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On a randomly chosen day, each of the 32 students in a class recorded the time, t / minutes to the nearest minute, they spent on their homework - Edexcel - A-Level Maths Statistics - Question 5 - 2011 - Paper 1
Question 5
On a randomly chosen day, each of the 32 students in a class recorded the time, t / minutes to the nearest minute, they spent on their homework. The data for the cla... show full transcript
Worked Solution & Example Answer:On a randomly chosen day, each of the 32 students in a class recorded the time, t / minutes to the nearest minute, they spent on their homework - Edexcel - A-Level Maths Statistics - Question 5 - 2011 - Paper 1
Step 1
Use interpolation to estimate the value of the median.
Answer
To find the median, we first need to determine the position of the median.
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Calculate the median position:
Position = \frac{N + 1}{2} = \frac{32 + 1}{2} = 16.5
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Now, we locate the 16.5th term in the cumulative frequency. The cumulative frequencies are:
- 10 – 19: 2
- 20 – 29: 6
- 30 – 39: 17
- 40 – 49: 28
- 50 – 59: 33
The 16.5th term falls between 30-39 and 40-49.
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Now, we use interpolation to estimate the median:
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Lower boundary of 30 - 39 = 30
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Frequency of 30 - 39 = 11
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Cumulative frequency before this class = 6
Median = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \cdot c
Where:
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L = 30
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CF (cumulative frequency before) = 6
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f = 11 (frequency of class)
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c (class width) = 10
Median = 30 + \left( \frac{16 - 6}{11} \right) \cdot 10 = 30 + 9.09 = 39.09
Therefore, the estimated median is approximately 39.1.
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Step 2
find the mean and the standard deviation of the times spent by the students on their homework.
Answer
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Calculate Mean:
Mean = \frac{\Sigma t}{N} = \frac{1414}{32} = 44.1875
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Calculate Standard Deviation:
To find the standard deviation, we use the formula:
( \sigma = \sqrt{\frac{\Sigma t^2}{N} - \left( \frac{\Sigma t}{N} \right)^2} )
( \sigma = \sqrt{\frac{69378}{32} - (44.1875)^2} )
Calculate each component:
- ( \frac{69378}{32} = 2161.1875 )
- ( (44.1875)^2 = 1955.625 )
Therefore,
( \sigma = \sqrt{2161.1875 - 1955.625} = \sqrt{205.5625} \approx 14.36 )
Thus, the mean is approximately 44.19 and the standard deviation is approximately 14.36.
Step 3
Comment on the skewness of the distribution of the times spent by the students on their homework.
Answer
The mean of the distribution is approximately 44.19, which is greater than the estimated median of 39.1. This indicates that the distribution of times spent is positively skewed.
In a positively skewed distribution, the tail on the right side is longer or fatter, meaning that a few students spent significantly more time on homework than the majority. Therefore, we conclude that the data is positively skewed based on this comparison.