500 candidates appeared for a certain test. The mean scores was 59 and variance was 25. Assuming the distribution to be normal. find 1.% of candidates scoring 64 and above. 2. No. of candidates having score between 54 and 69. ( Given : For a s.n.v.z, the area between x=0 to z=1 is 0.3414 and between z= 0toz=2 is 0.4773
sigma = sqrt(25) = 5
64 - 59 = 5
so we want to know what percent scored 1 sigma or more above mean.(z=1)
well .5 or 50% score between mean and infinity
and given is .3414 or 34.14% score between mean and z = 1
so
50-34.14 = 15.86 % or 79 score 64 or above
Now 69 is 2 sigma above (z=2) and 54 is 1 sigma below
so we want between z = -1 and z = +2
well we know that 34.14% are between z=-1 and mean and .4773 or 47.73^ are between mean and z=2
so
34.14 + 47.73 = 81.87 % are between
.8187 * 500 = 409