Answer: 0.9754
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Explanation:
p = 0.58 = probability HR rep says to follow up within two weeks
n = 7 = sample size
Let's calculate the binomial P(x) value for x = 0 which will tell us the probability of having a sample with 0 reps recommending the follow up within two weeks.
[tex]P(x) = (_n C _x)*(p)^{x}*(1-p)^{n-x}\\\\P(0) = (_{7} C _{0})*(0.58)^{0}*(1-0.58)^{7-0}\\\\P(0) = (1)*(0.58)^{0}*(0.42)^{7}\\\\P(0) \approx 0.00230539333249 \\\\P(0) \approx 0.002 305\\\\[/tex]
If we were to randomly select a sample of n = 7 HR managers, then there's roughly a 2.305% chance that x = 0 of that group will say to follow up within two weeks. This is when there's a 58% chance of each individual rep of saying "yes".
Repeat for x = 1 and you should find that P(1) = 0.022285 approximately.
Then we can say this:
[tex]P(\text{0 or 1})+P(\text{at least 2}) = 1\\\\P(\text{x=0 or x=1})+P(x \ge 2) = 1\\\\P(0)+P(1)+P(x \ge 2) = 1\\\\P(x \ge 2) = 1-P(0)-P(1)\\\\P(x \ge 2) \approx 1-0.002305-0.022285\\\\P(x \ge 2) \approx 0.97541\\\\P(x \ge 2) \approx 0.9754\\\\[/tex]
If we selected a sample of n = 7 HR reps, there's roughly a 97.54% chance that 2 or more reps will recommend following up within 2 weeks.
