Question

Suppose that six new employees, two men and four women, are assigned offices chosen at random from six available rooms. Suppose further that three of these offices have windows while the other three do not have windows. Assume that this process results in a random assignment of one office to each new employee. Let X denote the number of women who are assigned window offices.

a) What is the name of the distribution of X?

b) Notice that P(X = 0) = 0 since there are not enough men to assign all three window offices to men. Compute the probabilities: P(X = 1), P(X = 2), and P(X = 3).

c) Use the special formulas for this specific distribution to find E(X) and var(X).

Answer 1

Answer:-

Given That:-

Suppose that six new employees, two men and four women, are assigned offices chosen at random from six available rooms. Suppose further that three of these offices have windows while the other three do not have windows. Assume that this process results in a random assignment of one office to each new employee. Let X denote the number of women who are assigned window offices.

Given,

Total no. of employees= 6

Out of the 6 employees, no. of men= 2 and no. of women = 4

Total available rooms= 6

Out of the 6 rooms, no. of rooms with windows= 3 and no. of rooms without windows= 3

Now, the process results in a random assignment of a new office to each new employee, thus when a room is assigned to an employee, it is not available for assignment to another employee. This is the case of sampling without replacement.

Now, X denotes the no. of women who are assigned window offices. Then X can take values 1,2 or 3 because there are maximum 3 rooms with window and X can't take the value 0 since in that case 1 room with window will not get assigned because there are only 2 men.

a) What is the name of the distribution of X?

This is the case of hypergeometric distribution and the PMF is given by

P(X=x) = {4Cx * 2C(3-x)}/ 6C3 for x= 1,2,3

b) Notice that P(X = 0) = 0 since there are not enough men to assign all three window offices to men. Compute the probabilities: P(X = 1), P(X = 2), and P(X = 3).

P(X=1) = 4C1* 2C2 / 6C3

= 4/ 20

= 1/5

P(X=2) = 4C2* 2C1/ 6C3

= 12/20

= 3/5

P(X=3)= 4C3* 2C0/ 6C3

= 4/20

= 1/5

c) Use the special formulas for this specific distribution to find E(X) and var(X).

E(X) for hypergeometric distribution is given by nM/N. In this problem,

n= no. of window rooms= 3,

M= no. of women= 4,

N= total no. of employees= 6.

Thus,E(X) = 3*4/ 6

= 12/6

= 2

Var(X) for a hypergeometric distribution is given by NM(N-M)(N-n)/N^2(N-1). Thus Var(X)= 6*4*2*3/ 6^2*5

= 144/36*5

= 144/180

= 0.8

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