Using the normal approximation, what is the $z$-value of a sample difference of $\hat p_1−\hat p_2=−0.18$? A few concepts from my textbook that I do not understand:

  
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*If the Random, Normal, and Independent conditions are met, then is it true that $\hat p_1+2\hat p_2$ is approximately normally distributed?
  

My instinct is that it would remain normal, since both the mean and standard deviation of $\hat p_2$, which is itself normal, would be multiplied by a constant $-2$?

  
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*When Anne and Carl play tennis, Anne has a $50\%$ chance of winning each game, and when Billy and Carl play, Billy has a $40\%$ chance of winning each game. Anne and Billy each play Carl $25$ times. If $\hat p_1$ is Anne's sample proportion of wins and $\hat p_2$ is Billy's sample proportion of wins, then $\hat p_1−\hat p_2$ may be approximated by a normal distribution. Using this approximation, what is the $z$-value of a sample difference of $\hat p_1−\hat p_2=−0.18$?
  

I'm not sure how to approach this; should I use the probabilities in the two-sample Z formula and use the result of approximately $-1.286$?
Thanks!

I don't remember statistics quite well. But I think we can proceed as follows.  

Using this approximation, what is the $z$-value of a sample difference of
  $\hat p_1-\hat p_2 = -0.18$?

Denote the sample difference as
$$\hat d = \hat p_1-\hat p_2.$$
Then we are being asked to find the $z$ score of $\hat d$.
We see that
$$\mu = E[\hat d] = E[\hat p_1]-E[\hat p_2] = .5-.4 = .1$$
and that
$$\sigma^2 = \text{Var}(\hat d) = \text{Var}(\hat p_1)+\text{Var}(\hat p_2) = \frac{1}{25}(.5)(1-.5)+\frac{1}{25}(.4)(1-.6) = 0.0164.$$
Thus, under the normal and independent assumptions, we see that
$$\hat d \sim N(\mu = .1, \sigma^2 = 0.0164)$$
and hence
$$z = \frac{-0.18-\mu}{\sigma} = -2.186433$$
