Given the formula
[tex]\sigma^2=\mu(1-p)[/tex][tex]\begin{gathered} \text{where }\sigma=\text{standard deviation} \\ \mu=the\text{ average grade} \\ p=\text{ probability of passing the course} \end{gathered}[/tex][tex]\begin{gathered} \mu=50\text{ \%=}\frac{50}{100}=0.5 \\ \sigma=\text{ 15\%=}\frac{\text{15}}{100}=0.15 \end{gathered}[/tex][tex]\begin{gathered} 0.15^2=0.5(1-p) \\ 0.0225=0.5(1-p) \\ \text{Divide both sides by 0.5} \\ \frac{0.0225}{0.5}=1-p \\ 0.045=1-p \\ \text{collect like terms} \\ p=1-0.045 \\ p=0.955 \end{gathered}[/tex]Hence, the probability that the student passes the second professor's class is 0.955.
