Final answer:
To solve these probability problems, we will use the properties of the normal distribution. We are given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. We can find the probabilities of selecting a person with a certain IQ score or a group of people with mean IQ scores using the normal distribution and the Central Limit Theorem. The Central Limit Theorem applies when the sample size is large enough (typically n ≥ 30) and the population distribution is not extremely skewed.
Step-by-step explanation:
To solve these probability problems, we will use the properties of the normal distribution. We are given that IQ scores are normally distributed with a mean of 100 and a standard deviation of 15.
1. Probability of selecting one person with IQ less than 90:
We need to find the area under the curve to the left of the IQ score 90. This area represents the probability that a randomly selected person has an IQ less than 90. Using a standard normal table or a calculator, we find that this probability is approximately 0.2525.
2. Probability of selecting one person with IQ greater than 112:
We need to find the area under the curve to the right of the IQ score 112. This area represents the probability that a randomly selected person has an IQ greater than 112. Using a standard normal table or a calculator, we find that this probability is approximately 0.1189.
3. Probability of selecting one person with IQ between 98 and 108:
We need to find the area under the curve between the IQ scores 98 and 108. This area represents the probability that a randomly selected person has an IQ between 98 and 108. Using a standard normal table or a calculator, we can find the areas to the left of the IQ scores 98 and 108, and then subtract the smaller area from the larger area to find the desired probability.
4. Find the IQ score separating the bottom 77% from the top 23%:
Since the normal distribution is symmetric, we can find the z-score that corresponds to the bottom 13.5% of the distribution by subtracting 0.135 from 0.5, giving us 0.365. Using a standard normal table or a calculator, we find that the z-score corresponding to 0.365 is approximately -0.423. Now we can use this z-score to find the corresponding IQ score by multiplying it by the standard deviation and adding it to the mean: IQ score = -0.423 * 15 + 100 = 93.155. So, an IQ score of 93.155 separates the bottom 77% of the population from the top 23%.
5. Probability of mean IQ scores of a group of 32 people being less than 95:
To solve this problem, we will use the Central Limit Theorem. According to the Central Limit Theorem, when we take a large enough sample from a population, the distribution of the sample means will be approximately normal, regardless of the shape of the population distribution. In this case, we have a large enough sample size (n = 32), so we can assume that the distribution of the sample means will be approximately normal. The mean of the sample means will be equal to the population mean (100), and the standard deviation of the sample means will be equal to the population standard deviation divided by the square root of the sample size (15 / sqrt(32) ≈ 2.65). Now we can find the z-score corresponding to the sample mean IQ score of 95 using the formula z = (x - µ) / (σ / sqrt(n)), where x is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values, we find that the z-score is approximately (95 - 100) / (2.65) = -1.89. Using a standard normal table or a calculator, we can find the probability that the z-score is less than -1.89, which is approximately 0.0294.
6. Probability of mean IQ scores of a group of 40 people being greater than 106:
Similar to the previous question, we will use the Central Limit Theorem to solve this problem. The mean of the sample means will still be equal to the population mean (100), but the standard deviation of the sample means will now be equal to the population standard deviation divided by the square root of the sample size (15 / sqrt(40) ≈ 2.377). We can find the z-score corresponding to the sample mean IQ score of 106 using the formula z = (x - µ) / (σ / sqrt(n)). Plugging in the values, we find that the z-score is approximately (106 - 100) / (2.377) = 2.52. Using a standard normal table or a calculator, we can find the probability that the z-score is greater than 2.52, which is approximately 0.0059.
7. When does the Central Limit Theorem apply:
The Central Limit Theorem applies when the sample size is large enough (typically n ≥ 30) and the population distribution is not extremely skewed. When these conditions are met, the distribution of the sample means will be approximately normal, regardless of the shape of the population distribution.