Supplementary Exercise 8.67 of IPS7e
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Data: X ~ Binomial (50,p1) where p1=0.4
      Y ~ Binomial (60,p2) where p2=0.5
      X and Y assumed to be independent

(a)
p1_hat=X/50, E(p1_hat)=p1=0.4, sd(p1_hat)=sqrt(p1*(1-p1)/50)=0.06928
p1_hat=Y/60, E(p2_hat)=p2=0.5, sd(p2_hat)=sqrt(p2*(1-p2)/60)=0.06455

E(p1_hat-p2_hat) = E(p1_hat)-E(p2_hat) = 0.4-0.5 = -0.1
sd(p1_hat-p2_hat) = sqrt(sd(p1_hat)^2 + sd(p2_hat)^2) = sqrt(.06928^2 + .06455^2) = 0.09469 

Note that the standard deviations could also be called standard errors,
because they are standard deviations for the distribution of a statistic.

(c)
95% of a normal distribution is within +-1.96*sigma of the mean, where sigma
is the standard deviation. Therefore
  d = 1.96*0.09469 = 0.1856
The points marked on the curve should therefore be the mean (-0.1), and the lower bound
of -0.1-0.1856 = -0.286 and the upper bound of -0.1+0.1856 = 0.0856.