95.44% IQ scores are between 77 and 125
We're given,
Mean (a)=101
standard deviation (b)=12
To find : ![P[77 < x < 125]](https://tex.z-dn.net/?f=P%5B77%20%3C%20x%20%3C%20125%5D)
Using Empirical Rule:
∴ 
![P[\frac{77-101}{12} < \frac{x-a}{b} < \frac{125-101}{12}]\\](https://tex.z-dn.net/?f=P%5B%5Cfrac%7B77-101%7D%7B12%7D%20%3C%20%5Cfrac%7Bx-a%7D%7Bb%7D%20%3C%20%5Cfrac%7B125-101%7D%7B12%7D%5D%5C%5C)
![=P[-2 < z < 2]\\](https://tex.z-dn.net/?f=%3DP%5B-2%20%3C%20z%20%3C%202%5D%5C%5C)
![=P[Z < 2]-P[Z < -2]\\](https://tex.z-dn.net/?f=%3DP%5BZ%20%3C%202%5D-P%5BZ%20%3C%20-2%5D%5C%5C)
=95.44% (approx)
Learn more about Empirical rule of IQ calculation here:
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No way that is a real question you have to answer.
