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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

  • MAth -

    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
    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
    34.14 + 47.73 = 81.87 % are between
    .8187 * 500 = 409

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