Chapter 11 Data Analysis and Statistics - 11.3 Use Normal Distributions - 11.3 Exercises - Problem Solving - Page 762: 34d

$86.64\%$

$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$ Hence here the z-score is: $\frac{9-12}{2}=-1.5,\frac{15-12}{2}=1.5$ Then, $P(9\leq x\leq15)\approx P(z\leq15)-P(z\leq-1.5)=0.9332-0.0668=0.8664=86.64\%$