In this case
[tex]p=0.78[/tex][tex]n=80[/tex][tex]np=62.4[/tex][tex]q=1-p=0.22[/tex][tex]nq=17.6[/tex][tex]\text{Mean}=0.78=0.78[/tex][tex]\text{StandarDeviation}=\sigma=\sqrt[]{\frac{0.78\cdot0.22}{80}}\approx0.04631[/tex]And the probability that the sample proportion surviving for at least 3 years will be less than 70% is approximately 0.0421
The Z-score in this case is:
[tex]Z=\frac{0.78-0.7}{0.04631}=1.7273[/tex]