Intelligence quotients (IQs) on the Standford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16 . Find the z-score for an IQ of 124 . 1.50 -1.75 1.25 -1.50

To find the z-score for an IQ of 124 using the formula \( z = \frac{(X - \mu)}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation, we substitute our values: \( \mu = 100 \) and \( \sigma = 16 \). Thus, \( z = \frac{(124 - 100)}{16} = \frac{24}{16} = 1.5 \). So the z-score for an IQ of 124 is 1.50. Interestingly, the concept of IQ tests has a deep historical context, emerging in the early 1900s to identify students needing special assistance in French schools. The Stanford-Binet test, developed by Lewis Terman in 1916, was the first standardized test to measure intelligence, and it laid foundations for educational psychology. In real-world applications, understanding z-scores helps in various fields, from education to psychology. For instance, educators can gauge student performance relative to a population, enabling tailored interventions. Similarly, in hiring practices, employers use z-scores to assess applicants' aptitude against statistical benchmarks, helping identify strong candidates in a competitive job market.