Assume that when human resource managers are randomly selected, 44% say job applicants should follow up within two weeks. If 11 human resource managers are randomly selected, find the probability that fewer than 3 of them say job applicants should follow up within two weeks.
The probability is ________.
(Round to four decimal places as needed.)
Answer (2)
To solve this problem, we need to find the probability that fewer than 3 human resource managers out of 11 say job applicants should follow up within two weeks.
This situation can be modeled using a binomial distribution. Here's how you can approach the problem step-by-step:
Identify Parameters of the Binomial Distribution:
The number of trials (n) is 11 since we are selecting 11 managers.
The probability of success (p), which is the event where a manager says applicants should follow up within two weeks, is 0.44.
We are looking for the probability of having fewer than 3 successes.
Binomial Distribution Formula: The probability of getting exactly k successes in n trials is given by: P ( X = k ) = ( k n ) p k ( 1 − p ) n − k where ( k n ) is the binomial coefficient.
Calculate the Probability for Each Scenario:
For 0 successes: P ( X = 0 ) = ( 0 11 ) ( 0.44 ) 0 ( 0.56 ) 11 ≈ 0.0041
For 1 success: P ( X = 1 ) = ( 1 11 ) ( 0.44 ) 1 ( 0.56 ) 10 ≈ 0.0312
For 2 successes: P ( X = 2 ) = ( 2 11 ) ( 0.44 ) 2 ( 0.56 ) 9 ≈ 0.1007
Add Probabilities of Each Scenario: Add the probabilities of 0, 1, and 2 successes to find the probability of fewer than 3 successes: P ( X < 3 ) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) = 0.0041 + 0.0312 + 0.1007 = 0.1360
Therefore, the probability that fewer than 3 out of the 11 managers say job applicants should follow up within two weeks is approximately 0.1360 , rounded to four decimal places.
The probability that fewer than 3 HR managers out of 11 believe applicants should follow up within two weeks is approximately 0.1360. This result is calculated using the binomial distribution with parameters n = 11 and p = 0.44. The probabilities for 0, 1, and 2 successes were computed and summed to find this overall probability.
;
