The expected number of contracts is calculated using the formula for the expected value of a binomial distribution:

$E(X) = n \times p$

where:

• $n$ is the number of trials (in this case, the number of clients),
• $p$ is the probability of success (getting a sales contract with a client).

Given:

• $n = 21$ (the number of clients),
• $p = 0.22$ (the probability of getting a contract with any given client),

The expected number of contracts is:

$E(X) = 21 \times 0.22 = 4.62$

Thus, the expected number of contracts the sales representative will get this week is 4.6200.

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. What is the variance and standard deviation of the number of contracts?
2. What is the probability of getting exactly 5 contracts?
3. What is the probability of getting more than 4 contracts?
4. How would the expected value change if the probability of success increased to 0.30?
5. What is the cumulative probability of getting between 3 and 6 contracts?

Tip: The expected value in a binomial distribution is directly proportional to both the number of trials and the probability of success.