The problem asks for the probability that at least 7 out of 9 human resource managers say job applicants should follow up within two weeks, given that 51% of human resource managers hold this view. This can be modeled using the binomial distribution.

Step-by-step approach:

1. Parameters for the binomial distribution:

   • Number of trials $n = 9$ (the 9 human resource managers)
   • Probability of success on a single trial $p = 0.51$ (the probability that one HR manager says applicants should follow up)
   • We are asked to find the probability that at least 7 of them say so. Therefore, we need to compute $P(X \geq 7)$, where $X$ is the number of managers that agree.

2. Formula: The binomial probability mass function is given by: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ where $\binom{n}{k}$ is the binomial coefficient.

   However, since we need the probability for $X \geq 7$, we need to calculate: $P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9)$

3. Calculation: For each value of $k = 7, 8, 9$, we will use the formula above to compute the individual probabilities and sum them up.

Now, I'll calculate it.The probability that at least 7 out of 9 human resource managers say job applicants should follow up within two weeks is approximately 0.1001 (rounded to four decimal places).

Would you like more details on how this was calculated, or do you have any further questions?

Here are some related questions to explore:

1. How would the probability change if the success probability were different (e.g., 60%)?
2. What is the probability that exactly 7 out of 9 managers agree?
3. How would the calculation change if we needed to find the probability for at least 6 managers instead of 7?
4. Can we calculate the expected number of managers who would say applicants should follow up?
5. How does the variance of this binomial distribution affect the spread of probabilities?

Tip: In binomial distributions, using a cumulative distribution function (CDF) is a fast way to find probabilities for $X \geq k$ or $X \leq k$.