Question

Swim team Hanover High School has the best women's swimming team in the region. The $400$meter freestyle relay team is undefeated this year. In the $400$-meter freestyle relay, each swimmer swims $100$meters. The times, in seconds, for the four swimmers this season are approximately Normally distributed with means and standard deviations as shown. Assuming that the swimmer's individual times are independent, find the probability that the total team time in the $400$meter freestyle relay is less than $220$seconds.follow the four step process.

Swimmer	Mean	Std.dev
Wendy	55.2	2.8
Jill	58.0	3.0
Carmen	56.3	2.6
Latrice	54.7	2.7

Short answer

The probability that the total team time is less than$220$is$22\%$.

Step-by-step solution

Given Information

Given in the question that the details of the swimming team members from Hanover High School.

Calculate the mean and standard deviation 

Let $X_i,i=1,2,3,4$be the random variables showing the times in seconds for the four swimmers this season.

Given the mean and standard deviation of $X_i$are

$\begin{array}{r}{\mu }_{X_1}=55.2,{\sigma }_{X_1}=2.8\\ {\mu }_{X_2}=58.0,{\sigma }_{X_X}=3.0\\ {\mu }_{X_3}=56.3,{\sigma }_{X_3}=2.6\\ {\mu }_{X_4}=54.7,{\sigma }_{X_4}=2.7\end{array}$

The probability that the total team time in the $400$-meter freestyle relay is less than $220$seconds must be determined.

Consider the variable $T=X_1+X_2+X_3+X_4$

It showing the total times for the four swimmers in the season.

Let's find$P(T\le 220)$

Given that the swimmer's individual times are independent.

Mean of $T$

$$\begin{gathered} {\mu }_T={\mu }_{X_1}+X{\mu }_{X_2}+{\mu }_{X_3}+{\mu }_{X_4} \\ =55.2+58.0+56.3+54.7 \\ =224.2 \end{gathered}$$

Standard deviation of $T$

$$\begin{gathered} {\sigma }_T=\sqrt{{{\sigma }_{X_1}}^2+{{\sigma }_{X_2}}^2+{{\sigma }_{X_3}}^2+{{\sigma }_{X_4}}^2} \\ =\sqrt{{2.8}^2+{3.0}^2+{2.6}^2+{2.7}^2} \\ =5.56 \end{gathered}$$

Calculate the probability

The probability that the total team time in the $400$-meter freestyle relay is less than $220$seconds must be determined.

Let's standardize $T=220$

$$\begin{gathered} z=\frac{x-{\mu }_T}{{\sigma }_T} \\ =\frac{220-224.2}{5.56} \\ =-0.76 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

$$\begin{gathered} P(T\le 220)=P(z\le -0.76) \\ =P(z\ge 0.76) \\ =1-P(z\le 0.76) \\ =1-0.7764 \\ =0.2236 \end{gathered}$$