First we calculate z:
[tex]z=\frac{mean-testedValue}{\frac{Standardeviation}{\sqrt{n}}}=\frac{51.8-50}{\frac{7.4}{\sqrt{3012}}}=13.35[/tex]Now the p-value:
[tex]p=2P(z\ge13.35)=2*0=0[/tex]Because the p value is less than the alpha value 0.02, we can reject the claim.