# 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...

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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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**