95.44% IQ scores are between 77 and 125
We're given,
Mean (a)=101
standard deviation (b)=12
To find : [tex]P[77 < x < 125][/tex]
Using Empirical Rule:
∴ [tex]P[77 < x < 125][/tex] [tex]=[/tex] [tex]P[\frac{77-101}{12} < \frac{x-a}{b} < \frac{125-101}{12}]\\[/tex]
[tex]=P[-2 < z < 2]\\[/tex]
[tex]=P[Z < 2]-P[Z < -2]\\[/tex]
[tex]=0.9772-0.0228\\[/tex]
[tex]=0.9544[/tex] =95.44% (approx)
Learn more about Empirical rule of IQ calculation here:
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