Answer to Question #316455 in Statistics and Probability for kyla

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.