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AQA GCSE · Question 09.1 · Statistical Measures and Calculations
A tech company's website offers rewards according to how many days you visit it. The table shows the number of rewards achieved by a sample of 500 customers. Show, with working that, for this sample, median = interquartile range.
A tech company's website offers rewards according to how many days you visit it. The table shows the number of rewards achieved by a sample of 500 customers. Show, with working that, for this sample, median = interquartile range.
How to approach this question
1. **Calculate Frequencies**: Convert the percentages to actual frequencies by multiplying by the total sample size (500).
* 1 reward: 0.528 * 500 = 264
* 2 rewards: 0.272 * 500 = 136
* ...and so on for all categories.
2. **Calculate Cumulative Frequencies (CF)**: Add up the frequencies as you go down the table.
* CF for 1 reward = 264
* CF for 2 rewards = 264 + 136 = 400
* ...and so on.
3. **Find the Median**: Locate the position of the median value: (n+1)/2 = (500+1)/2 = 250.5. Find which category this position falls into using the CF.
4. **Find the Quartiles**:
* Lower Quartile (Q1) position: (n+1)/4 = 125.25. Find the category.
* Upper Quartile (UQ) position: 3(n+1)/4 = 375.75. Find the category.
5. **Calculate the Interquartile Range (IQR)**: IQR = UQ - Q1.
6. **Compare**: Check if the value you found for the median is equal to the value you calculated for the IQR.
Full Answer
First, we need to find the number of customers for each category and then the cumulative frequency. Total customers = 500.
- 1 reward: 52.8% of 500 = 0.528 * 500 = 264 customers. (CF = 264)
- 2 rewards: 27.2% of 500 = 0.272 * 500 = 136 customers. (CF = 264 + 136 = 400)
- 3 rewards: 10.4% of 500 = 0.104 * 500 = 52 customers. (CF = 400 + 52 = 452)
- 4 rewards: 7.6% of 500 = 0.076 * 500 = 38 customers. (CF = 452 + 38 = 490)
- 5 rewards: 2% of 500 = 0.02 * 500 = 10 customers. (CF = 490 + 10 = 500)
Now, find the median and quartiles:
- **Median (Q2)**: The position is (500+1)/2 = 250.5th value. This value falls in the "1 reward" category (as CF is 264). So, **Median = 1**.
- **Lower Quartile (Q1)**: The position is (500+1)/4 = 125.25th value. This value falls in the "1 reward" category. So, **LQ = 1**.
- **Upper Quartile (UQ)**: The position is 3*(500+1)/4 = 375.75th value. This value falls in the "2 rewards" category (as it's between 264 and 400). So, **UQ = 2**.
- **Interquartile Range (IQR)**: IQR = UQ - LQ = 2 - 1 = **1**.
Therefore, Median (1) = Interquartile Range (1).
This question requires finding the median and interquartile range from a frequency table and showing they are equal. The data is discrete.
**Step 1: Calculate frequencies and cumulative frequencies (CF)**
The total sample size is 500.
| Rewards | % | Frequency (f) | Cumulative Frequency (CF) |
|:-------:|:---:|:-------------:|:-------------------------:|
| 1 | 52.8| 264 | 264 |
| 2 | 27.2| 136 | 400 |
| 3 | 10.4| 52 | 452 |
| 4 | 7.6 | 38 | 490 |
| 5 | 2.0 | 10 | 500 |
**Step 2: Find the Median (Q2)**
The position of the median is the (n+1)/2-th value, which is (500+1)/2 = 250.5.
Looking at the CF column, the 250.5th value lies within the first group (up to the 264th value).
So, the **Median = 1**.
**Step 3: Find the Interquartile Range (IQR)**
First, find the lower quartile (Q1) and upper quartile (Q3).
- Q1 position = (n+1)/4 = (500+1)/4 = 125.25. The 125.25th value also lies within the first group. So, **Q1 = 1**.
- Q3 position = 3(n+1)/4 = 3(500+1)/4 = 375.75. The 375.75th value lies beyond the 264th value but before the 400th value, so it is in the second group. So, **Q3 = 2**.
- IQR = Q3 - Q1 = 2 - 1 = **1**.
**Step 4: Conclusion**
The Median is 1 and the IQR is 1. Therefore, median = interquartile range.
Common mistakes
✗ Using n/2, n/4, 3n/4 for positions instead of (n+1)/2 etc. for discrete data.
✗ Errors in calculating frequencies from percentages.
✗ Incorrectly identifying the category from the cumulative frequency.
Practice the full AQA GCSE Statistics Higher Tier Paper 1
42 questions · hints · full answers · grading
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