The King of the Hill triathlon race in Kinsale consists of a 750 metre swim, followed by a 20 kilometre cycle, followed by a 5 kilometre run - Leaving Cert Mathematics - Question 7 - 2014
Question 7
The King of the Hill triathlon race in Kinsale consists of a 750 metre swim, followed by a 20 kilometre cycle, followed by a 5 kilometre run.
Maire is analysing dat... show full transcript
Worked Solution & Example Answer:The King of the Hill triathlon race in Kinsale consists of a 750 metre swim, followed by a 20 kilometre cycle, followed by a 5 kilometre run - Leaving Cert Mathematics - Question 7 - 2014
Step 1
Use the summary statistics in the table to decide which histogram corresponds to each event.
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Answer
The histogram corresponding to the swim event is the middle histogram because it shows a distribution centered around the given mean of approximately 18.3 minutes. The histogram with the shortest duration corresponds to the run event, which has the least duration range. Thus, the leftmost histogram represents the cycle, as its mean is closest to that of the cycles shown.
Step 2
The mean and the median time for the run are approximately equal. Estimate this value from the corresponding histogram.
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Answer
From the histogram for the run, we can observe that both the mean and the median time appear to be close to 25 minutes, as most values cluster around this point.
Step 3
Estimate from the relevant histogram the standard deviation of the times for the swim.
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The standard deviation for the swim can be estimated visually from the histogram and appears to be around 3 minutes, given the spread of the data.
Step 4
When calculating the summary statistics, the software failed to find a mode for the data sets. Why do you think this is?
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The software likely failed to find a mode because the data for each event may not have a single value that appears most frequently. Instead, the distribution may be more continuous or uniform, without repeating values.
Step 5
Give a brief summary of the relationship between performance in the different events, based on the scatter diagrams.
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The scatter diagrams indicate a positive correlation between swim times and run times, suggesting that athletes who swim faster tend to run faster as well. There is also a strong positive correlation between run times and cycle times, indicating athletes who perform well in one event may also excel in another.
Step 6
The best-fit line for run-time based on swim-time is y = 0.53x + 15.2. The best-fit line for run-time based on cycle-time is y = 0.58x + 0.71. Brian did the swim in 17.6 minutes and the cycle in 35.7 minutes. Give your best estimate of Brian’s time for the run, and justify your answer.
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Answer
Using the equations provided, we estimate Brian’s run time using both:
From swim time: y=0.53(17.6)+15.2=24.528 minutes.
From cycle time: y=0.58(35.7)+0.71=21.416 minutes.
Thus, the average estimate for his run time would be (24.528+21.416)/2=22.972 minutes.
Step 7
Based on an assumption that the distribution of overall finishing times is approximately normal, use the empirical rule to complete the following sentence.
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By applying the empirical rule, we conclude that "95% of the athletes took between 67.5 and 108.7 minutes to complete the race." This is calculated based on two standard deviations from the mean finishing time.
Step 8
Using normal distribution tables, estimate the number of athletes who completed the race in less than 100 minutes.
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The z-score for 100 minutes is calculated as: z=10.3100−88.1=1.155. Referring to the normal distribution table, the probability of participants finishing in less than 100 minutes is approximately 0.8760, leading to an estimated number of athletes of 196.
Step 9
Assuming the athletes are cooperative and truthful, what is the probability that the second person she interviews will be the sixth person she approaches?
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The probability that a randomly selected athlete has a finishing time over 100 minutes is 0.124. Using this, we find the probability that the second occurrence happens on the sixth trial by modeling this scenario with a geometric distribution. The resulting probability is approximately 0.0453.
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