Santa moves quickly on Christmas Eve: normally distributed times for visits

Question

Here is a question from a recent exam I gave in a basic probability course--all dressed up for Christmas.

Even with all his magic, Santa has to move quickly on Christmas eve. He spends a brief (approximately normally distributed) amount of time at each house. At three quarters of the houses he can finish leaving toys within 76.75 seconds. Even more quickly, he can finish at a quarter of the houses within 63.25 seconds. What is the probability he takes 60 seconds or less at a (randomly chosen) house.

Answer 1

BY the symmetry of the normal distribution, we conclude that $\mu=\frac{76.75+63.25}{2}=70$. In a freely available z table, we read off that $p=\frac14=0.2500$ corresponds to $z\approx -0.63745$. Hence $\sigma \approx \frac{63.25-70}{-0.63745}$ and the desired 60 seconds correspond to $z=\frac{60-70}{\sigma}\approx \frac{60-70}{63.25-70}\cdot(-0.63745)\approx -0.944$. For this z-value, we read off $$p\approx 0.1726.$$

Answer 2

Your software value of $0.159$ is correct. As already stated in the previous answer, we easily get $\mu=70$. For the calculation of $\sigma $, however, the correct value of $z $ that corresponds to $p=0.25$ is $-0.6745$. Thus, from

$$\sigma= \frac{63.25-70}{-0.6745} $$

we get $\sigma \approx 10.0074$. So, the value of $60 $ corresponds to a $z$ value of

$$ \frac{60-70}{10.0074} \approx 0.9993$$

which in turn gives, using printed tables or online $z$-to-$p $ conversion tools,

$$p \approx 0.1588$$

in accordance with your software calculations.