Final answer:
The student is asked to calculate probabilities for individual and group IQ scores based on a normally distributed population with a mean (μ) of 100 and standard deviation (σ) of 15, specifically for scenarios exceeding a certain IQ value or the distribution of average and total IQs for a selected group.
Step-by-step explanation:
The question involves finding probabilities related to the normal distribution of IQ scores, where the mean (μ) is 100 and the standard deviation (σ) is 15. The normal distribution is represented as X~ N(100, 15).
(1) Probability that an adult's IQ is higher than 108
To find this, we calculate the z-score for an IQ of 108, subtract it from the mean, and then divide by the standard deviation. From z-tables or using software, the corresponding probability is found, which is one minus the cumulative probability up to the z-score.
(2) Probability that the average IQ of 5 adults is higher than 108
When finding the probability for the average IQ of a sample, the standard deviation changes. It becomes the original standard deviation divided by the square root of the sample size (σ/√n). We again calculate the z-score with this new standard deviation, and refer to the z-tables or software to find the probability.
(3) Probability that the total IQ of 5 adults is lower than 450
The total IQ would mean we're looking at the sum of the IQs. The mean total IQ would be 5 times the individual mean, and the standard deviation would be the individual standard deviation times the square root of the sample size. We find the z-score for a total IQ of 450 using these new parameters and use tables or software to find the corresponding probability.