Respuesta :
Answer:
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
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 question, we have that:
[tex]\mu = 96, \sigma = 15.7[/tex]
Probability that a randomly selected adult has an IQ greater than 123.4.
This is 1 subtracted by the pvalue of Z when X = 123.4. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{123.4 - 96}{15.7}[/tex]
[tex]Z = 1.745[/tex]
[tex]Z = 1.745[/tex] has a pvalue of 0.9595
1 - 0.9595 = 0.0405
4.05% probability that a randomly selected adult has an IQ greater than 123.4.
