Chapter 16, Problem 41E

(a)

To determine

To find: The mean and standard deviation for the cumulative time of the relayed team in this case can be found.

Answer to Problem 41E

Mean = 200.57

Standard Deviation = 0.461

Explanation of Solution

Given:

Swimmer	Mean	SD
1 (backstroke)	50.72	0.24
2 (breaststroke)	55.51	0.22
3 (butterfly)	49.43	0.25
4 (freestyle)	44.91	0.21

Suppose

B = Back Stroke Time Swimmer

R = breast stroke time swimmer

F = butterfly stroke time swimmer

S = free style time swimmer

Formula used:

  $\begin{array}{c}E\left(T\right)=E\left(B+R+F+S\right)\\ E\left(T\right)=\mathrm{var}\left(B+R+F+S\right)\\ SD\left(T\right)=\sqrt{\mathrm{var}\left(T\right)}\end{array}$

Calculation:

T be the time swimmer did all 4 strokes

  $\begin{array}{c}T=B+R+F+S\\ E\left(T\right)=E\left(B+R+F+S\right)\\ =50.72+55.51+49.43+44.91\\ =200.57\end{array}$

  $\begin{array}{c}E\left(T\right)=\mathrm{var}\left(B+R+F+S\right)\\ =\mathrm{var}\left(B\right)+\mathrm{var}\left(R\right)+\mathrm{var}\left(F\right)+\mathrm{var}\left(S\right)\\ ={\left(0.24\right)}^2+{\left(0.22\right)}^2+{\left(0.25\right)}^2+{\left(0.21\right)}^2\\ =0.2126\\ SD\left(T\right)=\sqrt{\mathrm{var}\left(T\right)}\\ =\sqrt{0.2126}\\ =0.461\end{array}$

Therefore, the mean and standard deviation is mention by 200.57 and 0.461 respectively.

(b)

To determine

To Explain: that at the conference championship the team is expected to swim better than that.

Answer to Problem 41E

0.991

Explanation of Solution

Given:

Swimmer	Mean	SD
1 (backstroke)	50.72	0.24
2 (breaststroke)	55.51	0.22
3 (butterfly)	49.43	0.25
4 (freestyle)	44.91	0.21

Suppose

B = Back Stroke Time Swimmer

R = breast stroke time swimmer

F = butterfly stroke time swimmer

S = free style time swimmer

Calculation:

probability that researcher can swim faster is mentioned by

  $\begin{array}{c}P\left(T>199.48\right)=P\left(Z>\frac{199.48-200.57}{0.461}\right)\\ =P\left(Z>-2.36\right)\\ =0.991\end{array}$

Graph:

  

Required probability to run faster is 0.991