Question

intelligence quotas on two different tests are normally distributed. test a has a mean of 100 and a standard deviation of 17. test b has a mean of 100 and a standard deviation of 16. use z - scores to determine which person has the higher iq: an individual who scores 124 on test a or an individual who scores 126 on test b.
which individual has the higher iq?
a. the individual who scores 124 on test a.
b. the individual who scores 126 on test b.
c. both individuals have the same iq.

Explanation:

Step1: Calculate z - score for Test A

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the score, $\mu$ is the mean and $\sigma$ is the standard deviation. For Test A, $\mu = 100$, $\sigma=17$ and $x = 124$. So $z_A=\frac{124 - 100}{17}=\frac{24}{17}\approx1.41$.

Step2: Calculate z - score for Test B

For Test B, $\mu = 100$, $\sigma = 16$ and $x = 126$. So $z_B=\frac{126-100}{16}=\frac{26}{16}=1.625$.

Step3: Compare z - scores

Since $1.625>1.41$ (i.e., $z_B > z_A$), the individual with the score on Test B has a relatively higher IQ.

Answer:

B. The individual who scores 126 on Test B.