SOLUTION:
We get the z score to be;
[tex]Z=\frac{\sqrt{n}(x-\mu)}{\sigma}[/tex]Inserting the values, we have;
[tex]\begin{gathered} z-\frac{\sqrt{35}(0.916-0.976)}{0.321} \\ \\ z=-1.1058 \end{gathered}[/tex]Using a z score table, we have;
[tex]\begin{gathered} P(\mu<0.916)=P(z<-1.1058) \\ \\ \\ =0.13441 \end{gathered}[/tex]Our judgement for this claim is;
Yes. The probability of this data is unlikely to have occurred by chance alone.