Final answer:
To find the probability that the average IQ of 15 people is between 98 and 102 points, we can use the Central Limit Theorem. The probability is approximately 0.7602, or 76.02%.
Step-by-step explanation:
To find the probability that the average IQ of 15 people is between 98 and 102 points, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is large enough.
First, we need to calculate the standard error of the sample mean. The standard error of the sample mean (SE) is equal to the standard deviation of the population (5 points) divided by the square root of the sample size (√15).
SE = 5 / √15 ≈ 1.29 points.
Next, we can convert the sample mean to a z-score using the formula Z = (X - μ) / SE, where X is the sample mean, μ is the population mean, and SE is the standard error of the sample mean. In this case, X = 100 (population mean) and SE ≈ 1.29.
Z1 = (98 - 100) / 1.29 ≈ -1.55
Z2 = (102 - 100) / 1.29 ≈ 1.55
We can then use a standard normal distribution table or calculator to find the probability corresponding to each z-score. The probability of the average IQ being between 98 and 102 points can be calculated as the difference between the cumulative probabilities of Z1 and Z2.
P(98 ≤ X ≤ 102) = P(Z1 ≤ Z ≤ Z2) ≈ P(-1.55 ≤ Z ≤ 1.55) ≈ 0.8801 - 0.1199 = 0.7602
Therefore, the probability that the average IQ of 15 people is between 98 and 102 points is approximately 0.7602, or 76.02%.