The Chief Examiner reported that this question was attempted by majority of the candidates and they performed very well in it. However, it was also observed that a few candidates did not use the assumed mean while calculating the mean. They lost some marks for doing so because they were required to use the assumed mean. Candidates were expected to obtain the class mark (x) which are the mid points of each class interval, obtain the deviation (d) of each class mark from the assumed mean, multiply each deviation with its corresponding class frequency and sum them up to obtain.
∫fd = 10. Candidates were also expected to square each deviation to obtain d², multiply each squared deviation with its corresponding class frequency and sum them up to obtain∫fd2= 19100. Using these results, candidates would then calculate the mean and variance as follows:
Mean = A + ∫fd where A = assumed mean = 45.5, ∫fd = 10, n = 40.
                         n
= 45.5 + 10 = 45.75
               40
Variance ∫fd2 _ (∫fd])² = 19100 _ (10)² = 477.44.
                                       n         (40)       (40)