Respuesta :
Answer:
a) In order to be eligible for Mensa, a person must have an IQ score of at least 130.81.
b) 2900 people are elegible for Mensa.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 100, \sigma = 15[/tex]
(a) What IQ score should one have in order to be eligible for Mensa?
Scores in at least the 98th percentile, so scores of at least X when Z has a pvalue of 0.98. So at least X when Z = 2.054.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.054 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = 2.054*15[/tex]
[tex]X = 130.81[/tex]
In order to be eligible for Mensa, a person must have an IQ score of at least 130.81.
(b) In a typical region of 145,000 people, how many are eligible for Mensa
Only 2% are elegible. So
0.02*145000 = 2900
2900 people are elegible for Mensa.
