Assume that when human resource managers are randomly​ selected, 56​% 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.

This problem can be solved using the binomial probability formula, as it involves a fixed number of trials (11 human resource managers), two possible outcomes (whether or not they say applicants should follow up within two weeks), and a constant probability of success (56%).

### Step 1: Define the binomial distribution
The binomial distribution is given by:

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

where:
- \( P(X = k) \) is the probability of exactly \( k \) successes,
- \( n \) is the number of trials (11 managers),
- \( p \) is the probability of success (0.56),
- \( k \) is the number of successes (managers who say applicants should follow up within two weeks),
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient.

### Step 2: Identify what we need
We are asked to find the probability that fewer than 3 human resource managers say applicants should follow up within two weeks. This means we need to find the sum of probabilities for \( X = 0 \), \( X = 1 \), and \( X = 2 \):

\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
\]

### Step 3: Calculate each probability
Now, we'll calculate the individual probabilities using the binomial probability formula for \( X = 0 \), \( X = 1 \), and \( X = 2 \).

For \( X = 0 \):
\[
P(X = 0) = \binom{11}{0} (0.56)^0 (0.44)^{11} = 1 \times (0.44)^{11}
\]

For \( X = 1 \):
\[
P(X = 1) = \binom{11}{1} (0.56)^1 (0.44)^{10} = 11 \times (0.56) \times (0.44)^{10}
\]

For \( X = 2 \):
\[
P(X = 2) = \binom{11}{2} (0.56)^2 (0.44)^{9} = \frac{11 \times 10}{2} \times (0.56)^2 \times (0.44)^9
\]

### Step 4: Sum the probabilities
Finally, we'll sum \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) to find the total probability.

Let me calculate that for you.The probability that fewer than 3 out of 11 human resource managers say job applicants should follow up within two weeks is approximately **0.0125**, or 1.25%.

Would you like more details or have any questions about the process?

Here are 5 related questions to further explore:
1. How does the probability change if 15 managers are selected instead of 11?
2. What if the probability of success changes to 70%? How does that affect the result?
3. How can we calculate the expected number of managers who say applicants should follow up within two weeks?
4. How does the binomial distribution approximate the normal distribution as the sample size increases?
5. Can you use a cumulative binomial distribution table for this problem, and how?

**Tip**: The binomial distribution is useful when you're dealing with independent trials, each with the same probability of success.

Assume that when human resource managers are randomly selected, 56% 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.

Find the probability that fewer than 3 out of 11 human resource managers say job applicants should follow up within two weeks. This problem involves binomial distribution, probability calculations, and uses the binomial probability theorem with a success probability of 56%. Suitable for Grades 10-12 students.

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