Probability of CDC data 
All of this is my work, for (c) I believe its $1 - pbinom(144, 138, 2.9)$ which yields $.0192 $
For (d) I know that when D is negative, that would mean the woman is taller than the man, so I think it is just $ 69.2/63.8$ ?
Please correct me if I am wrong, Thank you in advance!
 A: *

*Use pnorm; pbinom is for binomial random variables.

> 1-pnorm(72,69.2,2.9)
[1] 0.1671429



*Correct


*$$P\left(Z\ge\frac{144-133}{\sqrt{16.82}}\right)=P(Z\ge2.726249)$$
> 1-pnorm(2.726249)
[1] 0.003202933



*Correct


*$$P\left(Z\le\frac{0-5.4}{\sqrt{16.82}}\right)=P(Z\le -1.316682)$$
> pnorm(-1.316682)
[1] 0.09397262

A: (a) $P(X_1 > 72) = 0.1671.$  In R: I did not standardize because the CDF pnorm in R accepts
the actual means and standard deviations. [Irrelevant pbinom must be a typo--one I have sometimes made myself.]
1 - pnorm(72, 69.2, 2.9)
[1] 0.1671429

hdr = "Dist'n of Heights of Men"
curve(dnorm(x, 69.2, 2.9), 60, 80, lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=72, col="red")


(b) OK
(c) $P(S > 144) = 0.0037.$
1 - pnorm(144, 69.2+63.8, sqrt(2*2.9^2))
[1] 0.00365776

hdr = "Dist'n of Sums of Heights of Couples"
curve(dnorm(x, 69.2+63.8, sqrt(2*2.9^2)), 120, 150, 
      lwd=2, ylab="PDF", main=hdr)
  abline(h=0, col="green2"); abline(v=144, col="red")


(d) OK
(e) $P(D < 0) = 0.0940.$
pnorm(0, 69.2-63.8, sqrt(2*2.9^2))
[1] 0.09397269


hdr = "Dist'n of Man's Ht - Woman's Ht"
curve(dnorm(x, 69.2-63.8, sqrt(2*2.9^2)), -10, 20, 
            lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="red")


Also, by simulation of a million couples, which should give
a probability correct to about two places.
set.seed(2021)
m = rnorm(10^6, 69.2, 2.9)
w = rnorm(10^6, 63.8, 2.9)
mean(m < w)
[1] 0.094211

