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16 July, 23:58

A person must score in the upper of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. There are Mensa members in countries throughout the world (Mensa International website, January,). If IQ scores are normally distributed with a mean of and a standard deviation of, what score must a person have to qualify for Mensa (to whole number) ?

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  1. 17 July, 00:08
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    The person must have a score of 131 to be able to qualify for Mensa

    Step-by-step explanation:

    The complete and correct question is as follows;

    A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013). If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa (to whole number)

    Solution

    Firstly, we calculate the z score from the given left tailed area. Since our upper limit is 2%, left tailed area will be 98% which is simply 0.98

    Left tailed area = 0.98

    Then, using standard score table,

    z = 2.054

    Mathematically; x = u + z * s

    According to this question;

    u = mean = 100

    z = the critical z score = 2.054

    s = standard deviation = 15

    Substituting these values into the equation, we have

    x = 100 + 15 (2.054)

    x = 100 + 30.81 = 130.81

    This is 131 to whole number
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