Because each friend’s time is approximately normally distributed,

*the sum of their times is also approximately normally distributed*. We will do a normal approximation, but first we need to find the mean and standard deviation of the

*sum*. We learned how to do this in

Section 3.4.

Let the three friends be labeled \(X\text{,}\) \(Y\text{,}\) \(Z\text{.}\) We want \(P(X + Y + Z \lt 17.1)\text{.}\) The mean and standard deviation of the sum of \(X\text{,}\) \(Y\text{,}\) and \(Z\) is given by:

\begin{align*}
\mu_{sum} \amp = E(X+Y+Z) \amp \sigma_{sum}\amp = \sqrt{(SD_X)^2+(SD_Y)^2 + (SD_Z)^2}\\
\amp = E(X) + E(Y) + E(Z) \amp \amp = \sqrt{(0.11)^2+(0.13)^2+(0.12)^2}\\
\amp =4.6+4.8+4.5 \amp \amp = 0.208\\
\amp =17.5
\end{align*}

Now we can find the Z-score.

\begin{align*}
Z \amp = \frac{x_{sum}-\mu_{sum}}{\sigma_{sum}}\\
\amp =\frac{17.1-17.5}{0.208}\\
\amp =-1.92
\end{align*}

Finally, we want the probability that the sum is less than 17.5, so we shade the area to the left of \(Z = -1.92\text{.}\) Using the normal table or a calculator, we get

\begin{gather*}
P(Z \lt -1.92) = 0.027
\end{gather*}

There is a 2.7% chance that the friends will advance to the next level.