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Figure 2 shows a histogram for the variable t which represents the time taken, in minutes, by a group of people to swim 500m - Edexcel - A-Level Maths Statistics - Question 5 - 2007 - Paper 2
Question 5
Figure 2 shows a histogram for the variable t which represents the time taken, in minutes, by a group of people to swim 500m. (a) Complete the frequency table for t... show full transcript
Worked Solution & Example Answer:Figure 2 shows a histogram for the variable t which represents the time taken, in minutes, by a group of people to swim 500m - Edexcel - A-Level Maths Statistics - Question 5 - 2007 - Paper 2
Step 1
Complete the frequency table for t.
Answer
The frequency for the interval 18-25 can be found from the histogram, which shows a height of 7.5. Since the interval width is 7 (25-18), the area (frequency) for this interval is 7.5 * 7 = 52.5. Therefore, the frequency for the interval 18-25 is 52.5. Additionally, the interval between 25-40 has a height of 15 and hence a frequency of 15 * 15 = 225. The completed frequency table looks as follows:
| t | 5-10 | 10-14 | 14-18 | 18-25 | 25-40 |
|---|---|---|---|---|---|
| Frequency | 10 | 16 | 24 | 52.5 | 15 |
Step 2
Estimate the number of people who took longer than 20 minutes to swim 500m.
Answer
From the completed frequency table, the intervals longer than 20 minutes are 25-40. The frequency for this interval is 15. Adding up the frequencies for 18-25 gives approximately 52.5. Thus, the estimated number of people who took longer than 20 minutes is:
Total = 52.5 + 15 = 67.5.
Hence, the estimated total number of people is approximately 68.
Step 3
Find an estimate of the mean time taken.
Answer
To find the mean, we first calculate the midpoints for each interval:
- For 5-10, midpoint = 7.5
- For 10-14, midpoint = 12
- For 14-18, midpoint = 16
- For 18-25, midpoint = 21.5
- For 25-40, midpoint = 32.5
Now, we multiply each midpoint by its corresponding frequency:
total = (7.5 * 10) + (12 * 16) + (16 * 24) + (21.5 * 52.5) + (32.5 * 15)
The sum of frequencies is 100. Thus, the mean estimate becomes:
Mean = ( \frac{total}{\text{frequency}} = \frac{1891}{100} = 18.91 )
Step 4
Find an estimate for the standard deviation of t.
Answer
The standard deviation can be estimated using the formula:
[ \sigma = \sqrt{ \frac{\sum (f \cdot (x - \bar{x})^2)}{n} } ]
Where:
- f = frequency,
- x = midpoint,
- n = total frequency.
Calculating this value will involve computing ( f \cdot (x - \bar{x})^2 ) for each interval and summing these results before substituting into the formula.
Step 5
Find the median and quartiles for t.
Answer
To find the median, we locate the 50th percentile of the cumulative frequency distribution. The median lies between the cumulative frequencies of 50 and 51. For quartiles, identify the 25th and 75th percentiles using cumulative frequency. This will require summation of the frequencies leading up to the desired percentiles.
Assuming cumulative frequencies have been calculated accurately, the median can be approximately observed at the interval: 18-25, and quartiles can be calculated based on this approach.
Step 6
Evaluate this measure and describe the skewness of these data.
Answer
The skewness can be evaluated using:
[ \text{Skewness} = 3 \cdot \frac{(mean - median)}{standard deviation} ]
If skewness is positive, the data is right-skewed, indicating a longer tail for higher values. Analysis of the calculated mean, median, and standard deviation in conjunction with this formula will allow for a comprehensive understanding of data distribution.