The number of candidates n is 50 and the mean score

x

is 60. For there to be the maximum number of students who gained 100 marks the rest must score 0. Let y be the number who gained 100 and z be the number who gained 0 marks. Then (100y + 0z )/50 = 60 so y = 30. Therefore, 30 students scored 100 and 20 students scored 0.

The standard deviation is given by

s2 = å (xi- x   2   ) n = 1 n å xi2 - x   2   = 2400

so the standard deviation s = 20Ö6 = 49.0 to 3 s.f.

For a lower standard deviation it is not possible for so many students to score 100. Now suppose the standard deviation s = 8. If y candidates score 100 then, as the marks of the remaining 48 candidates contribute to the standard deviation, we know

s2 > 402y 50 ,

so we know 40² y < 3200 so y < 2.

If the standard deviation is a little higher, say 8.2. Then 40²y < 50×8.2² so y < 2.10125. As y is a whole number it could be 2 (but not more than 2).

If 2 students got 100, and the mean is 60, the sum of all the marks is 3000 and the sum of the scores of the remaining 48 students is 2800. Then 2800/48 = 58.3333.... so one set of results might be 47 students getting 58 marks, the 48th getting 74 marks, and 49th and 50th getting 100 marks. In this case the standard deviation is :

s2 = (2×1002 + 742 + 47×582)/50 - 602

which gives s = 8.4664.

Another set of possible scores with a lower standard deviation, around 8.2 or less, requires all candidates who don't score 100 to get scores as near to 60 as possible. Suppose q candidates score 59 and (48-q) score 58 then

59q + 58(48-q) = 2800

so q=16. In this case 2 candidates score 100, 16 score 59 and 32 score 58. 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.

If the standard deviation s = 4Ö2, as we know that the number of candidates who score 100 is always less than

50s2 402,

the number who could have scored 100 is less than

50 ×32 402 = 1

so it is impossible for any of the candidates to have scored 100.