Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.Marks Number of students0-10310-20520-30730-401040-501250-601560-70 1270-80680-90290-1008Now, we compute mean arithmetic mean by taking 55 as the assumed mean.Computative of Mean Marks\\\\((x_i)\\\\)Mid-value \\\\((f_i)\\\\)Frequency \\\\(u_i\\frac{x_i-55}{10}\\\\)\\\\(f_iu_i\\\\)0-1053-5-1510-20155-4-2020-30257-3-2130-403510-2-2040-504512-1-2050-6055150060-70651211270-8075621280-908523690-100958432Total \\\\(\\sum f_i=80\\\\)\\\\(\\sum f_iu_i\\\\)= -26 We have,N= \\\\(sum f_i= 80, \\sum f_iu_i= - 26\\\\), A= 55 and h= 10 \\\\(\ herefore \\overline {X} = A + h[\\frac{1}{N} \\sum f_iu_i]\\\\)\\\\(\\Rightarrow \\overline {X} = A+ h[ \\frac{1}{N} \\sum f_iu_i]\\\\)\\\\(\\Rightarrow \\overline {X} = 55+ 10 \ imes \\frac{-26}{80}= 55- 3.25 = 51. 75 \\\\)Marks.