To solve part d (the probability that there are exactly 6 right-handed and 6 left-handed people in the sample), we can model this as a binomial distribution problem, where the probability of success (being right-handed) is $p = 0.88$, the probability of failure (being left-handed) is $1 - p = 0.12$, and the sample size is $n = 12$.

The probability of exactly $k$ right-handed people in a sample of $n$ is given by the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}$

For this problem, we want to find $P(X = 6)$, where $X$ is the number of right-handed people. This means:

$P(X = 6) = \binom{12}{6} (0.88)^6 (0.12)^6$

Now let's calculate this.The probability that there are exactly 6 right-handed and 6 left-handed people in the sample is approximately $0.00128$ (rounded to five decimal places).

Would you like more details or have any further questions?

Here are five related questions to explore:

1. How would the probability change if the sample size increased to 20?
2. What is the probability of having exactly 8 right-handed people in the sample of 12?
3. Can we approximate this binomial distribution using a normal distribution?
4. How does the mean number of right-handers change if the sample size is reduced to 10?
5. What is the probability that more than 10 out of 12 people are right-handed?

Tip: The binomial distribution is ideal for modeling events with only two possible outcomes, such as success/failure or yes/no scenarios.