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if 6 surveyed are randomly selected without replacement for a follow up survey, find...

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if 6 surveyed are randomly selected without replacement for a follow up survey, find the probability that 3 of them said this

In a survey of 150 senior executives, 47% said that the most common job interview mistake is to have little or no knowledge of the company.

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The sample size is 6 out of 150. This is 4% < 5%. Therefore we can assume this as a binomial distribution. Otherwise we cant use binomial theory for instances without replacements.

The probability of saying this as common mistake = 0.47

The probability density function of a binomial distribution is given by,

`P(k;n,P) = ^nC_kP^k(1-P)^(n-k)`

The probability that 3 people said it

                          `= ^6C_3(0.47)^3(1-0.47)^(6-3)`

                          `= 0.3091`

Therefore there is a 30.91% chance that 3 people said it out of 6 randomly selected.

 

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