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intelligence quotas on two different tests are normally distributed. te…

Question

intelligence quotas on two different tests are normally distributed. test a has a mean of 100 and a standard deviation of 16. 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 125 on test a or an individual who scores 130 on test b. which individual has the higher iq? a. the individual who scores 125 on test a. b. the individual who scores 130 on test b. 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: \(\mu = 100\), \(\sigma=16\), \(x = 125\)
\(z_A=\frac{125 - 100}{16}=\frac{25}{16}=1.5625\)

Step2: Calculate z - score for Test B

For Test B: \(\mu = 100\), \(\sigma = 17\), \(x = 130\)
\(z_B=\frac{130 - 100}{17}=\frac{30}{17}\approx1.7647\)

Step3: Compare z - scores

Since \(1.7647>1.5625\) (i.e., \(z_B>z_A\))

Answer:

B. The individual who scores 130 on Test B.