Tossing a pair of coins
We have a white coin, for which P(Heads)=0.4 and a black coin for which P(Heads)=0.6. The flips of the same or of different coins are independent. For each of the following situations, determine whether the random variable N can be approximated by a normal. If yes, enter the mean and variance of N. If not, enter 0 in both of the corresponding answer boxes.
1. Let N be the number of Heads in 100 tosses of the white coin.
2. Let N be the number of Heads in 100 coin tosses. At each toss, one of the two coins is selected at random (either choice is equally likely), and independently from everything else.
3. Let N be the number of Heads in 50 tosses of the white coin followed by 50 tosses of the black coin (for a total of 100 tosses).
4. We select one of the two coins at random: each coin is equally likely to be selected. We then toss the selected coin 100 times, independently, and let N be the number of Heads.
mean and variances ???
1. mean = 40, variance = 24
2. mean = 50
3. mean = 50, variance = 24
4. mean = 0, variance = 0
the variance for question 2 is missing...Help us out, please!!
variance for q2?
2. mean 50, variance 25
1. In the first situation, where N is the number of Heads in 100 tosses of the white coin, N can be approximated by a normal distribution since the number of tosses is large enough. The mean of N is (0.4 * 100) = 40, and the variance is (0.4 * 0.6 * 100) = 24.
2. In the second situation, where N is the number of Heads in 100 coin tosses with random coin selection, N cannot be approximated by a normal distribution. Since the coin selection is random at each toss, the probability of Heads changes with each toss, making it difficult to model with a normal distribution. Therefore, both the mean and variance are 0.
3. In the third situation, where N is the number of Heads in 50 tosses of the white coin followed by 50 tosses of the black coin, N can be approximated by a normal distribution. The mean of N is (0.4 * 50) + (0.6 * 50) = 50, and the variance is (0.4 * 0.6 * 50) + (0.4 * 0.6 * 50) = 12.
4. In the fourth situation, where N is the number of Heads in 100 tosses of a randomly selected coin, N can be approximated by a normal distribution. Since each coin has an equal probability of being selected, the mean of N is (0.5 * 0.4 * 100) + (0.5 * 0.6 * 100) = 50, and the variance is (0.5 * 0.4 * 0.6 * 100) + (0.5 * 0.6 * 0.4 * 100) = 12.
To determine whether the random variable N can be approximated by a normal distribution, we can use the Central Limit Theorem (CLT) as a guideline. According to the CLT, if the sample size is sufficiently large and the underlying distribution is not heavily skewed, the distribution of the sum or average of independent random variables is approximately normal.
Now let's analyze each situation to determine if the random variable can be approximated by a normal distribution:
1. Let N be the number of Heads in 100 tosses of the white coin.
In this situation, the white coin is tossed 100 times, and we are interested in the count of Heads. Since the sample size is large (100 tosses), and the white coin is independently tossed each time, we can approximate N by a normal distribution.
The mean of N would be 100 * P(Heads) = 100 * 0.4 = 40.
The variance of N would be 100 * P(Heads) * (1 - P(Heads)) = 100 * 0.4 * 0.6 = 24.
2. Let N be the number of Heads in 100 coin tosses. At each toss, one of the two coins is selected at random, and independently from everything else.
In this situation, the coin chosen for each toss is random, which means the probability of getting Heads is not constant throughout the 100 tosses. Therefore, the random variable N cannot be approximated by a normal distribution. Enter 0 for both mean and variance.
3. Let N be the number of Heads in 50 tosses of the white coin followed by 50 tosses of the black coin (for a total of 100 tosses).
In this situation, we have two independent sets of coin tosses. The first 50 tosses are for the white coin, and the following 50 tosses are for the black coin. Since each set of tosses is independent, we can still use the central limit theorem to approximate N by a normal distribution.
The mean of N would be (50 * P(Heads, white)) + (50 * P(Heads, black)) = (50 * 0.4) + (50 * 0.6) = 20 + 30 = 50.
The variance of N would be (50 * P(Heads, white) * (1 - P(Heads, white))) + (50 * P(Heads, black) * (1 - P(Heads, black))) = (50 * 0.4 * 0.6) + (50 * 0.6 * 0.4) = 12 + 12 = 24.
4. We select one of the two coins at random, each coin is equally likely to be selected. We then toss the selected coin 100 times, independently, and let N be the number of Heads.
In this situation, the coin selected for each toss is random as well, just like in situation 2. Therefore, the random variable N cannot be approximated by a normal distribution. Enter 0 for both mean and variance.
To summarize:
1. Mean: 40, Variance: 24.
2. Mean: 0, Variance: 0.
3. Mean: 50, Variance: 24.
4. Mean: 0, Variance: 0.