Chapter 11 Data Analysis and Statistics - 11.3 Use Normal Distributions - 11.3 Exercises - Skill Practice - Page 760: 27

$0.5363$

$\text{z-score}=\frac{\text{data item-mean}}{\text{standard deviation}}$ Hence here the z-score is: $\frac{65-64}{7}\approx 0.1,\frac{45-64}{7}\approx -2.7$ Then, using the table: $P(45\lt x\leq 65)=P(x\leq 65)-P(x\leq45)\approx P(z\leq0.1)-P(z\leq-2.7)=0.5398-0.0035=0.5363$