Question

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.

Answer 1

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.

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).