Statistics/Population Sampling Question - can someone explain how to solve

The mean television viewing times for Americans is 15 hours per week. Suppose a sample of 60 Americans is taken to determine viewing habits. Assume population standard deviation for weekly viewing times is sigma = 4 hours.

1. What is the probability that the sample mean will be within one hour of the population mean?
2. What is the probability that the sample mean will be with 45 minutes of the population mean?

Let X = viewing times.
Since n = 60 is large, we know by the Central Limit Theorem that:
Xbar ~ Normal(15, 4^2/60)

1. P(|Xbar - 15| < 1)
= P(15 - 1 < Xbar < 15 + 1)
= P(-1/[4/sqrt(60)] < Z < 1/[4/sqrt(60)]) <--[standardize]
= P(-1.9365 < Z < 1.9365)
= P(Z < 1.9365) - P(Z < -1.9365)
= 0.9736 - 0.0264
= 0.9472

2. 45 minutes = 3/4 or 0.75 hours, so:
P(|Xbar - 15| < 0.75)
= P(15 - 0.75 < Xbar < 15 + 0.75)
= P(-0.75/[4/sqrt(60)] < Z < 0.75/[4/sqrt(60)]) <--[standardize]
= P(-1.4524 < Z < 1.4524)
= P(Z < 1.4524) - P(Z < -1.4524)
= 0.9268 - 0.0732
= 0.8536