The elongation of a steel bar under a particular load has been established to be normally distributed with a mean of $\mu = 0.05$ and a standard deviation of $ \sigma = 0.01$.
Find the probability that the elongation is
(a) above $0.1$ inch;
(b) between $0.025$ and $0.065$ inch.
My attempt:
$X:$ denote the elongation of the steel bar
$X$~$ N(.05, .01)$
(a) $P(X > 0.1)=1-P(X\leq 0.1) = 1-P(Z \leq (0.1-0.05)/.01)= 1-P(Z\leq5)$
Using R as $1-pnorm(5)$ I get the extremely low value of $2.866516e-07$ which makes me wonder if I'm using the correct approach.
(b) $P(0.025<x<0.065) = P(-2.5<Z<1.5)= pnorm(1.5)-pnorm(-2.5) = .927$
Are any of my answers correct or wrong?