Final answer:
IQ scores follow a standardized normal distribution with a mean of 100 and standard deviation of 15. Z-scores are used to find probabilities of scores within certain ranges, and calculation of z-scores is key to answering questions on proportions of the population within certain IQ categories.
Step-by-step explanation:
IQ Scores and Standard Normal Distribution
The question pertains to understanding the normal distribution applicable to IQ scores, which are standardized with a mean (μ) of 100 and a standard deviation (σ) of 15. To answer the various aspects of the question, we use z-scores, which are the number of standard deviations a particular score is from the mean. For example, to describe an IQ score of 115, since the standard deviation is 15, 115 is one standard deviation above the mean (100 + 15). In contrast, a score of 70 would be two standard deviations below the mean (100 - 2*15).
To find the proportion of the population or the probability of a person having an IQ score in different categories or ranges, we calculate z-scores using the formula z = (x - μ) /σ, where x is the score in question. This provides a standardized measure which can then be used to look up corresponding probabilities or proportions in a standard normal distribution table or utilize software-provided functions if required.
For example, a person with an IQ score greater than 120 would have a z-score of (120 - 100)/15 = 1.33. To find the proportion of people with IQ scores above 120, we would look up this z-score in the standard normal table or use relevant statistical functions to find the area under the curve to the right of the z-score of 1.33. For membership in MENSA, which requires being in the top 2 percent of all IQs, we would seek the z-score that corresponds to the 98th percentile of the standard normal distribution, which is approximately a z-score of 2.05, translating to an IQ score of 100 + (2.05 * 15) or about 130.75.
The middle 50 percent of IQ scores falls between the 25th and 75th percentiles. To find these values, we look for z-scores that correspond to these percentiles and convert them back to IQ scores using the mean and standard deviation of IQ scores.