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

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).

Answer 2

The true statement about the sample proportions of coins A and B is (b) No, 20 tosses for coin A is enough, but 20 tosses for coin B is not enough.

The given parameters are:

Coin A

$p = 0.5$ --- the probability of landing tails-side up

$n = 20$ --- number of toss

Coin B

$p = 0.8$ --- the probability of landing tails-side up

$n = 20$ --- number of toss

Start by calculating the expected number of times both coins land on either sides.

For coin A, we have:

Expected number of tails

$Tails = np$

$Tails = 20 \times 0.5$

$Tails = 10$

So, the expected number of heads would be:

$Head = n - Tails$

$Head = 20 - 10$

$Head = 10$

This means that, there will an expected equal amount of outcomes for the heads and tails of coin A.

Hence, 20 tosses is enough for coin A because the proportion is equally distributed and approximately normal.

For coin B, we have:

Expected number of tails

$Tails = np$

$Tails = 20 \times 0.8$

$Tails = 16$

So, the expected number of heads would be:

$Head = n - Tails$

$Head = 20 - 16$

$Head = 4$

This means that, there won't be an expected equal amount of outcomes for the heads and tails of coin B.

Hence, 20 tosses is not enough for coin B because the proportion is neither equally distributed nor approximately normal.

So, the correct option is (b)

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