Respuesta :
Using the normal distribution, it is found that there is a 0.4038 = 40.38% probability that the person has an IQ score between 92 and 108.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 100, \sigma = 15[/tex].
The probability that the person has an IQ score between 92 and 108 is the p-value of Z when X = 108 subtracted by the p-value of Z when X = 92, hence:
X = 108:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{108 - 100}{15}[/tex]
Z = 0.53
Z = 0.53 has a p-value of 0.7019.
X = 92:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{92 - 100}{15}[/tex]
Z = -0.53
Z = -0.53 has a p-value of 0.2981.
0.7019 - 0.2981 = 0.4038.
0.4038 = 40.38% probability that the person has an IQ score between 92 and 108.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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