1. **Problem statement:** We have a frequency distribution of amounts spent by customers in intervals and their frequencies. (a) We need to draw a histogram representing this data. (b) We want to find the value $x$ such that customers spending more than $x$ represent 20% of all customers. 2. **Histogram construction:** - The class intervals are: $0 < a \leq 10$, $10 < a \leq 20$, $20 < a \leq 40$, $40 < a \leq 60$. - Frequencies are: 34, 19, 22, 5 respectively. - The width of each class is: 10, 10, 20, 20. - Histogram bars have heights proportional to frequency density = $\frac{\text{frequency}}{\text{class width}}$. Calculate frequency densities: - For $0 < a \leq 10$: $\frac{34}{10} = 3.4$ - For $10 < a \leq 20$: $\frac{19}{10} = 1.9$ - For $20 < a \leq 40$: $\frac{22}{20} = 1.1$ - For $40 < a \leq 60$: $\frac{5}{20} = 0.25$ 3. **Finding $x$ for 20% customers spending more than $x$:** - Total customers = $34 + 19 + 22 + 5 = 80$ - 20% of 80 = $0.20 \times 80 = 16$ customers - We want the smallest $x$ such that customers spending more than $x$ are 16. 4. **Cumulative frequencies from highest to lowest intervals:** - $40 < a \leq 60$: 5 customers - $20 < a \leq 40$: 22 customers - $10 < a \leq 20$: 19 customers - $0 < a \leq 10$: 34 customers Starting from the top: - Customers spending more than 40: 5 (less than 16) - Customers spending more than 20: $5 + 22 = 27$ (more than 16) So $x$ lies between 20 and 40. 5. **Find exact $x$ using linear interpolation:** - Number of customers spending more than 40 = 5 - Number spending more than 20 = 27 - We want 16 customers, which is between 5 and 27. Let $x$ be the amount such that customers spending more than $x$ = 16. Using linear interpolation: $$16 = 5 + \frac{27 - 5}{20} (40 - x)$$ Simplify: $$16 - 5 = \frac{22}{20} (40 - x)$$ $$11 = 1.1 (40 - x)$$ $$\frac{11}{1.1} = 40 - x$$ $$10 = 40 - x$$ $$x = 40 - 10 = 30$$ **Final answer:** (a) Histogram bars with heights 3.4, 1.9, 1.1, 0.25 over intervals 0-10, 10-20, 20-40, 40-60. (b) The value of $x$ is $30$.