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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 determine the median position. Since there are 32 students, the median position is the ( \frac{32 + 1}{2} = 16.5 )th student. We need to find the cumulative frequency:
- 10 – 19: 2
- 20 – 29: 6 (2 + 4)
- 30 – 39: 17 (6 + 11)
- 40 – 49: 22 (17 + 5)
- 50 – 59: 24 (22 + 2)
- 60 – 69: 24 (not applicable)
- 70 – 79: 26 (24 + 2)
The 16.5th student falls in the interval 30 – 39. Using linear interpolation:
[ x = 30 + \left( 16.5 - 17 \right) \cdot 10/11 = 30 + \left( -0.5 \right) \cdot \frac{10}{11} \approx 30 - 4.55 = 35.45 ]
So the estimated median is approximately 35.45.
Step 2
find the mean and the standard deviation of the times spent by the students on their homework.
Answer
The mean is calculated using the formula:
[ \text{Mean} = \frac{\sum t}{n} = \frac{1414}{32} \approx 44.1875 ]
The calculation for the standard deviation involves finding the variance first:
[ \text{Variance} = \frac{\sum t^2}{n} - \left( \frac{\sum t}{n} \right)^2 ] [ = \frac{69378}{32} - \left( 44.1875 \right)^2 \approx 216.71 - 1955.33 \approx 14.7 ]
Thus, the standard deviation is:
[ \text{Standard deviation} = \sqrt{14.7} \approx 3.83 ]
Step 3
Comment on the skewness of the distribution of the times spent by the students on their homework.
Answer
To assess the skewness, we compare the mean and the median:
Since the mean (approximately 44.19) is greater than the median (approximately 35.45), the distribution is positively skewed. This can indicate that there are some higher outliers in the times spent by the students.