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. 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? If required, round your answers to nearest whole number.£

130.81

Step-by-step explanation:

Given that :

Mean, μ = 100

Standard deviation, σ = 15

To obtain the upper 2% of scores :

We find the Zscore (value) of the upper 2% from the normal probability distribution table ;

Zscore corresponding to the area in the left of (1 - 0.02) = 2.054

Using this with the Zscore formula :

Zscore = (x - μ) / σ

2.054 = (x - 100) / 15

2.054 * 15 = x - 100

30.81 = x - 100

30.81 + 100 = x

x = 130.81

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step-by-step explanation: