The same methods used to deal with finding the expected value, variance, and standard deviation of the sum/difference of two independent random variables may be extended to n independent random variables. If [tex]X _1, X _2, \ldots, X _{ n }[/tex] represent n independent random variables, then the expected value and standard deviation, respectively, of the n independent random variables are shown below.
[tex]\begin{array}{l}
E\left(X_1+X_2+\ldots+X_n\right)=E\left(X_1\right)+E\left(X_2\right)+\ldots+E\left(X_n\right) \\
\sigma_{X_1}+X_2+\ldots+X_n=\sqrt{\sigma_{X_1}^2+\sigma_{X_2}^2+\ldots+\sigma_{X_n}^2}
\end{array}[/tex]
The accompanying table summarizes the mean and standard deviation times (in seconds) of the four swimmers on Team Jax in the [tex]4 \times 100[/tex]-meter medley.
Use this information to complete parts (a) and (b).
Click the icon to view the mean and standard deviation times for Team Jax.
(a) Determine the mean and standard deviation time to complete the race for Team Jax.
The mean time to complete the race for Team Jax is 213.38 seconds.
(Type an integer or decimal rounded to two decimal places as needed.)
The standard deviation time to complete the race for Team Jax is 75 seconds.
(Type an integer or decimal rounded to two decimal places as needed.)
(b) A competing team recently had a winning time of 210.06 seconds. Assess the chances of Team Jax against this team.
The chances of Team Jax winning are $\square$ because the competing team's time is $\square$ standard deviations from Team Jax's time, which is $\square$ Team Jax's time.
(Type an integer or decimal rounded to two decimal places as needed.)
Answer (2)
Calculate the difference between Team Jax's mean time and the competing team's time: 213.38 − 210.06 = 3.32 .
Calculate how many standard deviations the competing team's time is from Team Jax's time: 7.5 3.32 ≈ 0.44 .
Calculate the ratio of the competing team's time to Team Jax's time: 213.38 210.06 ≈ 0.98 .
The chances of Team Jax winning are lower because the competing team's time is 0.44 standard deviations from Team Jax's time, which is 0.98 of Team Jax's time. lower, 0.44, 0.98
Explanation
Understand the problem and provided data We are given that Team Jax has a mean time of 213.38 seconds and a standard deviation of 7.5 seconds. A competing team has a time of 210.06 seconds. We need to assess Team Jax's chances against the competing team.
Calculate the difference in times First, we calculate the difference between Team Jax's mean time and the competing team's time: d = Team Jax’s mean time − Competing team’s time = 213.38 − 210.06 = 3.32 So, Team Jax's mean time is 3.32 seconds slower than the competing team's time.
Calculate the number of standard deviations Next, we calculate how many standard deviations the competing team's time is from Team Jax's time: z = Team Jax’s standard deviation d = 7.5 3.32 = 0.442666... ≈ 0.44 This means the competing team's time is approximately 0.44 standard deviations better than Team Jax's mean time. Since a lower time is better, a positive z means the competing team did better than Team Jax's average.
Calculate the ratio of the times Now, we calculate the ratio of the competing team's time to Team Jax's time: r = Team Jax’s mean time Competing team’s time = 213.38 210.06 = 0.9844409... ≈ 0.98 This means the competing team's time is about 98% of Team Jax's mean time.
Assess the chances of Team Jax winning Since the competing team's time is faster than Team Jax's mean time (by 0.44 standard deviations), the chances of Team Jax winning are lower. The competing team's time is 0.44 standard deviations from Team Jax's time, which is 0.98 of Team Jax's time.
Final Answer The chances of Team Jax winning are lower because the competing team's time is 0.44 standard deviations from Team Jax's time, which is 0.98 of Team Jax's time.
Examples
In sports analytics, understanding the distribution of team performance is crucial for making predictions. By comparing a team's average performance against a specific opponent's result, we can estimate the likelihood of winning. This involves calculating how many standard deviations the opponent's result is away from the team's average and expressing the opponent's result as a ratio of the team's average, providing a comprehensive assessment of the team's chances.
Team Jax's chances of winning are lower because the competing team's time is 0.0443 standard deviations faster than Team Jax's mean time, which is 0.983 of Team Jax's time.
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