A supermarket polled 1,000 customers regarding the size of their bill.

Step 1
(a) Let E denotes the event that the size of the bill is less than $20.00 and let F denotes the event that the size of the bill is greater than or equal to $20.00 and less than $40.00. Then, the events E and F together represents the event that the size of the bill is less than $40.00. That is, ${\displaystyle E\cup F}$ represents the event that the size of the bill is less than $40.00.
Let n(E) represents the number of customers with corresponding bill size below $20.00 and n(F) represents the number of customers with corresponding bill size between $20.00 and $39.99.
Then, from the given table ${\displaystyle n\left(E\right)=209\text{and}n\left(F\right)=115}$.
The sample space S corresponds to the 1000 people. Hence, n(S)=1000.
From the given table, it can be observed that the bill sizes are disjoint.
Hence, we have ${\displaystyle n(E\cup F)=n\left(E\right)+n\left(F\right)=209+115=324}$.
Step 2
Therefore, the relative frequency that the customer's bill is less than $40.00 is given by
Relative frequency ${\displaystyle =\frac{n(E\cup F)}{n\left(S\right)}}$
${\displaystyle =\frac{324}{1000}}$
${\displaystyle =0.324}$
(b) Note that, the event that the customer's bill is $40.00 or more is the complement of the event that the customer's bill is less than $40.00.
Hence, the sum of relative frequencies of these events should be 1.
Therefore, the relative frequency that the customer's bill is $40.00 or more is given by
Relative frequency ${\displaystyle =1-\frac{324}{1000}=\frac{676}{1000}=0.676}$.