Chapter 8 - Section 8.1 - Assess Your Understanding - Applying the Concepts - Page 412: 30c

$P(70 \leq X \leq 75) = 0.2183$. This implies that about 2 out of 100 samples of size 50 will have a mean between 70 and 75 dollars inclusive.

Here we use: n = 50 $z = \frac{70-67}{4.95} =0.61$ $z = \frac{75-67}{4.95} =1.62$ $P(70 \leq X \leq 75) = P(0.61 < z < 1.62)$ P(z < 1.62) - P(z < 0.61) = 0.9474 - 0.7291 = 0.2183