Question

Medley in the $4\times 100$ medley relay event, four swimmers swim $100$ yards, each using a different stroke.

A college team preparing for the conference championship looks at the times their swimmers have posted and creates a model based on the following assumptions:
$\bullet$ The swimmers' performances are independent.
$\bullet$Each swimmer's times follow a Normal model.
$\bullet$The means and standard deviations of the times (in seconds) are as shown:

$$\begin{array}{|ccc|}\hline \text{Swimmer}& \text{Mean}& \text{SD}\\ 1\text{(backstroke)}& 50.72& 0.24\\ 2\text{(breaststroke)}& 55.51& 0.22\\ 3\text{(butterfly)}& 49.43& 0.25\\ 4\text{(freestyle)}& 44.91& 0.21\\ \hline\end{array}$$
a) What are the mean and standard deviation for the relay team’s total time in this event?
b) The team’s best time so far this season was 3:19.48. (That’s 199.48 seconds.)

Do you think the team is likely to swim faster than this at the conference championship? Explain.

Answer 1

a) To determine the mean and standard deviation for the team's total time in the event.

According to formula, $E\left(X\pm Y\right)=E\left(X\right)\pm E\left(X\right)$ where $E$ is called expectation value or the mean define by ${\mu }_x=E\left(X\right)=\underset{i=1}{\overset{n}{\sum p_i{x}_i}}$.

The mean of four players are $50.72,55.51,49.43\text{and}44.91$.

Let the numbers be represented as $E\left(X_1\right)=50.72,E\left(X_2\right)=55.51,E\left(X_3\right)=49.43\text{and}E\left(X_4\right)=44.91$.

The mean of the team of four players is:

$\begin{array}{rcl}E\left(Team\right)& =& E\left(X_1+X_2+X_3+X_4\right)\\ & =& E\left(X_1\right)+E\left(X_2\right)+E\left(X_3\right)+E\left(X_4\right)\\ & =& 50.72+55.51+49.43+44.91\\ & =& 200.57\end{array}$

Hence the mean of the team of four players is $200.57$ seconds.

Now, find the standard deviation use the relation $\sigma \left(X\right)=\sqrt{Var\left(X\right)}\text{or}Var\left(X\right)={\sigma }^2\left(X\right)$ and $Var\left(X\pm Y\right)=Var\left(X\right)\pm Var\left(Y\right)$ where $\sigma$ is standard deviation and $Var$ is variance.

According to our assumption the standard deviation of the four players are given as $\sigma \left(X_1\right)=0.24,\sigma \left(X_2\right)=0.22,\sigma \left(X_3\right)=0.25\text{and}\sigma \left(X_4\right)=0.21$.

Now find the variance of the team of four players as follows:

$\begin{array}{rcl}Var\left(Team\right)& =& Var\left(X_1\right)+Var\left(X_2\right)+Var\left(X_3\right)+Var\left(X_4\right)\\ & =& {\sigma }^2\left(X_1\right)+{\sigma }^2\left(X_2\right)+{\sigma }^2\left(X_3\right)+{\sigma }^2\left(X_4\right)\\ & =& {\left(0.24\right)}^2+{\left(0.22\right)}^2+{\left(0.25\right)}^2+{\left(0.21\right)}^2\\ & =& 0.0576+0.0484+0.0625+0.0441\\ & =& 0.2126\end{array}$

The standard deviation of the team of four players is:

$\begin{array}{rcl}\sigma \left(Team\right)& =& \sqrt{Var\left(Team\right)}\\ & =& \sqrt{0.2126}\\ & =& 0.461\end{array}$

Hence the standard deviation of the team is $\begin{array}{rcl}& & 0.461\end{array}$ seconds.

b) Determine if the team can swim faster than their best timing $199.48$ seconds.

Here, need to find the probability as follows:

$\begin{array}{rcl}P\left(X_1+X_2+X_3+X<199.48\right)& =& P\left(X<199.48\right)\\ & =& P\left(Z<\frac{x-\mu }{\sigma }\right)\\ & =& P\left(Z<\frac{199.48-200.57}{0.461}\right)\\ & =& P\left(Z<-2.364\right)\\ & =& 0.009\end{array}$

Hence, there are one percent chance the team can swim faster than their best timing $199.48$ seconds.