Questions: Assume that when human resource managers are randomly selected, 55% say job applicants should follow up within two weeks. If 30 human resource managers are randomly selected, find the probability that exactly 21 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.)

Solution Steps

We are tasked with finding the probability that exactly \( x = 21 \) out of \( n = 30 \) human resource managers believe that job applicants should follow up within two weeks, given that the probability of a manager supporting this view is \( p = 0.55 \).

The probability of exactly \( x \) successes in \( n \) trials for a binomial distribution can be calculated using the formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( \binom{n}{x} \) is the binomial coefficient, calculated as \( \frac{n!}{x!(n-x)!} \),
• \( p \) is the probability of success,
• \( q = 1 - p \) is the probability of failure.

Substituting the values into the formula:

• \( n = 30 \)
• \( x = 21 \)
• \( p = 0.55 \)
• \( q = 1 - 0.55 = 0.45 \)

The calculation yields:

\[ P(X = 21) = \binom{30}{21} \cdot (0.55)^{21} \cdot (0.45)^{9} \]

After performing the calculations, we find:

\[ P(X = 21) \approx 0.0382 \]

Final Answer