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

$83.51\%$

$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$ Hence here the z-score is: $\frac{7-12}{2}=-2.5,\frac{14-12}{2}=1$ Then, using the table: $P(7\leq x\leq14)\approx P(z\leq1)-P(z\leq-2.5)=0.8413-0.0062=0.8351=83.51\%$