Question
Assume that when Human Resource managers are randomly selected, 53% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that 4 of them say job applicants should follow up within two weeks.
149
likes745 views
Answer to a math question Assume that when Human Resource managers are randomly selected, 53% say job applicants should follow up within two weeks. If 6 human resource managers are randomly selected, find the probability that 4 of them say job applicants should follow up within two weeks.
Murray
4.592 Answers
1. Calculate the binomial coefficient \( \binom{6}{4} \):
\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15
2. Calculate \( (0.53)^4 \):
(0.53)^4=0.53\times0.53\times0.53\times0.53\approx0.07890481
3. Calculate \( (1-0.53)^{6-4} = (0.47)^2 \):
(0.47)^2 = 0.47 \times 0.47 = 0.2209
4. Put these into the formula:
P(X=4)=15\times0.07890481\times0.2209\approx0.26145
So, the probability that exactly 4 out of 6 HR managers say job applicants should follow up within two weeks is approximately:
0.2615
2. Calculate \( (0.53)^4 \):
3. Calculate \( (1-0.53)^{6-4} = (0.47)^2 \):
4. Put these into the formula:
So, the probability that exactly 4 out of 6 HR managers say job applicants should follow up within two weeks is approximately:
Frequently asked questions (FAQs)
What is the limit of (x sin(1/x)) as x approaches 0?
+
What is 0.35 as a percent?
+
What are the components of a unit vector in the direction of vector v(2i + 3j - 4k)?
+
New questions in Mathematics