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Sagot :
Using the normal distribution, it is found that 1851 people would have an IQ less than 115.
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 proportion of IQ scores less than 115 is the p-value of Z when X = 115, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{115 - 100}{15}[/tex]
Z = 1
Z = 1 has a p-value of 0.8413.
Out of 2200 people:
0.8413 x 2200 = 1851.
More can be learned about the normal distribution at https://brainly.com/question/27643290
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