In the 4 × 100 relay event,each of four runners runs 100 metres.A university team is preparing for a competition.The means and standard deviations of the times (in seconds) of their four runners are as shown in the table: $112.190.12212.370.14312.010.08411.730.07\begin{array} { c | c | c } \text { Runner } & \text { Mean } & \text { SD } \\\hline 1 & 12.19 & 0.12 \\2 & 12.37 & 0.14 \\3 & 12.01 & 0.08 \\4 & 11.73 & 0.07\end{array}$ What are the mean and standard deviation of the relay team's total time in this event? Assume that the runners' performances are independent.

A) ? = 48.3 sec,? = 12.64 sec
B) ? = 48.3 sec,? = 0.045 sec
C) ? = 24.15 sec,? = 0.21 sec
D) ? = 48.3 sec,? = 0.21 sec
E) ? = 24.15 sec,? = 12.64 sec

Question

In the 4 × 100 relay event,each of four runners runs 100 metres.A university team is preparing for a competition.The means and standard deviations of the times (in seconds) of their four runners are as shown in the table:

112.190.12212.370.14312.010.08411.730.07\begin{array} { c | c | c } \text { Runner } & \text { Mean } & \text { SD } \\\hline 1 & 12.19 & 0.12 \\2 & 12.37 & 0.14 \\3 & 12.01 & 0.08 \\4 & 11.73 & 0.07\end{array}

What are the mean and standard deviation of the relay team's total time in this event? Assume that the runners' performances are independent.

A) ? = 48.3 sec,? = 12.64 sec
B) ? = 48.3 sec,? = 0.045 sec
C) ? = 24.15 sec,? = 0.21 sec
D) ? = 48.3 sec,? = 0.21 sec
E) ? = 24.15 sec,? = 12.64 sec

Jasmine Irwin · Accepted Answer

Final Answer : D, Explanation :  The mean of the relay team's total time would simply be the sum of the means of the individual runners, which is:μ = 12.1 + 12.2 + 12.5 + 11.5 = 48.3 secTo find the standard deviation of the total time, we need to use the formula:σ_total = sqrt(σ_1^2 + σ_2^2 + σ_3^2 + σ_4^2)where σ_1, σ_2, σ_3, and σ_4 are the standard deviations of the individual runners.Plugging in the given values, we get:σ_total = sqrt(0.06^2 + 0.1^2 + 0.07^2 + 0.08^2) ≈ 0.21 secTherefore, the best choice is D, with μ = 48.3 sec and σ = 0.21 sec.

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