Q:

A test score of 48.4 on a test having a mean of 66 and a standard deviation of 11. Find the ​z-score corresponding to the given value and use the ​z-score to determine whether the value is significant. Consider a score to be significant if its ​z-score is less than -2.00 or greater than 2.00. Round the ​z-score to the nearest tenth if necessary. A. -1.6; not significant B.-17.6; significant C. -1.6, significant D. 1.6; not significant

Accepted Solution

A:
Answer:The z-score is -1.6 and it is not significant .Step-by-step explanation:Given : Test score = 48.4             Mean = 66             Standard deviation = 11To Find :Find the ​z-score corresponding to the given value and use the ​z-score to determine whether the value is significant.Solution:Formula : [tex]z=\frac{x-\mu}{\sigma}[/tex][tex]\mu =66\\x=48.4\\\sigma =11[/tex]Substitute the values [tex]z=\frac{48.4-66}{11}[/tex] [tex]z=-1.6[/tex]So, the z-score is -1.6Now we are given that Consider a score to be significant if its ​z-score is less than -2.00 or greater than 2.00. Since -1.6>-2 and -1.6<2So, It is not significant Hence the z-score is -1.6 and it is not significant .