1. **Problem Statement:** We have a box and whisker diagram representing the number of sick days taken by employees in a year. The key values are: - Minimum: 2 - Lower quartile (Q1): 6 - Median (Q2): between 10 and 15 (exact value not given) - Upper quartile (Q3): near 15 - Maximum: 20 We need to find: (a.i) Minimum number of sick days. (a.ii) Lower quartile. (a.iii) Median. (b) Evaluate Paul's claim about the percentages of employees taking fewer than 6 and more than 11 sick days. 2. **Step-by-step solution:** **(a.i) Minimum number of sick days:** - The minimum is the smallest value in the data set, shown by the left whisker end. - From the diagram, minimum = $2$. **(a.ii) Lower quartile (Q1):** - The lower quartile is the left edge of the box. - Given as $6$. **(a.iii) Median (Q2):** - The median is the line inside the box. - It lies between $10$ and $15$. - Since exact value is not given, we can say median is approximately between $10$ and $15$. **(b) Paul's claim:** - Paul claims the percentage of employees with fewer than $6$ sick days is smaller than those with more than $11$ sick days. - The lower quartile $Q1=6$ means $25\%$ of employees took fewer than $6$ sick days. - The median is between $10$ and $15$, so $50\%$ took fewer than median. - To find percentage with more than $11$ sick days, note that $11$ is just above the median. - Since median splits data into two halves, more than $11$ sick days is less than $50\%$. - Also, the box extends to about $15$ (upper quartile), so $25\%$ of employees took more than $15$ sick days. - Therefore, the percentage with more than $11$ sick days is between $25\%$ and $50\%$. - Since $25\%$ took fewer than $6$ days and more than $11$ days is more than $25\%$, Paul's claim is correct. 3. **Summary:** - Minimum = $2$ - Lower quartile = $6$ - Median between $10$ and $15$ - Paul's claim is correct because the percentage of employees with fewer than $6$ sick days ($25\%$) is smaller than the percentage with more than $11$ sick days (between $25\%$ and $50\%$).