Question

alex scored 90 during the first periodic exam in mathematics and 88 during the second periodic exam.The scores in first periodic exam ha a mean μ=83 and a standard deviation Ó=9.Scores in the second periodic exam ha av mean μ=8p and the standard deviation Ó=8.In which periodic exam was his standing better, assuming that the scores in his periodic exam are normally distributed?

Answer 1

To answer this question we need to find the z score for each exam.

$z_1=\frac{x_1-\mu_2}{\sigma_1}=\frac{90-83}{9}=0.7778$

$z_2= \frac{x_2-\mu_2}{\sigma_2} = \frac{ 88−80}{8} ​ =1$

We see that the highest z-score is 1, which means that Alex was his standing better in the second periodic exam.