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*∫fd**2*= 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)

* *