Given that the sample mean is μ=74 and sample standard deviation σₓ=[tex] \frac{σ}{ \sqrt{n}} =\frac{12}{\sqrt{36}} =2 [/tex].
Thge Z- score is [tex] Z=\frac{76.8-74}{2} =1.4 [/tex].
Now we use the Z-score to calculate the required probability.
Refer to standard normal distribution table.
The required probability is
[tex] P(X>76.8) =P(Z>1.4)=1-P(Z<1.4)=0.0808[/tex]
