The maximum who can score 100 occurs if the rest score as few marks as possible, so if p candidates score 100 and the rest score zero then 100p=50×60 so p=30. The standard deviation is given by

s2 = å (xi- x   2   ) n = 30×402 + 20×602 50 = 2400

so s = 20Ö6 = 50.0 to 3 s.f.

If p candidates score 100 then, as

å

(xi-60)2 = 50 × s2

, if s = 8, we know 40² p < 3200 so p < 2. Clearly p cannot be equal to 2 because the marks of the remaining 48 candidates will contribute to the standard deviation but we now suppose the standard deviation is a little higher, say 8.2. Then 40²p < 50×8.2² so p < 2.10125. As p is a whole number it is at most 2.

If 2 candidates score 100 then the remaining 48 score 2800 between them for the mean to be 60. To minimise the variance they should all score marks as close to 60 as possible. Dividing 2800 by 48 gives 58.3 to 3 s.f. Suppose q candidates score 59 and (48-q) score 58 then

59q + 58(48-q) = 2800

so q=16. Hence the mean mark is 60 if 2 candidates score 100, 16 score 59 and 32 score 58. In this case the standard deviation is given by

s2 = 2×402 + 16×12 + 32×22 50 = 66.88.

So the standard deviation is 8.178 to 3 d.p.

The number of candidates who score 100 is less than 50s²/ 40² so if s = 4Ö2 this is less than 50×32/ 1600 = 1 so it is impossible for any candidates to have scored 100 because the scores of the other 49 candidates also contribute to the standard deviation.

So with a standard deviation of 4Ö2 (approx 5.657 to 3 d.p) or less no candidates could have scored 100.