Normal Approximation to Difference of Two Binomial RVs- Where is this specific calculation wrong

Question

Components $A$ and $B$ are produced (some do not work). Let $X_A, X_B$ be the stochastic variable of the number of working components. We produce 1200 A components and 1300 B components. The working components distribute as follows: $$X_A \sim Bin(1200,0.95)$$ $$X_B \sim Bin(1300, 0.90)$$ The normal distribution approximation to the binomial distribution works fine here, I get: $$X_A \stackrel{\text{aprx}}{\sim} N(1140,7.55)$$ $$X_B \stackrel{\text{aprx}}{\sim} N(1170,10.82).$$

I know this is correct.

Now, the problem is to calculate the probability that more A components than B components are working..

I let $X_A - X_B = D \sim N(-30,13.2).$ I want to know: $$P(D > 0) = 1 - P(D \leq 0) = 1-\Phi(\frac{-30}{13.2}) = 1-\Phi(-2.27)$$ $$=1-(1-\Phi(2.27)) = \Phi(2.27) = 0.98840.$$

The answer is supposed to be $1-0.98840 = 0.0116.$ Where am I wrong?

Answer 1

When transforming D to a standard normal random variable, you should be doing $X - \mu \over \sigma$ $=$ $0 - (-30) \over 13.2$ $=$ $30 \over 13.2$. So now, $P(D \le 0)$ $=$ $1 - \Phi(2.27)$. If you think about this intuitively, the expected number of working A components is less than the expected number of working B components, so the probability of more A than B components functioning should be less than 0.5.

Answer 2

First, just by logic $E(X_A) < E(X_B),$ so the $P(D = X_A - X_B > 0)$ must be smaller than $1/2.$

The mistake is in the standardization, $\Phi(-30/13.2)$ should be $\Phi(30/13.2).$ In particular, $-(-30) = 30.$