A marketing firm is considering making up to three new hires. Given its specific needs, the firm feels that there is a 60% chance of hiring at least two candidates. There is only a 5% chance that it will not make any hires and a 10% chance that it will make all three hires. a. What is the probability that the firm will make at least one hire? b. Find the expected value and the standard deviation of the number of hires.
Answers
Given : A marketing firm is considering making up to three new hires. Chances of hiring Different number oof person is given
To find : a. What is the probability that the firm will make at least one hire? b. Find the expected value & Standard deviation (SD)
Solution:
5% chance that it will not make any hires
=> Probability of Hiring 0 Candidates P(0) = 0.05
10% chance that it will make all three hires
=> Probability of Hiring 3 Candidates P(3) = 0.1
60% chance of hiring at least two candidates.
=> P(2) + P(3) = 0.6
=> P(2) + 0.1 = 0.6
=> P(2) = 0.5
P(2) = Probability of Hiring 2 Candidates
P(0) + P(1) + P(2) + P(3) = 1
=> 0.05 + P(1) + 0.5 + 0.1 = 1
=> P(1) = 0.35
P(1) = Probability of Hiring 1 Candidates
probability that the firm will make at least one hire
= 1 - firm will make no Hire
= 1 - 0.05
= 0.95
or P(1) + P(2) + P(3) = 0.35 + 0.5 + 0.1 = 0.95
probability that the firm will make at least one hire = 0.95
Expected value = ∑ P(x) * x
P(0) * 0 + P(1) *1 + P(2)*2 + P(3) * 3
= 0 + 0.35 + 1 + 0.3
= 1.65
Expected value = 1.65
SD² = ∑ ( x - expected Value)² * P(x)
= (-1.65)²(0) + (-0.65)²(0.35) + (0.35)²(0.5) + (1.35)²(0.1)
= 0 + 0.147875 + 0.06125 + 0.18225
= 0.391375
=> SD = √ 0.391375
=> SD = 0.626
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Answer:The previous answer was right but I found while doing this problem for school with slightly different numbers that for the Standard Deviation for P(0). He said P(0)= (0-expected value)^2 * 0 when in truth it is P(0)=(0-expected value)^2* whatever the probability of having no new hires is. Step-by-step explanation: