# Average frequency example 3

Subject: Advertising and Public Relations

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/apr/#tee/practice/numbers/frequency3

Canonical: https://deep-learning-study-room.vercel.app/read/apr/tee/practice/numbers--frequency3/

Extra drills; no claim that these exact questions or allocated marks appeared in TEE

### Faculty-note worked example. In a four-week campaign, 6,000 people receive 5 exposures each, 6,000 people receive 7 exposures each, 3,000 people receive 3 exposures each. The groups are distinct with no overlap. Find gross exposures, unique reach and average frequency. [3 practice marks]

#### Answer

Faculty-note worked example: the values come from APR M2 notes p28. The question has been rewritten to teach the steps; this is not verbatim PYQ wording. Any practice-mark allocation is author-set.

## Given

| Group | Distinct people | Exposures each |
| --- | --- | --- |
| 1 | 6000 | 5 |
| 2 | 6000 | 7 |
| 3 | 3000 | 3 |

## To find

Gross exposures, unduplicated reach and average frequency.

## Formula and why



$$
I=\sum_i n_i f_i,\qquad n=\sum_i n_i,\qquad\bar f=\frac{I}{n}
$$



Multiply each group size by its frequency; then add. The reach sum is valid because the question explicitly says the groups do not overlap.

## Step-by-step solution

#### Step 1. Count exposures for group 1



$$
6000\times5=30000
$$



#### Step 2. Count exposures for group 2



$$
6000\times7=42000
$$



#### Step 3. Count exposures for group 3



$$
3000\times3=9000
$$



#### Step 4. Add all group exposures



$$
30000+42000+9000=81000
$$



#### Step 5. Count distinct people



$$
6000+6000+3000=15000
$$



#### Step 6. Divide exposures by people reached



$$
\bar f=\frac{81000}{15000}=5.4
$$



**Final answer:** Gross exposures = 81,000; reach = 15,000 people; average frequency = 5.4.

#### Practical interpretation · What the answer means

5.4 is the average number of opportunities among reached people. It does not mean each individual had exactly that many exposures.