Question

intelligence quotas on two different tests are normally distributed test a has a mean of 100 and a standard deviation of 15 test b has a mean of 100 and a standard deviation of 17 use z - scores to determine which person has the higher iq an individual who scores 121 on test a or an individual who scores 122 on test b which individual has the higher iq? a. the individual who scores 122 on test b b. the individual who scores 121 on test a c. both individuals have the same iq

Explanation:

Step1: Calculate z - score for Test A

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For Test A: \(\mu = 100\), \(\sigma=15\), \(x = 121\)
\(z_A=\frac{121 - 100}{15}=\frac{21}{15}=1.4\)

Step2: Calculate z - score for Test B

For Test B: \(\mu = 100\), \(\sigma = 17\), \(x = 122\)
\(z_B=\frac{122-100}{17}=\frac{22}{17}\approx1.29\)

Step3: Compare z - scores

Since \(1.4>1.29\) (i.e., \(z_A>z_B\))

Answer:

B. The individual who scores 121 on Test A