The problem provides a table showing the weekly sales of 1000 traders at a flea market, categorized into different ranges. The table includes the frequency and frequency density for each sales range. We are asked to find the value of $x$, the frequency for the sales range $400 \le S < 450$. Then we must find the value of $y$, the frequency density for the sales range $200 \le S < 300$. Finally, we need to determine the modal class.

1. Problem Description

The problem provides a table showing the weekly sales of 1000 traders at a flea market, categorized into different ranges. The table includes the frequency and frequency density for each sales range. We are asked to find the value of $x$, the frequency for the sales range $400 \le S < 450$. Then we must find the value of $y$, the frequency density for the sales range $200 \le S < 300$. Finally, we need to determine the modal class.

2. Solution Steps

(a) (i) To find the value of $x$, we use the relationship between frequency, frequency density, and class width:

Frequency Density = Frequency / Class Width

The class width for $400 \le S < 450$ is $450 - 400 = 50$. The frequency density is given as 1.

2. Therefore,

$1.2 = x / 50$

$x = 1.2 * 50 = 60$

(a) (ii) To find the value of $y$, we use the same relationship:

Frequency Density = Frequency / Class Width

For the class $200 \le S < 300$, the frequency is 380 and the class width is $300 - 200 = 100$.

$y = 380 / 100 = 3.8$

(b) (i) The modal class is the class with the highest frequency density. Comparing the frequency densities:

$50 \le S < 100$: 1.6

$100 \le S < 200$: 3.6

$200 \le S < 300$: 3.8

$300 \le S < 400$: 1.2

$400 \le S < 450$: 1.2

The highest frequency density is 3.8, which corresponds to the sales range $200 \le S < 300$.

3. Final Answer

(a) (i) $x = 60$

(a) (ii) $y = 3.8$

(b) (i) $200 \le S < 300$