We have a red coin, for which P(Heads)=0.4, a green coin, for which P(Heads)=0.5 and a yellow 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 300 tosses of the red coin.
2. Let N be the number of Heads in 300 coin tosses. At each toss, one of the three coins is selected at random (either choice is equally likely), and independently from everything else.
3. Let N be the number of Heads in 100 tosses of the red coin followed by 100 tosses of the yellow coin (for a total of 300 tosses).
mean and variances ???
above answers are partially wrong
here right ones:
1. mean = 120, variance = 72
2. mean = 150, variance = 75
3. mean = 150, variance = 73
4. mean = 0, variance = 0
To determine whether the random variable N can be approximated by a normal distribution, we need to check if the conditions for the Central Limit Theorem (CLT) are satisfied. The Central Limit Theorem states that the sum or average of a large number of independent and identically distributed random variables tends to follow a normal distribution, regardless of the shape of the original distribution.
1. Situation 1:
Here, we have N as the number of heads in 300 tosses of the red coin. Since each toss is independent and has the same probability of success (0.4 for heads), N follows a binomial distribution with parameters n = 300 (number of tosses) and p = 0.4 (probability of success).
The conditions for applying the Central Limit Theorem to a binomial distribution require np and n(1-p) both to be greater than or equal to 5. In this case, np = 300 * 0.4 = 120 and n(1-p) = 300 * (1 - 0.4) = 180, which satisfy this criterion. Therefore, we can approximate the random variable N by a normal distribution.
To find the mean and variance of N, we use the formula for the mean and variance of a binomial distribution:
Mean (μ) = np = 300 * 0.4 = 120
Variance (σ^2) = np(1-p) = 300 * 0.4 * (1 - 0.4) = 72
So, for Situation 1, the random variable N can be approximated by a normal distribution with a mean of 120 and a variance of 72.
2. Situation 2:
In this situation, N represents the number of heads in 300 coin tosses, where each toss selects one of the three coins (red, green, or yellow) at random, with equal probability. Since the coin choice is random for each toss, this results in a mixture of binomial distributions.
To determine if we can approximate N by a normal distribution, we need to consider the number of heads for each coin toss individually, rather than the total number of heads. For each coin toss, there are three possible outcomes (heads with different probabilities for each coin). Therefore, the distribution of N is not binomial and does not satisfy the conditions for the Central Limit Theorem.
Hence, for Situation 2, we cannot approximate the random variable N by a normal distribution, and the mean and variance are both 0.
3. Situation 3:
In this situation, N represents the number of heads in 100 tosses of the red coin followed by 100 tosses of the yellow coin, for a total of 300 tosses. Similar to Situation 1, we have a binomial distribution, but with two sets of 100 tosses each.
Applying the conditions for the Central Limit Theorem to each set separately:
For the first set of 100 tosses (red coin), np = 100 * 0.4 = 40 and n(1-p) = 100 * (1 - 0.4) = 60, which satisfy the criteria.
For the second set of 100 tosses (yellow coin), np = 100 * 0.6 = 60 and n(1-p) = 100 * (1 - 0.6) = 40, which also satisfy the criteria.
Therefore, for Situation 3, we can approximate the random variable N by a normal distribution.
To find the mean and variance of N, we sum the means and variances of each set of coin tosses:
Mean (μ) = np_red + np_yellow = 40 + 60 = 100
Variance (σ^2) = np_red(1-p_red) + np_yellow(1-p_yellow) = 40 * (1 - 0.4) + 60 * (1 - 0.6) = 64
Thus, for Situation 3, the random variable N can be approximated by a normal distribution with a mean of 100 and a variance of 64.
1. In this situation, since we have the same coin being tossed 300 times, the random variable N follows a binomial distribution, not a normal distribution. Therefore, N cannot be approximated by a normal distribution. Enter 0 in both answer boxes.
2. In this situation, the coin being chosen randomly at each toss introduces a mixture of three different binomial distributions. Since this is a mixture distribution and not a single binomial distribution, N cannot be approximated by a normal distribution. Enter 0 in both answer boxes.
3. In this situation, we have two different coins being tossed in separate batches of 100 tosses each. Since the batches are independent from each other, the random variable N follows a binomial distribution with different coin probabilities for each batch. Therefore, N cannot be approximated by a normal distribution. Enter 0 in both answer boxes.
To determine whether the random variable N can be approximated by a normal distribution, we need to check if the conditions for using the normal approximation are satisfied:
1. Let N be the number of Heads in 300 tosses of the red coin.
For this situation, we have a fixed number of trials (300 tosses of the red coin). Since the number of trials is sufficiently large (n = 300), we can use the normal approximation.
Mean = n * p = 300 * 0.4 = 120
Variance = n * p * (1 - p) = 300 * 0.4 * (1 - 0.4) = 72
2. Let N be the number of Heads in 300 coin tosses. At each toss, one of the three coins is selected at random, and independently from everything else.
In this situation, the coin used at each toss is chosen randomly. The probability of heads varies from toss to toss. Therefore, the conditions for using the normal approximation are not satisfied.
Mean = 0
Variance = 0
3. Let N be the number of Heads in 100 tosses of the red coin followed by 100 tosses of the yellow coin (for a total of 300 tosses).
Here, we have a mixture of two different coins, first red and then yellow, for a total of 300 tosses. Since the probability of heads changes after the first 100 tosses, the conditions for using the normal approximation are not satisfied.
Mean = 0
Variance = 0
So, the answers are as follows:
1. Mean = 120, Variance = 72
2. Mean = 0, Variance = 0
3. Mean = 0, Variance = 0
1. mean = 120, variance = 72
2. mean = 150, variance = 73
3. mean = 150, variance = 72
4. mean = 0, variance = 0