Median Iqr Df2548

1. **State the problem:** We are given daily sales data for collars and leashes over a week. We need to find the missing values in the summary table: the interquartile range (IQR) for collars and the median for leashes. 2. **Recall definitions:** - The median is the middle value when data is ordered. - The interquartile range (IQR) is the difference between the third quartile ($Q_3$) and the first quartile ($Q_1$): $$\text{IQR} = Q_3 - Q_1$$ 3. **Organize the data:** - Collars data sorted: $$17, 21, 22, 25, 26, 30, 34$$ - Leashes data sorted: $$27, 29, 30, 32, 34, 35, 39$$ 4. **Find the median for collars (already given as 25):** - The middle value of 7 data points is the 4th value: $$25$$ 5. **Find the interquartile range for collars:** - First quartile $Q_1$ is the median of the lower half (first 3 values): $$17, 21, 22$$ - Median of these is the 2nd value: $$21$$ - Third quartile $Q_3$ is the median of the upper half (last 3 values): $$26, 30, 34$$ - Median of these is the 2nd value: $$30$$ - Calculate IQR: $$\text{IQR} = 30 - 21 = 9$$ 6. **Find the median for leashes:** - Middle value of 7 data points is the 4th value: $$32$$ 7. **Check the interquartile range for leashes (given as 6):** - First quartile $Q_1$ is median of lower half: $$27, 29, 30$$ median is $$29$$ - Third quartile $Q_3$ is median of upper half: $$34, 35, 39$$ median is $$35$$ - IQR is $$35 - 29 = 6$$ which matches the given value. **Final answers:** - Collars interquartile range: $$9$$ - Leashes median: $$32$$