**Answer:**

Is the number of tosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal?

b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.

**Step-by-step explanation:
**

a) Data and Calculations:

The probability of coin A landing tails-side up = 0.5

The proportion of times coin A lands tails-side up (PA) = 20 * 0.5 = 10

Therefore, the probability of coin A landing heads-side up = 0.5 (1 - 0.5)

And the proportion of times that coin A lands heads-side up = 20 * 0.5 = 10.

The proportion on either side is equally distributed.

This is why 20 tosses for coin A is enough, since the sample proportions PA is approximately normal, symmetric, and equally distributed. There will be equal amounts of 10 tosses (0.5 *20) for either heads-side up or tails-side up.

For coin B, the probability of landing tails-side up = 0.8

The proportion of times coin B lands tails-side up (PB) = 20 * 0.8 = 16

Therefore, the probability of coin B landing heads-side up = 0.2 (1 - 0-.8)

The proportion on either side is not equally distributed, but skewed.

This is why 20 tosses for coin B is not enough, since the sample proportions PB is not approximately normal, symmetric, and equally distributed. There will be 16 tosses landing tails-side up (0.8*20) and only 4 tosses landing heads-side up (0.2*20).