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The histogram shows information about the times taken by some students to finish a puzzle - Edexcel - GCSE Maths - Question 18 - 2018 - Paper 2

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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:

  1. 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 ext{Frequency} = ext{Density} \times \text{Width} = 0.8 \times 5 = 4

  2. 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 ext{Frequency} = 0.6 \times 10 = 6

  3. For the interval (15 < n ≤ 25), the density is about 1.2, so: extFrequency=1.2×10=12 ext{Frequency} = 1.2 \times 10 = 12

  4. For the interval (25 < n ≤ 30), the frequency density is around 0.4, thus: extFrequency=0.4×5=2 ext{Frequency} = 0.4 \times 5 = 2

  5. Finally, for the interval (30 < n ≤ 50), the density is about 0.2, yielding: extFrequency=0.2×20=4 ext{Frequency} = 0.2 \times 20 = 4

The completed frequency table is:

Time taken (n minutes)Frequency
0 < n ≤ 54
5 < n ≤ 156
15 < n ≤ 2512
25 < n ≤ 302
30 < n ≤ 504

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:

  1. Calculate the total frequency: extTotalfrequency=4+6+12+2+4=28 ext{Total frequency} = 4 + 6 + 12 + 2 + 4 = 28

  2. The lower quartile corresponds to the first 25% of the data, which is: Q1=14×28=7Q1 = \frac{1}{4} \times 28 = 7

  3. 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).

  1. 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+(74104)×(155)Q1 = 5 + \left(\frac{7 - 4}{10 - 4}\right) \times (15 - 5) Q1=5+(36)×10=5+5=10Q1 = 5 + \left(\frac{3}{6}\right) \times 10 = 5 + 5 = 10

Thus, an estimate for the lower quartile is approximately 10 minutes.

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