Given that the z-score of science is 2.25, the proportion of values greater than this is calculated as,
[tex]P(z>2.25)=P(z>0)-P(0Since the normal curve is symmetric, the area after z=0 is 0.5,[tex]P(z>2.25)=0.5-\emptyset(2.25)[/tex]From the Standard Normal Distribution Table,
[tex]\emptyset(2.25)=0.4878[/tex]Substitute this value,
[tex]\begin{gathered} P(z>2.25)=0.5-0.4878 \\ P(z>2.25)=0.0122 \end{gathered}[/tex]Thus, 0.0122 proportion of the normal distribution corresponds to z-score values greater than the child’s z-score on the science test.
