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Assuming that a probability distribution of the number of job interviews is described,...

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surfer3 | (Level 1) Valedictorian

Posted May 28, 2012 at 9:03 PM via web

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Assuming that a probability distribution of the number of job interviews is described, find their mean and standard deviation.

Based on information from MRINetwork, some job applicants are required to have several interviews before a decision is made. the number of required interviews and the corresponding probabilities are 1(0.09);2(0.31);3(0.37);4(0.12);5(0.05);6(0.05)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 29, 2012 at 2:02 AM (Answer #1)

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The probability that an applicant has to face one interview is 1 is given by 0.09. For 2 it is (0.31), for 3 it is (0.37), for 4 it is(0.12), for 5 it is (0.05) and for 6 it is (0.05).

The mean number of interviews that a candidate has to have is 1*(0.09)+2*(0.31)+3*(0.37)+4*(0.12)+5*(0.05)+6*(0.05) = 2.85

The standard deviation of the number of interviews that have to be taken is `sigma` = `sqrt(sum_(i=1)^N p_i*(x_i - mu)^2)`

` mu = sum(p_i*x_i)` = `1*(0.09)+2*(0.31)+3*(0.37)+4*(0.12)+5*(0.05)+6*(0.05) = 2.85`

`sigma` = `sqrt(sum_(i=1)^N p_i*(x_i - mu)^2)` = sqrt(0.09(1 - 2.85)^2 + 0.31(2 - 2.85)^2 + 0.37(3- 2.85)^2 + 0.12(4 - 2.85)^2 + 0.05(5 - 2.85)^2 + 0.05*(6 - 2.85)^2)

=> `sqrt(1.426) = 1.194`

The mean number of interviews is 2.85 and the standard deviation is 1.194

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