Answer:
A person must score at least 130.825 to qualify for Mensa
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Top 2%
Scores of x and higher, in which X is found when Z has a pvalue of 0.98. So it is X when Z = 2.055.
A person must score at least 130.825 to qualify for Mensa