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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 class... 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
First, calculate the cumulative frequency:
| Time, t | Number of students | Cumulative Frequency |
|---|---|---|
| 10 – 19 | 2 | 2 |
| 20 – 29 | 4 | 6 |
| 30 – 39 | 11 | 17 |
| 40 – 49 | 5 | 22 |
| 50 – 59 | 5 | 27 |
| 60 – 69 | 5 | 32 |
| 70 – 79 | 2 | 34 |
Since the total number of students is 32, the median will be the average of the 16th and 17th values.
From the cumulative frequency, the 16th term falls in the class 30 - 39. Since there are 11 students in the 30 - 39 range, we need to interpolate:
- Lower boundary = 30
- Cumulative frequency before this class = 6
- Frequency of the class = 11
Let x be the median:
Thus, the estimated median is approximately 39.1.
Step 2
Find the mean and the standard deviation of the times spent by the students on their homework.
Answer
To find the mean:
Mean = ( \frac{\sum{t}}{n} = \frac{1414}{32} \approx 44.1875 ) (rounded to 4 decimal places).
For the standard deviation:
- First, calculate the mean of squared times:\n( \sum{t^2} = 69378 )
- Use the formula for standard deviation:
( \sigma = \sqrt{\frac{\sum{t^2}}{n} - \left(\frac{\sum{t}}{n}\right)^2} )
( \sigma = \sqrt{\frac{69378}{32} - \left(\frac{1414}{32}\right)^2} \approx \sqrt{2161.1875 - 1940.2656} \approx \sqrt{220.9219} \approx 14.87 ) (rounded to 2 decimal places).
Thus, the mean is approximately 44.19 and the standard deviation is approximately 14.87.
Step 3
Comment on the skewness of the distribution of the times spent by the students on their homework.
Answer
The mean (approximately 44.19) is greater than the median (approximately 39.1). This suggests that the distribution of times spent on homework is positively skewed. In a positively skewed distribution, more students spent less time on homework with a few students spending significantly more time, pulling the mean higher than the median.