Answer:
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
---> a is less than b
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---> a is greater than or equal to b