The histogram shows information about the times taken by some students to finish a puzzle - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 2
Question 18
The histogram shows information about the times taken by some students to finish a puzzle.
(a) Complete the frequency table for this information.
Time taken (n min... show full transcript
Worked Solution & Example Answer:The histogram shows information about the times taken by some students to finish a puzzle - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 2
Step 1
Complete the frequency table for this information.
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Answer
To complete the frequency table, we need to calculate the frequencies based on the histogram data:
For the interval (0 < n ≤ 5), the histogram indicates a frequency density of approximately 0.8. The width of this interval is 5 minutes. Thus, the frequency is:
extFrequency=extDensity×Width=0.8×5=4
For the interval (5 < n ≤ 15), the histogram shows a frequency density of approximately 0.6 for a width of 10 minutes:
extFrequency=0.6×10=6
For the interval (15 < n ≤ 25), the density is about 1.2, so:
extFrequency=1.2×10=12
For the interval (25 < n ≤ 30), the frequency density is around 0.4, thus:
extFrequency=0.4×5=2
Finally, for the interval (30 < n ≤ 50), the density is about 0.2, yielding:
extFrequency=0.2×20=4
The completed frequency table is:
Time taken (n minutes)
Frequency
0 < n ≤ 5
4
5 < n ≤ 15
6
15 < n ≤ 25
12
25 < n ≤ 30
2
30 < n ≤ 50
4
Step 2
Find an estimate for the lower quartile of the times taken to finish the puzzle.
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Answer
To find the lower quartile (Q1), we need to determine the position of Q1 in the cumulative frequency:
Calculate the total frequency:
extTotalfrequency=4+6+12+2+4=28
The lower quartile corresponds to the first 25% of the data, which is:
Q1=41×28=7
Now, we find where this falls in the cumulative frequency:
For the interval (0 < n ≤ 5): Cumulative frequency = 4
For the interval (5 < n ≤ 15): Cumulative frequency = 10
For the interval (15 < n ≤ 25): Cumulative frequency = 22
Since Q1 = 7 falls in the second interval (5 < n ≤ 15).
Now, we use linear interpolation to estimate the exact value:
Where the cumulative frequency just before 7 is 4 and just after is 10, we can interpolate:
Q1=5+(10−47−4)×(15−5)Q1=5+(63)×10=5+5=10
Thus, an estimate for the lower quartile is approximately 10 minutes.