We can do this by checking if $n\hat{p}$ and $n(1-\hat{p})$ are both greater than 10. In this case, $n\hat{p}=120*0.69=82.8$ and $n(1-\hat{p})=120*(1-0.69)=37.2$. Since both are greater than 10, it is appropriate to use the normal distribution.
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