‘Find the z-score for an IQ test score of 142 when the mean is 100 and the slandard deviation is 15 0 ~2.8 2.8 0 127 18.78’
The Correct Answer and Explanation is:
To find the z-score for an IQ test score of 142 when the mean is 100 and the standard deviation is 15, we use the formula: z=x−μσz = \frac{x – \mu}{\sigma}z=σx−μ
Where:
- xxx is the individual score (142),
- μ\muμ is the mean of the distribution (100),
- σ\sigmaσ is the standard deviation (15).
Substituting the values: z=142−10015z = \frac{142 – 100}{15}z=15142−100 z=4215z = \frac{42}{15}z=1542 z=2.8z = 2.8z=2.8
So, the z-score for an IQ score of 142 is 2.8.
Explanation:
A z-score is a measure that describes a value’s position relative to the mean of a group of values. Specifically, it tells you how many standard deviations the value is away from the mean. A positive z-score indicates that the value is above the mean, while a negative z-score would indicate that the value is below the mean.
In this case:
- The mean IQ score is 100, which is considered the average.
- The standard deviation of 15 tells us the average amount by which individual scores vary from the mean.
- A score of 142 is 42 points higher than the mean (100), and when divided by the standard deviation (15), it gives us a z-score of 2.8.
This means that an IQ score of 142 is 2.8 standard deviations above the average IQ score. This is considered quite a high score, placing the individual well above the average population in terms of IQ performance.
