Two coins, A and B, each have a side for heads and a side for tails. When coin A is tossed, the probability it will land tails-side up is 0.5. When coin B is tossed, the probability it will land tails-side up is 0.8. Both coins will be tossed 20 times. Let A represent the proportion of times coin A lands tails-side up, and let PB represent the proportion of times coin B lands tails-side up. Is the number of fosses for each coin enough for the sampling distribution of the difference in sample proportions PA-PB to be approximately normal? a. Yes, 20 tosses for each coin is enough.
b. No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.
c. No, 20 tosses for coin A is not enough, but 20 tosses for coin B is enough.
d. No, 20 tosses is not enough for either coin.
e. There is not enough information given to determine it 20 tosses is enough.

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 (PA) = 0.5

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

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 (PB) = 0.8

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

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).