Question

a job placement agency advertised that last year its clients, on average, had a starting salary of \\$40,000. assuming that average refers to the mean, which of the following claims must be true based on this information?

note: more than one statement could be true. if none of the statements is true, mark the appropriate box.

last year all of their clients had a starting salary of at least \\$40,000.
two years ago some of their clients had a starting salary of at least \\$40,000.
last year at least one of their clients had a starting salary of more than \\$37,000.
last year, the number of their clients who had a starting salary of less than \\$40,000 was equal to the number of their clients who had a starting salary of more than \\$40,000.
last year at least one of their clients had a starting salary of exactly \\$40,000.
none of the above statements are true.

Explanation:

Analyze the given mean salary

Using the Sample Mean and Mean Properties knowledge points

$$ \mu = \$40,000 $$

The arithmetic mean is the sum of all starting salaries divided by the total number of clients \(N\).

Evaluate statement 1 and statement 2

Using the Statistical Interpretation knowledge point

• Statement 1: "Last year all of their clients had a starting salary of at least \$40,000."

If all clients earned at least \$40,000, and even one earned more, the mean would exceed \$40,000. If some earned less, this statement is false. It is not guaranteed to be true.

• Statement 2: "Two years ago some of their clients had a starting salary of at least \$40,000."

The advertisement only provides data for "last year." No information is given about "two years ago." Thus, this is not guaranteed.

Evaluate statement 3

Using the Mean Properties and Statistical Interpretation knowledge points

• Statement 3: "Last year at least one of their clients had a starting salary of more than \$37,000."

Suppose for contradiction that all clients had a starting salary of \$37,000 or less:

$$ x_i \le 37,000 \implies \mu = \frac{1}{N}\sum_{i=1}^{N} x_i \le 37,000 $$

Since the actual mean is \(\$40,000 > \$37,000\), at least one client must have earned more than \$37,000. This statement must be true.

Evaluate statement 4 and statement 5

Using the Statistical Interpretation knowledge point

• Statement 4: "Last year, the number of their clients who had a starting salary of less than \$40,000 was equal to the number of their clients who had a starting salary of more than \$40,000."

This describes the median, not the mean. For skewed distributions, these counts are rarely equal.

• Statement 5: "Last year at least one of their clients had a starting salary of exactly \$40,000."

The mean can be \$40,000 without any individual value being exactly \$40,000 (e.g., two clients earning \$30,000 and \$50,000).

Answer:

• Last year all of their clients had a starting salary of at least $40,000.
• Two years ago some of their clients had a starting salary of at least $40,000.
• Last year at least one of their clients had a starting salary of more than $37,000. (Correct answer)
• Last year, the number of their clients who had a starting salary of less than $40,000 was equal to the number of their clients who had a starting salary of more than $40,000.
• Last year at least one of their clients had a starting salary of exactly $40,000.
• None of the above statements are true.