Respuesta :
Answer:
65.68% probability that a randomly selected adult has an IQ between 90 and 120
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]
Find the probability that a randomly selected adult has an IQ between 90 and 120
This is the pvalue of Z when X = 120 subtracted by the pvalue of Z when X = 90. So
X = 120
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 100}{15}[/tex]
[tex]Z = 1.33[/tex]
[tex]Z = 1.33[/tex] has a pvalue of 0.9082
X = 90
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{90 - 100}{15}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514
0.9082 - 0.2514 = 0.6568
65.68% probability that a randomly selected adult has an IQ between 90 and 120
