Classical

June 2015 - Paper 3

46:  

Consider the following statements :

(a) In the Game theory, the saddle point is the largest value in its row and the smallest in its column.

(b) The classical EOQ model is very sensitive to changes in the variables.

(c) In LP problem, an optimal solution does not necessarily use all the limited resources.

(d) In PERT, it is assumed that the distribution of completion time for a project is normal distribution.

Indicate the true answers.

A.

Statements (c) and (d)

B.

Statements (a), (c) and (d)

C.

Statements (a), (b) and (d)

D.

All are true

 
 

Option: A

Explanation :


47:  

Consider the following statements :

(a) When there is no saddle point in a competitive game, the players adopt mixed strategies.

(b) The EOQ model of inventory control can also be applied to the production process, with appropriate modifications.

(c) If the price of an item of inventory increases, the EOQ will also increase proportionately.

(d) In an LP problem, an equation is more restrictive than an inequality.

Indicate the correct statements.

A.

All of the above

B.

(a), (b) and (d)

C.

(a) and (b)

D.

None of the statements is true

 
 

Option: B

Explanation :


48:  

Consider the following statements :

(a) In the single - channel queuing model, if μ = λ, the queue length would be infinite.

(b) In the simplex solution procedure for an LP problem, the shadow prices of the resources are identified with the variables which are included in the optimal solution.

(c) In the case of gradual receipt of the ordered quantity, the EOQ will be larger than the EOQ in the case of instantaneous receipt.

(d) Time is the only variable in PERT

Indicate the correct answer.

A.

Statements (a) and (b) are true, others are not

B.

Statements (b) and (c) are true, others are not

C.

Statements (a), (c) and (d) are true, but (b) is false

D.

All the statements are true

 
 

Option: C

Explanation :


49:  

A Random variable X has the following probability distribution :

X                :                - 2 0       2

Probability :             0.25 0.50 0.25

The mean and variance of X shall be:

A.

0.50 and 1

B.

0 and 1

C.

0.50 and 2

D.

0 and 2

 
 

Option: D

Explanation :


50:  

Match the following, considering the state of conclusion for each of the situation at α = 0.05.

List – I

List – II

(a) H0 : P1 – P2 = 0, Ha : P1 – P2 < 0; P = 0.0233

(i) reject H0

(b) H0 : P1 – P2 = 0, Ha : P1 – P2 > 0; P = 0.0233

(ii) accept H0

(c) H0 : P1 – P2 = 0, Ha : P1 – P2 ≠ 0; P = 0.0233

(iii) no decision

(d) H0 : P1 = P2 , Ha : P1 ≠ P2 ; P = 0.0233

 

Codes :

A.

(ii) (i) (i) (ii)

B.

(i) (iii) (i) (i)

C.

(ii) (iii) (ii) (i)

D.

(i) (ii) (ii) (i)

 
 

Option: D

Explanation :




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