This problem follows a binomial probability distribution since each human resource manager either agrees or disagrees with the statement.

We use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

where:

• $n = 8$ (total HR managers surveyed),
• $k = 2$ (number of managers who say applicants should follow up),
• $p = 0.54$ (probability that a manager agrees),
• $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ (binomial coefficient).

Step 1: Compute the binomial coefficient

$\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28$

Step 2: Compute the probability

$P(X = 2) = 28 \times (0.54)^2 \times (1 - 0.54)^{8-2}$

$= 28 \times (0.2916) \times (0.46)^6$

$= 28 \times 0.2916 \times 0.046655$

$= 28 \times 0.013610$

$= 0.3811$

Final Answer:

$\mathbf{0.3811}$

(Rounded to four decimal places)

Would you like a breakdown of any part?

Related Questions:

1. What is the probability that at most 3 managers say applicants should follow up?
2. How do we determine if a probability distribution is binomial?
3. What is the expected number of HR managers who say applicants should follow up?
4. How would the probability change if the sample size increased to 10 managers?
5. How does changing the success probability ($p$) affect the binomial distribution?

Tip:

For large sample sizes ($n$), the binomial distribution can be approximated using the normal distribution when $np$ and $n(1-p)$ are both greater than 5.