Final answer:
To find the probability that a sample of 3 individuals will have an average IQ score greater than 120, we can use the Central Limit Theorem and calculate the z-score. The probability turns out to be approximately 0.0019, which is not one of the given options.
Step-by-step explanation:
To find the probability that a sample of 3 individuals will have an average IQ score greater than 120, we can use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means becomes approximately normal as the sample size increases, regardless of the shape of the population distribution.
Since the sample size is small (3), the distribution may not be exactly normal, but it is reasonable to approximate it as normal because the population standard deviation is known. We can standardize the sample mean using the formula:
z = (x - μ) / (σ / √n)
Where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, the population mean (μ) is 105, the population standard deviation (σ) is 20, and the desired sample mean (x) is 120. Plugging in these values, we get:
z = (120 - 105) / (20 / √3) ≈ 2.89
We can then calculate the probability using the z-score.
Looking up the z-score in a standard normal distribution table, the probability of obtaining a z-score of 2.89 or greater is approximately 0.0019. Therefore, the probability that a sample of 3 individuals will have an average IQ score greater than 120 is approximately 0.0019, which is not one of the given options.