We are given the following information:

• The probability that a randomly selected HR manager says job applicants should follow up within two weeks is $p = 0.41$.
• We have 10 HR managers selected, so $n = 10$.
• We need to find the probability that at least 8 of them say job applicants should follow up within two weeks.

This is a binomial probability problem, where the probability of a certain number of successes (HR managers who say applicants should follow up) is calculated based on the binomial distribution formula:

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

where:

• $X$ is the number of successes (in this case, the number of HR managers),
• $n$ is the total number of trials (HR managers selected),
• $k$ is the number of successes we're interested in (here, $k \geq 8$),
• $p = 0.41$ is the probability of success.

We are asked to find the probability that at least 8 HR managers (out of 10) say applicants should follow up within two weeks, i.e., $P(X \geq 8)$.

This can be broken down as:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

Let me calculate this probability.The probability that at least 8 out of 10 randomly selected human resource managers say that job applicants should follow up within two weeks is approximately $0.0146$, or 1.46%.

Do you need further details or explanations?

Here are 5 related questions to expand your understanding:

1. What is the probability that exactly 8 HR managers say applicants should follow up?
2. How would the probability change if we asked for at least 7 managers instead of 8?
3. Can we approximate this result using the normal distribution?
4. How does the sample size (number of HR managers) affect the binomial probability?
5. What would be the probability if the success rate was 50% instead of 41%?

Tip: The binomial distribution is useful when dealing with a fixed number of trials and only two possible outcomes (success/failure).