Respuesta :
Answer:
9.18% probability that a randomly selected adult has an IQ greater than 134.1
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 = 102.6, \sigma = 23.6[/tex]
Find the probability that a randomly selected adult has an IQ greater than 134.1.
This is 1 subtracted by the pvalue of Z when X = 134.1. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{134.1 - 102.6}{23.6}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082.
1-0.9082 = 0.0918
So there is a 9.18% probability that a randomly selected adult has an IQ greater than 134.1
