The cumulative frequency table shows information about the times, in minutes, taken by 40 people to complete a puzzle - Edexcel - GCSE Maths - Question 11 - 2019 - Paper 1
Question 11
The cumulative frequency table shows information about the times, in minutes, taken by 40 people to complete a puzzle.
Time (m minutes)
20 < m ≤ 40 5
20 < m ≤ 60 ... show full transcript
Worked Solution & Example Answer:The cumulative frequency table shows information about the times, in minutes, taken by 40 people to complete a puzzle - Edexcel - GCSE Maths - Question 11 - 2019 - Paper 1
Step 1
On the grid below, draw a cumulative frequency graph for this information.
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Answer
To construct the cumulative frequency graph, we will plot the cumulative frequency against the upper class boundaries of the corresponding time intervals. The points to plot are:
(40, 5)
(60, 25)
(80, 38)
(100, 40)
After plotting these points, connect them with a smooth curve, ensuring to label the graph accurately.
Step 2
Use your graph to find an estimate for the interquartile range.
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Answer
To find the interquartile range, we need to determine the first quartile (Q1) and the third quartile (Q3) from the graph. From the cumulative frequency of 40, calculate:
Q1: Locate the point at the 10th percentile (0.25 * 40 = 10) on the vertical axis and trace horizontally to find the corresponding time on the horizontal axis.
Q3: Locate the point at the 30th percentile (0.75 * 40 = 30) and do the same.
The interquartile range is given by Q3 - Q1.
Step 3
Use your graph to find an estimate for the probability that this person took between 50 minutes and 90 minutes to complete the puzzle.
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Answer
To calculate this probability from the graph, first determine the cumulative frequencies at 50 minutes and 90 minutes by tracing up from these time values to the graph- follow horizontally and back down to read the frequencies. The probability is then calculated as:
P(50<m<90)=40C(90)−C(50)
where C(90) and C(50) are the cumulative frequencies corresponding to 90 and 50 minutes, respectively.