statistic problem

If the variance is 16 then the standard deviation is 4

49 is one standard deviation above the mean.

50% of scores are above the mean and

68%/2 = 34% are between the mean and 1 standard dev above the mean.

SO

(50-34)% = 16% of scores are above 49    

This assumes the distribution is a normal (Gaussian) one, though the question doesn't say so.  If it were a symmetrical triangular distribution with those parameters, for example, none of the displayed solutions would be right.

I wonder if there's a symmetrical distribution of some sort for which (b) would be the right answer!

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I didn't think of that Alan, yes there must be a distribution for which B is correct. 

C and D are more than 50% so they will never  be correct!

Thank you all