 = 100p - 180q + 3480? 2520 = 100p - 180q? 252 = 10p - 18q ------(2)Solving (1) and (2) we get? q = 1? The value of q = 1.