Respuesta :
Answer:
The person must score at least [tex]X = \mu + Z\sigma[/tex], in which Z has a p-value of [tex]1 - \frac{p}{100}[/tex], considering p the upper percentage the person must score, [tex]\mu[/tex] is the mean IQ score for the population and [tex]\sigma[/tex] is the standard deviation.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 p-value, we get the probability that the value of the measure is greater than X.
In this question:
Mean [tex]\mu[/tex], standard deviation [tex]\sigma[/tex]
What score must a person have to qualify for Mensa?
Score of at least X, given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]X - \mu = Z\sigma[/tex]
[tex]X = \mu + Z\sigma[/tex]
In which Z has a p-value of [tex]1 - \frac{p}{100}[/tex], considering p the upper percentage the person must score.
