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.

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.