Answer:
22.86% probability that the persons IQ is between 110 and 130
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]
If one person is randomly selected what is the probability that the persons IQ is between 110 and 130
This is the pvalue of Z when X = 130 subtracted by the pvalue of Z when X = 110.
X = 130
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 100}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
X = 110
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{110 - 100}{15}[/tex]
[tex]Z = 0.67[/tex]
[tex]Z = 0.67[/tex] has a pvalue of 0.7486
0.9772 - 0.7486 = 0.2286
22.86% probability that the persons IQ is between 110 and 130
