Answer
$n=217$
Work Step by Step
We expect to survey a large number of people. In this case, $t_{\frac{α}{2}}\approx z_{\frac{α}{2}}$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$t_{0.025}\approx z_{0.025}=1.96$
$E=1$ (within 1 hour)
$s=7.5$
$n=(\frac{t_{\frac{α}{2}}.s}{E})^2$
$n=(\frac{1.96\times7.5}{1})^2$
$n=216.09$
Round up:
$n=217$
