aligned & Mean deviation about the mean for the following data is & array |l|c|c|c|c|c| Class Interval & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 Frequency & 1 & 2 & 3 & 2 & 1 array aligned

Step-by-step solution

Find the class midpoints ( (x_i ) ) : The midpoint for a class interval is calculated as: (x_i= Lower limit + Upper limit 2 ) - (x₁= 0+6 2 =3 ) - (x₂= 6+12 2 =9 ) - (x₃= 12+18 2 =15 ) - (x₄= 18+24 2 =21 ) - (x₅= 24+30 2 =27 ) Thus, (x_i=3,9,15,21,27 ). Compute the mean (( x ): ) The mean is calculated as: ( x = f_i x_i f_i ) - ( f_i=1+2+3+2+1=9 ) - ( f_i x_i=(1 3)+(2 9)+(3 15)+(2 21)+(1 27) ) ( gathered f_i x_i=3+18+45+42+27=135 x = 135 9 =15 gathered ) Find ( |x_i- x | ) and (f_i |x_i- x | ) : ( |x_i- x |= Absolut