Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.4

Choose the correct answer

Question 1.
The transportation problem is said to be unbalanced if
(a) Total supply ≠ Total demand
(b) Total supply = Total demand
(c) m = n
(d) m + n – 1
Solution:
(a) Total supply ≠ Total demand

Question 2.
In a non – degenerate solution number of allocation is
(a) Equal to m + n – 1
(b) Equal to m + n + 1
(c) Not equal to m + n – 1
(d) Not equal to m + n + 1
Solution:
(a) Equal to m + n – 1

Question 3.
In a degenerate solution number of allocations is
(a) Equal to m + n – 1
(b) Not equal to m + n – 1
(c) Less then m + n – 1
(d) Greater then m + n – 1
Solution:
(c) Less then m + n – 1

Question 4.
The Penalty in VAM represents difference between the first
(a) Two largest costs
(b) Largest and Smallest costs
(c) smallest two costs
(d) None of these
Solution:
(c) smallest two costs

Question 5.
Number of basic allocation in any row or column in an assignment problem can be
(a) Exactly one
(b) At least one
(c) At most one
(d) None of these
Solution:
(a) Exactly one

Question 6.
North – West Corner refers to
(a) Top left corner
(b) Top right corner
(c) Bottom right corner
(d) Bottom left corner
Solution:
(a) Top left corner

Question 7.
Solution for transportation problem using method is nearer to an optimal
(a) NWCM
(b) LCM
(c) VAM
(d) Row Minima
Solution:
(c) VAM

Question 8.
In an assignment problem the value of iedsion variable x_ij is
(a) 1
(b) 0
(c) 1 or 0
(d) none of them
Solution:
(c) 1 or 0

Question 9.
If number of sources is not equal to number of destinations, the assignment problem is called
(a) balanced
(b) unsymmetric
(c) symmetric
(d) unbalanced
Solution:
(d) unbalanced

Question 10.
The purpose of a dummy row or column in an assignment problem is to
(a) prevent a solution from becoming degenerate
(b) balance between total activities and total resources
(c) provide a means of representing a dummy problem
(d) None of the above
Solution:
(b) balance between total activities and total resources

Question 11.
The solution for an assignment problem is optimal if
(a) each row and each column has no assignment
(b) each row and each column has atleast one assignment
(c) each row and each column has atmost one assignment
(d) each row and each column has exactly one assignment
Solution:
(d) each row and each column has exactly one assignment

Question 12.
In an assignment problem involving four workers and three jobs, total number of assignments possible are
(a) 4
(b) 3
(c) 7
(d) 12
Solution:
(b) 3

Question 13.
Decision theory is concerned with
(a) analysis of information that is available
(b) decision making under certainty
(c) selecting optimal decisions in sequential problem
(d) All of the above
Solution:
(d) All of the above

Question 14.
A type of decision – making environment is
(a) certainty
(b) uncertainty
(c) risk
(d) all of the above
Solution:
(d) all of the above

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.3

Question 1.
Given the following pay – off matrix (in rupees) for three strategies and two states for nature.

Select a strategy using each of the following rule (i) Maximin (ii) Minimax
Solution:

(i) Max min
Max ( 40, -20, -40) = 40. Since the maximum pay-off is 40, the alternative S, is selected.

(ii) Minimax
min (60, 10, 150) = 10, Since the minimum payoff is 10. the alternative S₂ is selected.

Question 2.
A farmer wants to decide which of three crops he should plant on his 100 – acre farm. The profit from each is dependent on the rainfall during the growing season. The farmer has categorized the amount of rainfall as high medium and low. His estimated profit for each is shown in the table.

In the farmer wishes to plant only crop, decide which should be his best crop using (i) Maximin (ii) Minimax
Solution:

(i) Maximin
Max ( 3500, 4500, 2000) = 4500. Since the maximum pay-off is 4500, the alternative ‘Medium’, is selected.

(ii) Minimax
min (8000, 5000, 5000) = 5000, Since the minimum pay-off is 5000. the alternatives both ‘Medium’ and ‘Low’ are selected.

Question 3.
The research department of Hindustan Ltd. has recommended to pay marketing department to launch a shampoo of three different types. The marketing types of shampoo to be launched under the following T estimated pay-offs for various level of sales.

What will be the marketing manager’s decision if (i) Maximin and (ii) Minimax principle applied?
Solution:

(i) Maximin
Max ( 10, 5, 3) = 10. Since the maximum pay of is 10, “Egg Shampoo”, is selected.

(ii) Minimax
min (30, 40, 55) = 30, Since the minimum pay-off is 30. the alternative “Egg Shampoo” is selected

Question 4.
Following pay – off matrix, which is the optimal decision under of the following rule (i) maximin (ii) minimax

Solution:

(i) Maximin
Max ( 5, 7, 9, 8) = 5. Since the maximum pay-off is 5, the alternative A₁ is selected.

(ii) Minimax
min (14, 11, 11, 13) = 11, Since the minimum pay-off is 11. the alternative alternatives A₂ and A₃ are selected.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.2

Question 1.
What is the Assignment problem?
Solution:
Suppose that we have ‘m1 jobs to be performed on ‘n’ machines. The cost of assigning each job to each machine is C_ij (i = 1, 2, … n and j = 1, 2, … , n). Our objective is to assign the different jobs to the different machines (one job per machine) to minimize the overall cost. This is known as an assignment problem.

Question 2.
Give mathematical form of assignment problem.
Solution:
Consider the problem of assigning n jobs to n machines (one job to one machine). Let C_ij be the cost of assigning i^th job to the j^th machine and x_ij represents the assignment of i^th job to the j^th machine.

x_ij is missing in any cell means that no assignment is made between the pair of job and machine.(i.e) x_ij = 0.
x_ij is present in any cell means that an assignment is made their. In such cases x_ij = 1
The assignment model can written in LPP as follows
Minimize Z = \(\sum_{i=1}^{m}\) \(\sum_{j=1}^{n}\) C_ij x_ij
Subject to the constrains
\(\sum_{i=1}^{n}\) x_ij = 1, j = 1, 2, …. n
\(\sum_{j=1}^{n}\) x_ij = 1, i = 1,2,….n and x_ij =0 (or) 1 for all i, j

Question 3.
What is the difference between Assignment Problem and Transportation Problem?
Solution:

Question 4.
Three jobs A, B, and C one to be assigned to three machines U, V, and W. The processing cost for each job machine combination is shown in the matrix given below. Determine the allocation that minimizes the overall processing cost.
(cost is in Rs per unit)

Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Look for atleast one zero in each row and each column.
Here each and every row and columns having exactly one zero No need for step 2 go to step 3.

Step 3.

Mark the zero by □ Mark other zeros in its column by X.
Since each row and each column contains exactly one assignment, all three machines have been assigned a job.

The Optimal assignment (minimum) cost = 46

Question 5.
A computer centre has got three expert programmers. The centre needs three application programmes to be developed. The head of the computer centre, after studying carefully the programmes to be developed, estimates the computer time in minutes required by the experts to the application programme as follows.

Assign the programmers to the programme in such a way that the total computer time is the least.
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero, mark the zero by □. Mark other zeros in its column by X.

Step: 4
Now examine the columns with exactly one zero marks the zero by □. Mark other zeros in its row by X.

Thus all three assignments have been made. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs 280

Question 6.
A departmental head has four subordinates and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. His estimates of the time man would take to perform each task is given below.

How should the tasks to allocated to subordinates so as to minimize the total man-hours?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero Mark the zero by □. Mark other zeros in its row by X.

Step 4.
Now examine the columns with exactly one zero. Mark the zero by □ Mark other zeros in its row by X.

Step 5.
Cover all the zeros of table 4 with three lines, since three assignments were made check (✓) row S since it has no assignment.

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in table 5. Take the smallest element. This is 1 (one) our case. By subtracting 1 from the uncovered cells.

[Adding 1 to elements (Q, S, R) that line at the intersection of two lines]

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignment.

Step 8.
Determine an assignment.

Thus all the four assignment have been made. The optimal assignment schedule and total time is

The optimum time (minimum) = 41 Hrs.

Question 7.
Find the optimal solution for the assignment problem with the following cost matrix.

Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and sub tract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero. Mark the zero by □ Mark other zeros in its column by X

Thus all the four assignments have been made. The optimal assignment schedule and total cost.

The Optimum cost (minimum) = Rs 37

Question 8.
Assign four trucks 1, 2, 3 and 4 to vacant spaces A, B, C, D, E and F so that distance travelled is minimized. The matrix below shows the distance.

Solution:
Since the number of columns is less than the number of rows, the given assignment problem is unbalanced one. To balance it, introduce two dummy columns with all the entries zeros.
The revised assignment problem is

Here only 4 tasks can be assigned to 4 vacant spaces.
Step 1.
It is not necessary, since each row contains zero entry. Go to step 2.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Since each row contains more than one zeros. Go to step 4.

Step 4.
Examine the columns with exactly one zero, mark the zero by □ Mark other zeros in its rows by X.

Step 5.
Here all the four assignments have been made we can assign d₁ for D then we will get d₂ for E.

The optimal assignment schedule and total distance is

∴ The Optimum Distant (minimum) = 12 units

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

Question 1.
What is transportation problem?
Solution:
The objective of transportation problem is to determine the amount to be transported from each origin to each destinations such that the total transportation cost is minimized.

Question 2.
Write mathematical form of transportation problem.
Solution:
Let there be m origins and n destinations. Let the amount of supply at th i th origin is a_i. Let the demand at j th destination is b_j.
The cost of transporting one unit of an item from origin i to destination j is C_ij and is known for all combination (i,j). Quantity transported from origin i to destination j be x_ij.

The objective is to determine the quantity x_ij to be transported over all routes (i,j) so as to minimize the total transportation cost. The supply limits at the origins and the demand requirements at the destinations must be satisfied.
The above transportation problem can be written in the following tabular form:

Now the linear programming model representing the transportation problem is given by
The objective function is Minimize Z = \(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\), \(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) c_ij x_ij
Subject to the constraints
\(\sum_{\mathbf{j}=\mathbf{1}}^{\mathbf{n}}\) = x_ij = a_i, i = 1, 2 …….. m (supply constraints)
\(\sum_{\mathbf{i}=\mathbf{1}}^{\mathbf{m}}\) = x_ij = b_j, i = 1, 2 …….. n (demand constraints)
x_ij ≥ 0 for all i, j (non- negative restrictions)

Question 3.
What are feasible solution and non-degenerate solutions to the transportation problem?
Solution:
Feasible Solution: A feasible solution to a transportation problem is a set of non-negative values x_ij (i = 1, 2,.., m, j = 1, 2, …n) that satisfies the constraints.
Non-degenerate basic feasible Solution: If a basic feasible solution to a transportation problem contains exactly m + n – 1 allocation in independent positions, it is called a Non-degenerate basic feasible solution. Here m is the number of rows and n is the number of columns in a transportation problem.

Question 4.
What do you mean by a balanced transportation problem?
Solution:
In a transportation problem if the total supply equals the total demand (Σa_i = Σb_j) then it is said to be balanced transportation problem.

Question 5.
Find an intial basic feasible solution of the following problem using north west corner rule.

Solution:
Here total Supply 19 + 37 + 34 = 90
Total demand = 16 + 18 + 31 + 25 = 90
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem
∴ we can final an initial basic feasible solution to due given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (Q₁, D₁). Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column’s exhausted.
(i.e) x₁₁ = min (19, 16) = 16

Now the cell in the north west corner is (O₁, D₂) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
(i.e) x₁₂ = min (3, 18) = 3

Reduced transportation table is

Now the cell in the north west corner is (O₂, D₂)
x₂₂ = min (37, 15) = 15

Now the cell in the north west corner is (O₂, D₃)
x₂₃ = min (22, 31) = 9

Now the cell in the north west corner is (O₃, D₃)
x₃₃ = min (34, 9) = 9

Thus we have the following allocations

Transportation schedule:
O₁ → D₁, O₁, → D₂, O₂ → D₂; O₂ → D₃ O₃ → D₃; O₃ → D₄
= (16 × 5) + (3 × 3) + (15 × 7) + (22 × 9) + (9 × 7) + (25 × 5)
= 80 + 9 + 105 + 198 + 63 + 125
= 580

Question 6.
Determine an intial basic feasible solution of the following transportation problem by north west corner method.

Solution:
Here total capacity (a_i) = 30 + 40 + 50 = 120
Total demand (b_j) = 35 + 28 + 32 + 25 = 120
(i.e) Total capacity = Total demand
∴ The given problem is balanced transportation.
∴ We can find an initial basic feasible solution to the given problem.

From the above table we can choose the cell in the North west corner. Here the cell is (chennai, Bangalore)
x₁₁ = min (30, 35) = 30
Reduced transportation table is

Now the cell in the North west corner is (Madurai, Bangalore)
x₂₁ = min(40, 5) = 5

From the above table we can choose the cell in the North west corner. Here the cell is (Nasik, Madurai)
x₂₂ = min (35, 28) = 28

From the above table we can choose the cell in the North west corner. Here the cell is (Bhopal, Madurai)
x₂₂ = min (7, 32) = 7

From the above table we can choose the cell in the North west corner. Here the cell is (Trichy, Bhopal)
x₃₃ = min (50, 25) = 25
Reduced transportation table is

x₃₄ = min (25, 25) = 25
Thus we have the following allocations

Transportation Schedule:
Chennai → Bangalore; Madurai → Bangalore;
Madurai → Naisk; Madurai → Bhopal
Trichy → Bhopal; Trichy → Delhi
The total transportation cost =
(30 × 6) + (5 × 5) + (28 × 11) + (7 × 9) + (25 × 7) + (25 × 13)
= 180 + 25 + 308 + 63 + 175 + 325
= 1076

Question 7.
Obtain an initial basic feasible solution to the following transportation problem by using least-cost method.

Solution:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total demand = 30 + 25 + 45 = 100
(i.e) Total supply = Total demand
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cos is 4 corresponds to the cell (O₂, D₃)
Allocate min (35, 45) = 35 units to this cell

The least cost corresponds to the cell (O₁, D₃)
Allocate min (25, 10) = 10 units to this cell

The least cost is 6 corresponds to the cell (O₃, D₂)
Allocate min (40, 25) = 25 units to this cell

The least cost is 7 corresponds to the cell (O₃, D₁)
Allocate min (15, 30) = 15

Thus we have the following allocations

Transportation Schedule:
O₁ → D₁; O₁, → D₃; O₂ → D₃; O₃ → DI; O₃ → D₂
Total Transportation cost
= (15 × 9) + (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6)
= 135 + 50 + 140 + 105 + 150
= 580

Question 8.
Explain vogel’s approximation method by obtaining initial feasible solution of the following transportation problem.

Solution:
Here Σ ai = 6 + 1 + 10 = 17
Σ bj = 7 + 5 + 3 + 2 = 17
Σ ai = Σ bj
(i.e) Total supply = Total Demand
∴ The given problem is balanced transportation problem
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns.

Choose the largest difference. Here the largest difference is 6 which corresponds to column D₄
In this column choose the least cost. Here the least cost corresponds to (O₂, D₄)
Allocate min (1, 2) = 1 unit to this cell the reduced transportation table is

choose the largest difference 5 which corresponds to column D₂. Here the least cost corresponds to (O₁, D₂).
Allocate min (6, 5) = 5 units in this cell

Choose the largest difference 5 which corresponds to row O₁. Here the least cost corresponds to (O₁, D₁)
Allocate min (1, 7) = 1 unit in this cell

Choose the largest difference 4 which corresponds to row O₃. Here least cost corresponds to (10, 6) = 6 units in this cell.

Choose the largest difference 6 which corresponds to row O₃. Here the least cost corresponds to (O₃, D₄).
Allocate min (4, 1) = 1

Thus we have the following allocations

Transportation schedule
O₁ → D₁; O₁ → D₂; O₂ → D₄;
O₃ → D₁; O₃ → D₃; O₃ → D₄
Total transportation cost:
= (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 9.
Consider the following transportation problem.

Determine initital basic feasible solution by VAM
Solution:
Given Transportation problem is

Here Σ ai = Σ bj = 100
∴ The given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First let us find the difference (penalty) between the first two smallest costs in each row and column and write them in brackets against the respective rows and columns

Choose the largest difference. Here the difference is 3 which corresponds to D₂
In this column choose the least cost. Here the least cos corresponds to (O₃, D₂)
Allocate min (20, 40) = 20 units to this cell

Choose the largest difference is 4 which corresponds to column D₃. In this column choose the least cost. Here the least cost corresponds to (O₁, D₃).
Allocate min (30, 20) = 20 units to this cell

Choose the largest different is 3 which corresponds to column D₂. In this column choose the least cost. Here the least cost corresponds to (O₂, D₂)
Allocate min (50, 20) = 20 units to this cell

Choose the largest difference is 2 which corresponds to column D₄. In this column choose the least cost. Here the least cost corresponds to (O₂, D₄).
Allocate min (30, 10) = 10 units to this cell

Allocate min (20, 30) = 20 units to this cell

Here
x₁₁ = 10
x₁₃ = 20
x₂₁ = 20
x₂₂ = 20
x₂₄ = 10
x₃₂ = 20
Transportation schedule
O₁ → D₁; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost
= (10 × 5) + (20 × 3) + (20 × 4) + (20 × 5) + (10 × 4) + (20 × 2)
= 50 + 60 + 80 + 100 + 40 + 40
= 370

Question 10.
Determine basic feasible solution to the following transportation problem using North west Corner rule.

Solution:
Here total supply = 4 + 8 + 9 = 21
Total demand = 3 + 3 + 4 + 5+ 6 = 21
(i.e) Total supply = Total demand
∴ The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (P, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.

Form the above table we can choose the cell in North west corner. Here the cell is (P,B)
x = min (1, 3) = 1

From the above table we can choose the cell in north west corner. Here the cell is (Q, B)
x = min (2, 8) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (Q, C)

From the above table, we can choose the cell in North west corner. Here the cell is (Q, D)
x = min (2, 5) = 2

From the above table, we can choose the cell in North west corner. Here the cell is (R, D)
x = min (9, 3) = 3

Thus we have the following table

Transportation Schedule:
P → A; P → B; Q → B; Q → C; Q → D R → D; R → E
Total transportation cost:
= (3 × 2) + (1 × 11) + (2 × 4) + (4 × 7) + (2 × 2) + (3 × 8) + (6 × 2)
= 6 + 11 + 8 + 28 + 4 + 24 + 72
= 153

Question 11.
Find the initial basic feasible solution of the following transportation problem:

Using
(i) North West corner tule
(ii) Least Cost method
(iii) Vogel’s approximation method
Solution:
(i) North west corner rule:
Here the total supply = 10 + 10 + 10 = 30
Total demand = 7 + 12 + 11 = 30
(i.e) Total supply = Total demand
The given problem is balanced transportation problem.
∴ we can find an initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North west corner. Here the cell is (A, I)
x₁₁ = min (7, 10) = 7

From the above table we can choose the cell in the north west corner. Here the cell is (B, I)
x = min (3, 12) = 3

From the above table we can choose the cell in the North west corner. Here the cell is (B, II)
x = min (9, 10) = 9

Here the cell in the North west corner is (C, II)
x = min (11, 1) = 1

Thus we have the following allocations

Transportation schedule:
A → I; B → I; B → II; C → II; C → III
Total transportation cost:
= (7 × 1) + (3 × 0) + (9 × 4) + (1 × 1) + (10 × 5)
= 7 + 0 + 36 + 1 + 50
= 94

(ii) Least cost method:
The given transportation table is

Here Total supply = Total demand = 30
∴ The given problem is a balanced transportation problem.
Hence there exists a feasible solution to the given problem.
The least cost is 0 corresponds to the cell (B, I)
Allocate min (12, 10) = 10 units to this cell

The least cost 1 corresponds to the cell (C, II)
Allocate min (11, 10) = 10 units to this cell

Here the least cost 2 corresponds to the cell (B, III)
Allocate min (2, 10) = 2 units to this cell.

Here the least cost is 5 corresponds to the cell (C, III)

Transportation schedule:
A → III; B → I; B → III; C → II; C → III
Total transportation cost:
= (7 × 6) + (10 × 10) + (2 × 2) + (10 × 1) + (1 × 5)
= 42 + 0 + 4 + 10 + 5
= 61

(iii) Vogel’s approximation method:
Here Σ ai = Σ bj = 30
(i.e) Total supply = Total demand
∴ This given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.


A → I; B → I; B → III; C → I; C → II
Total transportation cost:
= (7 × 1) + (2 × 0) + (10 × 2) + (1 × 3) + (10 × 1)
= 7 + 0 + 20 + 3 + 10
= 40

Question 12.
Obtain an initial basic feasible solution to the following transportation problem by north west corner method.

Solution:
Here the total available = 250 + 300 + 400 = 950
Total Required = 200 + 225 + 275 + 250 = 950
(i.e) Total Available = total required
∴ The given problem is balanced transportation problem.
we can find an initial basic feasible solution to the given problem.
From the above table, we can choose the cell in the North west corner. Here the cell is (A, D).
x₁₁ = min (250, 200) = 200

From the above table we can choose the cell in North west corner. Here the cell is (A, E)
x = min (50, 225) = 50

From the above table, the north west corner cell is (B, E)
x = min (300, 175) = 175

From the above table, the north west corner cell is (B, F)
x = min (125, 275) = 125

Here the north west corner cell is (C, F)
x = min (400, 150) = 150

Transportation schedule:
A → D; A → E; B → E; B → F; C → G
Total Transportation cost:
= (200 × 11) + (50 × 13) + (175 × 18) + (125 × 14) + (150 × 13) + (250 × 10)
= 2200 + 650 + 3150 + 1750 + 1950 + 2500
= 12,200

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 10 Operations Research Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 10 Operations Research Miscellaneous Problems

Question 1.
The following table summarizes the supply, demand and cost information for four factors S₁, S₂, S₃, S₄ Shipping goods to three warehouses D₁, D₂, D₃.

Find an initial solution by using north west corner rule. What is the total cost for this solution?
Solution:
The given transportation table is

Here total supply = 5 + 8 + 7+14 = 34
Total amount =7 + 9 + 18 = 34
(i.e) Total supply = Total demand.
∴ The given problem is balanced transformation problem.
We can find an initial basic feasible solution to the given problem.
First allocation:



Transportation schedule:
S₁ → D₁; S₂ → D₁; S₂ → D₂;
S₃ → D₂; S₃ → D₃; S₄ → D₃
The transportation cost:
= (5 × 2) + (2 × 3) + (6 × 3) + (3 × 4) + (4 × 7) + (14 × 2)
= 10 + 6 + 18 + 12 + 28 + 28
= 102

Question 2.
Consider the following transportation problem

Determine an initial basic feasible solution using (a) Least cost method (b) Vogel’s approximation method.
Solution:
(a) Least cost method
Given transportation table is

Total Availability = Total Requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₂ = 10; x₁₃ = 20; x₂₁ = 30;
x₂₂ = 10; x₂₄ = 10; x₃₂ = 20
Transportation Scheme:
O₁ → D₂; O₁ → D₃; O₂ → D₁;
O₂ → D₂; O₂ → D₄; O₃ → D₂
The transportation cost:
=(10 × 8) + (20 × 3) + (30 × 4) + (10 × 5) + (10 × 4) + (20 × 2)
= 80 + 60 + 120 + 50 + 40 + 40
= 390

(ii) Vogel’s approximation method:
Here Σai = Σbj = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₁ = 30; x₁₃ = 20; x₂₂ = 20;
x₂₄ = 10; x₃₂ = 20
Transportation Scheme:
O₁ → D₁; O₂ → D₃; O₂ → D₂;
O₂ → D₄; O₃ → D₂;
Total transportation cost:
= (30 × 5) + (20 × 3) + (20 × 5) + (10 × 4) + (20 × 2)
= 150 + 60 + 100 + 40 + 40
= 390

Question 3.
Determine an initital basic feasible solution to the following transportation problem by using (i) North West Corner rule (ii) least cost method.

Solution:
(i) North West Corner rule:
The given transportation table is

Total supply = 25 + 35 + 40 = 100
Total Requirement = 30 + 25 + 45 + 100
(i.e) Total supply = Total requirement.
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:


Transformation schedule:
S₁ → D₁; S₂ → D₁; S₂ → D₂;
S₃ → D₃; S₃ → D₃;
The transformation cost:
= (25 × 9) + (5 × 6) + (25 × 8) + (5 × 4) + (40 × 9)
= 225 + 30 + 200 + 20 + 360 = 835

(ii) Least cost method
The given transportation table is

(i.e) Total supply = Total requirement = 100
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:


Transportation schedule:
S₁ → D₃; S₂ → D₃; S₃ → D₁
S₃ → D₂;
The transportation cost:
= (10 × 5) + (35 × 4) + (15 × 7) + (25 × 6) + (40 × 9)
= 50 + 140 + 105 + 150 + 360 = 805

Question 4.
Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

Solution:
The given transportation table is

Here Σai = Σbj =17
(i.e) Total supply = Total Demand
∴ The given problem is balanced transformation problem.
Hence there exists a feasible solution to the given problem
First allocation:



Here
x₁₁ = 1; x₁₂ = 5; x₂₄ = 1;
x₃₁ = 6; x₃₃ = 15;
Transportation schedule:
O₁ → D₁; O₁ → D₂; O₂ → D₄
O₃ → D₁; O₃ → D₃
The transportation cost:
=(1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)
= 2 + 15 + 1 + 30 + 45 + 9
= 102

Question 5.
A car hire company has one car at each of five depots a,b,c,d and e. A customer in each of the fine towers A, B, C, D and E requires a car. The distance (in miles) between the depots (origins) and the towers (destinations) where the customers are given the following distance matrix.

How should the cars be assigned to the customers so as to minimize the distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3. (Assignment)
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Now Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 5.
Cover all the zeros of table 4 with three lives. Since three assignments were made please note that check [✓] Row C and E which have no assignment.

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element in each row and subtract from the uncovered cells, depots

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.
depots

Step 8.
Determine an assignment

Here all the five assignments have been made. The optimal assignment schedule and total distance is

∴ The optimum Distance (minimum) 575 kms

Question 6.
A natural truck – rental service has a surplus of one truck in each of the cities 1, 2, 3, 4, 5 and 6 and a deficit of one truck in each of the cities 7, 8, 9, 10, 11 and 12. The distance (in kilometers) between the cities with a surplus and the cities with

How should the truck be dispersed so as to minimize the total distance travelled?
Solution:
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1.
Select the smallest element in each row and subtract this from all the elements in its row.

Step 2.
Select the smallest element in each column and subtract this from all the elements in its column.

Step 3.
Examine the rows with exactly one zero, mark the zero by □ mark other zeros, in its column by X

Step 4.
Examine the Columns with exactly one zero. If there is exactly one zero, mark that zero by □ mark other zeros in its rows by X

Step 5.
Cover all the zeros of table 4 with five lines. Since three assignments were made

Step 6.
Develop the new revised tableau. Examine those elements that are not covered by a line in Table 5. Take the smallest element. This is l(one) in our case. By subtracting 1 from the uncovered cells.

Step 7.
Go to step 3 and repeat the procedure until you arrive at an optimal assignments.

Step 8.
Determine an assignment

Here all the six assignments have been made. The optimal assignment schedule and total distance is

∴The optimum Distance (minimum) = 125 kms

Question 7.
A person wants to invest in one of three alternative investment plans: Stock, Bonds and Debentures. It is assumed that the person wishes to invest all of the funds in a plan. The pay – off matrix based on three potential economic conditions is given in the following table

Solution:

(i) Maximin
Max (3000, 1000, 6000) = 6000. Since the maximum pay of is 6000, the alternative ‘Debentures’, is selected.

(ii) Minimax
Min (10000, 8000, 6000) = 6000, Since the minimum pay-off is 6000. the alternative ‘Debentures’ is selected.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2

Question 1.
Mention two branches of statistical inference?
Solution:
The two branches of statistical inference are estimation and testing of hypothesis.

Question 2.
What is an estimator?
Solution:
Any sample statistic which is used to estimate an unknown population parameter is called an estimator ie., an estimator is a sample statistic used to estimate a population parameter.

Question 3.
What is an estimate?
Solution:
When we observe a specific numerical value of our estimator, we call that value is an estimate. In other words, an estimate is a specific value of a statistic.

Question 4.
What is point estimation?
Solution:
Point estimation involves the use of sample data to calculate a single value which is to serve as a best estimate of an unknown population parameter. For example the mean height of 145 cm from a sample of 15 students is‘a point estimate for the mean height of the class of 100 students.

Question 5.
What is interval estimation?
Solution:
Generally, there are situation where point estimation is not desirable and we are interested in finding limits within which the parameter would be expected to lie is called an interval estimation.

Question 6.
What is confidence interval?
Solution:
Let us choose a small value of a which is known as level of significance (1% or 5%) and determine two constants says c₁ and c₂ such that p(c₁ < θ < c₂|t) = 1 – α

The quantities c₁ and c₂ so determined are known as the confidence Limits and the interval [c₁, c₂] within which the unknown value of the population parameter is expected to lie is known as a confidence interval.(1 – α)is called as confidence coefficient.

Question 7.
What is null hypothesis? Give an example.
Solution:
According to prof R. A. fisher, “Null hypothesis is the hypothesis which is tested for possible rejection under the assumption that it is true”, and it is denoted by H₀.

For example: If we want to find the population mean has a specified value µ₀, then the null hypothesis H₀ is set as follows H₀ : µ = µ₀

Question 8.
Define alternative hypothesis.
Solution:
A null hypothesis is a type of hypothesis, that proposes that no statistical significance exists in a set of given observations. For example, let the average time to cook a specific dish is 15 minutes. The null hypothesis would be stated as “The population mean is equal to 15 minutes”, (i.e) H₀ : µ = 15

Any hypothesis which is complementary to the null hypothesis is called as the alternative hypothesis and is usually denoted by H₁.

For example: If we want to test the null hypothesis that the population has specified mean µ i.e., H₀ : µ = µ₀ then the alternative hypothesis could be anyone among the following:
(i) H₁ : µ ≠ µ₀ (µ > or µ < µ₀)
(ii) H₁ : µ > µ₀
(iii) H₁ : µ < µ₀

Question 9.
Define critical region.
Solution:
A region corresponding to a test statistic in the sample space which tends to rejection of H₀ is called critical region or region of rejection.

Question 10.
Define critical value.
Solution:
The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depends upon.
(i) The level of significance
(ii) The alternative hypothesis whether it is two-tailed or single-tailed

Question 11.
Define the level of significance.
Solution:
The level of significance is defined as the probability of rejecting a null hypothesis by the test when it is really true, which is denoted as α. That is P(Type 1 error) = α. For example, the level of significance 0.1 is related to the 90% confidence level.

Question 12.
What is a type I error
Solution:
There is every chance that a decision regarding a null hypothesis may be correct or may not be correct. The error of rejecting H₀ when it is true is called type I error.

Question 13.
What is single tailed test.
Solution:
When the hypothesis about the population parameter is rejected only for the value of sample statistic falling into one of the tails of the sampling distribution, then it is known as a one-tailed test. Here H₁: µ > µ₀ and H₁: µ < µ₀ is known as a one-tailed alternative.

Question 14.
A sample of 100 items, draw from a universe with mean value 64 and S.D 3, has a mean value 63.5. Is the difference in the mean significant?
Solution:
sample size n = 100 ; sample mean \(\bar { x}\) = 63.5
sample SD S = 3;
population mean µ = 64 population SD σ = 3
Null Hypothesis H₀ : µ = 64 (the sample has been drawn from the population mean µ = 64 and SD σ = 3)
Alternative Hypothesis H₁ : µ ≠ 64 (two tail) i.e the sample has not been drawn from the population mean µ = 64 and SD σ = 3
The level of significance α = 5% = 0.05
Test statistic
z = \(\frac { 63.5-64}{\frac{3}{\sqrt{100}}}\) = \(\frac { -0.5 }{(\frac{3}{10})}\) = \(\frac { -0.5 }{0.3}\) = -1.667
|z| = 1.667
∴ calculated z = 1.667
critical value at 5% level of
significance is z_{\(\frac { α }{2}\)} = 1.96
Inference:
At 5% level of significance Z < Z_{\(\frac { α }{2}\)} since the calculated value is less than the table value the null hypothesis is accepted.

Question 15.
A sample of 400 individuals is found to have a mean height of 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height of 67.39 inches and standard deviation 1.30 inches?
Solution:
sample size n = 400; sample mean \(\bar { x}\) = 67.47 inches
sample SD S = 1.30 inches population mean
µ = 67.39 inches
population SD σ = 1.30 inches
Null Hypothesis H₀ : µ = 67.39 inches (the sample has been drawn from the population mean µ = 67.39 inches; population SD σ = 1.30 inches)
Alternative Hypothesis H₁ = µ ≠ 67.39 inches(two tail)
i.e the sample has not been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches
The level of significance α = 5% = 0.05
Test static:

Thus the calculated and the significant value or Z_{\(\frac { α }{2}\)} = 1.96
table value comparing the calculated and table values
Z_{\(\frac { α }{2}\)} (i.e.,) 1.2308 < 1.96
Inference: since the calculated value is less than value i.e Z > Z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is accepted Hence we conclude that the data doesn’t provide us any evidence against the null hypothesis. Therefore, the sample has been drawn from the population mean µ = 67.39 inches and SD σ = 1.30 inches.

Question 16.
The average score on a nationally administered aptitude test was 76 and the corresponding standard deviation was 8. In order to evaluate a state’s education system, the scores of 100 of the state’s students were randomly selected. These students had an average score of 72. Test at a significance level of 0.05 if there is a significant difference between the state scores and the national scores.
Solution:
sample size n = 100
sample mean \(\bar { x}\) = 72
sample SD S = 8
population mean µ = 76
under the Null hypothesis H₀ : p = 76
Against the alternative hypothesis H₀ : µ ≠ 76 (two mail)
Level of significance µ = 0.05
Test statistic:

since alternative hypothesis is of two tailed test we can take |Z| = 5.
∴ critical value 5% level of significance is z > z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is less than table value i.e z > z_{\(\frac { α }{2}\)} at 5% level of significance the null hypothesis H0 is rejected Therefore, we conclude that there is significant difference between the sample mean and population mean µ = 76 and SD σ = 8.

Question 17.
The mean breaking strength of cables supplied by a manufacturer is 1,800 with a standard deviation 100. By a new technique in the manufacturing process it is claimed that the breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can you support the claim at 0.01 level of significance.
Solution:
Sample size n = 50
Sample mean \(\bar { x}\) = 1800
Sample SD S = 100
population mean µ = 1850
Null hypothesis H₀ : µ = 1850
Level of significance µ = 0.01

= -3.5355
∴ z = -3.536
calculated value |z| = 3.536
Critical value at 1% level of significance is z_{\(\frac { α }{2}\)} = 2.58
Inference:
Since the calculated value is greater than table (ie) z > z_{\(\frac { α }{2}\)} at 1% level of significance, the null hypothesis is rejected, therefore we conclude that we is rejected, Therefore we conclude that we can not support we conclude that we can support the claim of 0.01 of significance.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.1

Question 1.
Define Time series.
Solution:
When quantitative data are arranged in the order of their occurrence, the resulting series is called the Time Series.

Question 2.
What is the need for studying time series?
Solution:
Time series analysis is one of the statistical methods used to determine the patterns in data collected for a period of time. Generally, each of us should know about the past data to observe and understand the changes that have taken place in the past and current time. One can also identify the regular or irregular occurrence of any specific feature over a time period in a time series data.

Question 3.
State the uses of time series.
Solution:

Question 4.
Mention the components of the time series.
Solution:
There are four types of components in a time series. They are as follows
(i) Secular Trend
(ii) Seasonal variations
(iii) Cyclic variations
(iv) Irregular variations

Question 5.
Define secular trend.
Solution:
Secular Trend: It is a general tendency, of time series to increase or decrease, or stagnates during a long period of time. An upward tendency is usually observed in the population of a country, production, sales, prices in industries, the income of individuals, etc., A downward tendency is observed in deaths, epidemics, prices of electronic gadgets, water sources, mortality rate, etc…

Question 6.
Write a brief note on seasonal variations
Solution:
As the name suggests, tendency movements are due to nature which repeats themselves periodically in every season. These variations repeat themselves in less than one year time. It is measured in an interval of time. Seasonal variations may be influenced by natural force, social customs and traditions. These variations are the results of such factors which uniformly and regularly rise and fall in the magnitude. For example, selling of umbrellas’ and raincoat in the rainy season, sales of cool drinks in the summer season, crackers in the Deepawali season, purchase of dresses in a festival season, sugarcane in Pongal season.

Question 7.
Explain cyclic variations.
Solution:
Cyclic Variations: These variations are not necessarily uniformly periodic in nature. That is, they may or may not follow exactly similar patterns after equal intervals of time. Generally, one cyclic period ranges from 7 to 9 years and there is no hard and fast rule in the fixation of years for a cyclic period. For example, every business cycle has a Start- Boom-Depression- Recover, maintenance during booms and depressions, changes in government monetary policies, changes in interest rates.

Question 8.
Explain the stratified random sampling with a suitable example.
Solution:
These variations do not have a particular pattern and there is no regular period of time of their occurrences. These are accidental changes which are purely random or unpredictable. Normally they are short-term variations, but their occurrence sometimes has its effect so intense that they may give rise to new cyclic or other movements of variations. For example floods, wars, earthquakes, Tsunami, strikes, lockouts, etc.

Question 9.
Define seasonal index.
Solution:
Seasonal Index for every season (i.e) months, quarters, or year is given by
Seasonal Index (S.I) = \(\frac{\text { Seasonal Average }}{\text { Grand average }}\) × 100
Where seasonal average is calculated for month, (or) quarter depending on the problem and Grand Average (G) is the average of averages.

Question 10.
Explain the method of fitting a straight line.
Solution:
(i) The straight-line trend is represented by the equation Y = a + bX …………. (1)
Where Y is the actual value, X is time, a, b are constants.
(ii) The constants ‘a and ‘b’ are estimated by solving the following two normal Equations
ΣY = n a + b ΣX ………… (2)
ΣXY = a ΣX + b ΣX² …………… (3)
Where ‘n’ = a number of years given in the data.
(iii) By taking the mid-point of the time as the ori-gin, we get ΣX = 0
(iv) When ΣX = 0, the two normal equations reduces to
ΣY = na + b(0) ; a = \(\frac { ΣY }{n}\) = \(\bar { Y}\)
ΣXY = a(0) + bΣX² ; b = \(\frac { ΣXY }{ΣX^2}\)
The constant ‘a’ gives the mean of Y and ‘b gives the rate of change (slope),
(v) By substituting the values of ‘a and ‘b’ in the trend equation (1), we get the Line of Best Fit.

Question 11.
State the two normal equations used in fitting a straight line.
Solution:
The normal equations used in fitting a straight line are
ΣY = na + b ΣX and ΣXY = a ΣX + b ΣX²
Where n = number of years given in the data,
X = time
Y = actual value
a, b = constants

Question 12.
State the different methods of measuring trends.
Solution:
Following are the methods by which we can measure the trend.
(i) Freehand or Graphic Method
(ii) Method of Semi-Averages
(iii) Method of Moving Averages

Question 13.
Compute the average seasonal movement for the following series

Solution:
Computation of seasonal. Index by the method of simple averages.

Question 14.
The following figures relates to the profits of a commercial concern for 8 years

Find the trend of profits by the method of three yearly moving averages.
Solution:

Question 15.
Find the trend of production by the method of a five-yearly period of moving average for the following data:

Solution:
I Computation of five – yearly moving averages

Question 16.
The following table gives the number of small- scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.

Solution:

Question 17.
The annual production of a commodity is given as follows:

Fit a straight line trend by the method of least squares
Solution:

Therefore the required equation of the straight line
Y = a + bx
Y = 169.428 + 3.285 X
⇒ Y = 169.428 + 3.285 (x – 1998)

Question 18.
Determine the equation of a straight line which best fits the following data

Compute the trend values for all years from 2000 to 2004
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 54 + 5.4X
Y = 54 + 5.4 (x – 2002)
The trend value can be obtained by when x = 2000
Yt = 54 + 5.4 (2000 – 2002)
Y = 54+ 5.4 (-2)
= 54 – 10.8
= 43.2
When x = 2001
Yt = 54 + 5.4 (2001 – 2002)
Y = 54 + 5.4 (-1)
= 54 – 5.4
= 48.6
When x = 2002
Yt = 54 + 5.4 (2002 – 2002)
V = 54 + 5.4 (0)
= 54
When x = 2003
Yt = 54 + 5.4 (2003 – 2002)
Y = 54 + 5.4 (1)
= 54 + 5.4
= 59.4
When x = 2004
Yt = 54 + 5.4 (2004 – 2002)
Y = 54 + 5.4(2)
= 54 + 10.8
= 64.8

Question 19.
The sales of a commodity in tones varied from January 2010 to December 2010 as follows: in year 2010 Sales (in tones)

Fit a trend line by the method of semi-average.
Solution:
Since the number of years is even (twelve). We can equally divide the given data it two equal parts and obtain averages of the first six months and the last six.

Question 20.
Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:

Question 21.
Use the method of monthly averages to find T the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:
Computation of Seasonal Index by the method of simple averages

Question 22.
The following table shows the number of salesmen working for a certain concern:

Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997
Solution:
Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by
Y = a + bX
Y = 48.8 + 2X
(i.e) Y = 48.8 + 2 (x – 1994)
When x = 1992
Yt = 48.8 + 2 (1992 – 1994)
Y = 48.8 + 2 (-2)
= 48.8 – 4
= 44.8
When x = 1993
Yt = 48.8 + 2 (1993 – 1994)
Y = 48.8 + 2 (-1)
= 48.8 – 2
= 46.8
When x = 1994
Yt = 48.8 + 2 (1994 – 1994)
Y = 48.8 + 2(0)
= 48.8
When x = 1995
Yt = 48.8 + 2 (1995 – 1994)
Y = 48.8 + 2(1)
= 50.8
When x = 1996
Yt = 48.8 + 2 (1996 – 1994)
Y = 48.8 + 2 (2)
= 48.8 + 4
= 52.8
When x = 1997
Yt = 48.8 + 2 (1997 – 1994)
Y = 48.8 + 2 (3)
= 48.8 + 6
= 54.8

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.3

Question 1.
Define Statistical Quality Control.
Solution:
Statistical quality control (SQC) refers to the use of statistical methods in the monitoring and maintaining of the quality of products and services. This method is used to determine the tolerance limits for accepting a production process.

Question 2.
Mention the types of causes for variation in a production process.
Solution:
There are two causes of variation which affects the quality of a product, namely
1. Chance Causes (or) Random causes
2. Assignable Causes

Question 3.
Define Chance Cause.
Solution:
These are small variations which are natural and inherent in the manufacturing process. The variation occurring due to these causes is beyond human control and cannot be prevented or eliminated under any circumstances, the minor causes which do not affect the quality of the products to an extent are called as Chance Causes (or) Random causes. For example Rain, floods, power cuts, etc.,

Question 4.
Define Assignable cause.
Solution:
The R chart is used to show the variability or dispersion of the samples taken from the given process. The average range is given by \(\overline{\mathrm{R}}=\frac{\sum R}{n}\), where R = x_max – x_min for each ‘n’ samples. For samples of size less than 20, the range provides a good estimate of σ. Hence to measure the variance in the variable, range chart is used.

Question 5.
What do you mean by product control?
Solution:
Product Control means that controlling the quality of the product by critical examination through sampling inspection plans. Product Control aims at a certain quality level to be guaranteed to the customers. It attempts to ensure that the product sold does not contain a large number of defective items. Thus it is concerned with classification of raw materials, semi-finished goods or finished goods into acceptable on rejectable products.

Question 6.
What do you mean by process control?
Solution:
In-Process Control the proportion of defective items in the production process is to be minimized and it is achieved through the technique of control charts.

Question 7.
Define a control chart.
Solution:
A control chart is essentially a graphic device for presenting data so as to directly reveal the frequency and extent of variations from established standards or goals. Control charts are simple to construct and easy to interpret and they tell the manager at a glance whether or not the process is in control, i.e within the tolerance limits.

Question 8.
Name the control charts for variables.
Solution:
(i) Charts for mean (\(\bar { X}\))
(ii) Charts for Range (R)

Question 9.
Define mean chart.
Solution:
The mean chart (\(\bar { X}\) chart) is used to show, the quality average of the samples taken from the given process. The \(\bar { X}\) charts are usually required for decision making to accept or reject the process.
Procedure for \(\bar { X}\)
i. Let X₁, X₂, X₃, etc. be the samples selected each containing “n” observations usually (n = 4.5 or 6)

ii. Calculate mean for each samples \(\bar { X}\)₁, \(\bar { X}\)₂, \(\bar { X}\)₃, ……… by using \(\bar { X}\)_i = \(\frac { ΣX_i }{n}\), i = 1, 2, 3, 4, ….
where Σx_i = total f “n” values included in the sample X₁.

iii. Find the mean (\(\bar { \bar X}\)) of the sample means
\(\bar { \bar X}\) = \(\frac { Σ \bar X }{number of samples}\)
where Σ\(\bar { X}\) = total of all the sample means.

Question 10.
Define R chart.
Solution:
The R chart is used to show the variability or dispersion of the samples taken from the given process. R charts are also required for decision making to accept of reject the process.
Procedure for R-Charts
Calculate R = Xmax – Xmin
Let R₁, R₂, R₃ ………….. be the ranges of the “n” samples. The average range is given by
\(\bar { X}\) = \(\frac { ΣR }{n}\)

Question 11.
What are the uses of statistical quality control?
Solution:
(i) The role of statistical quality control is to collect and analyse relevant data for the purpose of detecting whether the process is under control or not.
(ii) The value of quality control lies in the fact that assignable causes in a process can be quickly detected. Infact the variations are often discovered before the product becomes defective.
(iii) Statistical quality control is only diagnostic. It tells us whether the standard is being maintained or not.
(iv) This technique is used in almost all production industries such as automobile textile, electrical equipment, biscuits, both soaps, chemicals, Petroleum products, etc.
(v) The purpose for which SQC are used in two fold namely (a) process control (b) product control.
The main purpose of SQC is to device statistical techniques which would help in elimination of assignable causes and bring the production process under control.

Question 12.
Write the control limits for the mean chart.
Solution:
The calculation of control limits for x chart in two different cases is

Question 13.
Write the control limits for the R chart.
Solution:
The calculation of control limits for R chart in two different cases are

The values of A₂, D₂ and D₄ are given in the table.

Question 14.
A machine is set to deliver packets of a given j weight. Ten samples of size five each were recorded. Below are given relevant data:

Solution:
Calculate the control limits for mean chart and the range chart and then comment on the state of control, (conversion factors for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 16.2 + (0.58)(7.4)
= 16.2 + 4.292
= 20.492
= 20.49
CL = \(\bar { \bar X}\) = 16.2
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 16.2 – (0.58) (7.4)
= 16.2 – 4.292
= 11.908
= 11.91
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(7.4)
= 15.651
= 15.65
CL = \(\bar { R}\) = 7.4
LCL = D₃ \(\bar { R}\) = (0)(7.4) = 0

Question 15.
Ten samples each of size five are difawn at regular intervals from a manufacturing process. The sample means (X) and their ranges (R) are given below:

Calculate the control limits in respect of \(\bar { X}\) chart. (Given A₂ = 0.58, D₃ = 0 and D₄ = 2.115 ) Comment on the state of control.
Solution:

UCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 + (0.58)(6.8)
= 46.2 + 3.944
= 50.144
= 50.14
CL = \(\bar { \bar X}\) = 46.2
LCL = \(\bar { \bar X}\) + A₂ \(\bar { R}\)
= 46.2 – (0.58)(6.8)
= 46.2 – 3.944
= 42.256
= 42.26
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= (2.115)(6.8)
= 14.382
= 14.38
CL = \(\bar { R}\) = 6.8
LCL = D₃\(\bar { R}\) = (0)(6.8) = 0

Question 16.
Construct X and R charts for the following data:

Solution:


\(\bar { R}\) = \(\frac { 144 }{8}\) = 18
UCL = \(\bar { \bar x}\) + A₂ \(\bar { R}\)
= 37.71 + (0.58)(18)
= 37.71 + 10.44
= 48.15
CL = \(\bar { \bar x}\) = 37.71
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 37.71 – (0.58)(18)
= 37.71 – 10.44
= 27.27
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (18) = 38.07
CL = \(\bar { R}\) = 18
LCL = D₃ \(\bar { R}\) = 0(18) = 0

Question 17.
The following data show the values of sample mean (\(\bar {x}\)) and its range (R) for the samples of size five each. Calculate the values for control limits for mean, range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 + (0.58)(6.3)
= 10.66 + 3.654 = 14.314
= 14.31
CL = \(\bar { \bar x}\) = 10.66
LCL = \(\bar { \bar x}\) – A₂ \(\bar { R}\)
= 10.66 – (0.58)(6.3)
= 10.66 – 3.654
= 7.006
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = 2.115 (6.3)
= 13.3245
= 13.32
CL = \(\bar { R}\) = 6.3
LCL = D₃ \(\bar { R}\) = 0(6.3) = 0
Conclusion: Since all the points of sample range is within UCL of R chart, the process is in control.

Question 18.
A quality control inspector has taken ten ” samples of size four packets each from a potato chips company. The contents of the sample are given below, Calculate the control limits for mean and range chart.

(Given for n = 5, A₂ = 0.58, D₃ = 0 and D₄ = 2.115)
Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 12.5 + (0.58)(0.37)
= 12.5 + 0.2146 = 12.7146
= 12.71
CL = \(\bar { \bar X}\) = 12.5
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\) = 12.5 – (0.58) (0.37)
= 12.5 – 0.2146 = 12.2854
= 12.29
The control limits for Range chart is
UCL = D₄ \(\bar { R}\) = (2.115)(0.37) = 0.78255
= 0.78
CL = \(\bar { R}\) = 0.37
LCL = D₃\(\bar { R}\) = (0)(0.37) = 0

Question 19.
The following data show the values of sample means and the ranges for ten samples of size 4 each. Construct the control chart for mean and range chart and determine whether the process is in control.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 +(0.73)(20.1)
= 30.1 + 14.673 = 44.773
= 44.77
CL = \(\bar { \bar X}\) = 30.1
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 30.1 – (0.73) (20.1)
= 30.1 – 14.673 = 15.427
= 15.43
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(20.1) = 45.828
= 45.83
CL = \(\bar { R}\) = 20.1
LCL = D₃ \(\bar { R}\) = 0(20.1) = 0

Question 20.
In a production process, eight samples of size 4 are collected and their means and ranges are given below. Construct mean chart and range chart with control limits.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 + (0.73) (3.12)
= 13.25 + 2.2776 = 15.5276
= 15.53
CL = \(\bar { \bar X}\) = 13.25
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 13.25 – (0.73)(3.12)
= 13.25 – 2.2776 = 10.972
= 10.97
The control limits for Range chart is
UCL = D₄ \(\bar { R}\)
= 2.28(3.12) = 7.11984
= 7.12
CL = \(\bar { R}\) = 3.12
LCL = D₃\(\bar { R}\) = 0(3.12) = 0

Question 21.
In a certain bottling industry the quality control inspector recorded the weight of each of the 5 bottles selected at random during each hour of four hours in the morning.

Solution:

UCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 + (0.58)(4)
41 + 2.32 = 43.32
CL = \(\bar { \bar X}\) = 41
LCL = \(\bar { \bar X}\) – A₂ \(\bar { R}\)
= 41 – (0.58)(4)
= 41 – 2.32
= 38.68
The control limits for range chart is
UCL = D₄ \(\bar { R}\) = 2.115(4)
= 8.46
CL = \(\bar { R}\) = 4
LCL = D₂ \(\bar { R}\) = 0(4) = 0
Conclusion: Since all the points of the sample mean and Range are within the control limits, the process is in control.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3

Choose the correct Answer:

Question 1.
A ……………… may be finite or infinite according as the number of observation or items in it is finite or infinite
(a) Population
(b) census
(c) parameter
(d) none of these
Solution:
(a) population

Question 2.
A …………….. of statistical individuals in a population is called a sample.
(a) Infinite set
(b) finite subset
(c) finite set
(d) entire set
Solution:
(b) finite subset

Question 3.
A finite subset of statistical individuals in a population is called ………………
a) a sample
(b) a population
(c) universe
(d) census
Solution:
(a) a sample

Question 4.
Any statistical measure computed from sample data is known as ……………..
(a) Parameter
(b) random sample
(c) Infinite measure
(d) uncountable
Solution:
(b) random sample

Question 5.
A ………………. is one where each item in the universe has an equal chance of known opportu¬nity of being selected
(a) Parameter
(b) random sample
(c) statistic
(d) entire data
Solution:
(b) random sample

Question 6.
A random sample is a sample selected in such a way that every item in the population has an equal chance of being included
(a) Harper
(b) fisher
(c) karl pearson
(d) Dr. yates
Solution:
(a) Harper

Question 7.
Which one of the following is probability sampling
(a) Purposive sampling
(b) judgement sampling
(c) sample random sampling
(d) Convenience sampling
Solution:
(c) sample random sampling

Question 8.
In simple random sampling of drawing any unit, the probability of drawing any unit at the draw is ?
(a) \(\frac { n }{N}\)
(b) \(\frac { 1 }{N}\)
(c) \(\frac { N }{n}\)
(d) n
Solution:
(b) \(\frac { 1 }{N}\)

Question 9.
In ……………. the heterogeneous groups divided into homogeneous groups
(a) Non-probability sample
(b) a sample random sample
(c) a stratified random sample
(d) Systematic sample
Solution:
(c) a stratified random sample

Question 10.
Errors in sampling are of
(a) Two types
(b) three types
(c) four types
(d) five types
Solution:
(a) Two types

Question 11.
The method of obtaining the most likely value of the population parameter using statistic is called
(a) estimate
(b) estimate
(c) biased estimate
(d) standard error
Solution:
(d) standard error

Question 12.
An estimator is a sample statistic used to estimate a
(a) population parameter
(b) biased estimate
(c) sample size
(d) census
Solution:
(a) population parameter

Question 13.
…………… is a relative property, which states that one estimate is efficient relative to another.
(a) efficiency
(b) sufficiency
(c) unbiased
(d) consistency.
Solution:
(a) efficiency

Question 14.
If probability p[|\(\bar { θ }\) – θ|< ∈|< ∈|] 1 → µ as n → α for any positive then \(\bar { θ }\) is said to estimator of θ
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(d) Consistent

Question 15.
An estimator is said to be ………….. if it contains all the information in the data about the parameter it estimates.
(a) efficient
(b) sufficient
(c) unbiased
(d) consistent
Solution:
(b) sufficient

Question 16.
An estimate of a population parameter given by two numbers between which the parameter would be expected to lie called an ………….. interval estimate of the parameter
(a) point estimate
(b) interval estimate
(c) standard error
(d) confidence
Solution:
(b) interval estimate

Question 17.
A ……………… is a statement or an assertion about the population parameter
(a) hypothesis
(b) statistic
(c) sample
(d) census
Solution:
(a) hypothesis

Question 18.
Type I error is
(a) Accept H₀ when it is true
(b) Accept H₀ when it is false
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(c) Reject H₀ when it is true

Question 19.
Type II error is?
(a) Accept H₀ when it is wrong
(b) Accept H₀ when it is when it is true
(c) Reject H₀ when it is true
(d) Reject H₀ when it is false
Solution:
(a) Accept H₀ when it is wrong

Question 20.
The standard error of sample mean is?
(a) \(\frac { σ }{\sqrt{2n}}\)
(b) \(\frac { σ }{n}\)
(c) \(\frac { σ }{√n}\)
(d) \(\frac { σ^2 }{√n}\)
Solution:
(c) \(\frac { σ }{√n}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

Question 1.
Explain the types of sampling.
Solution:
The different types of sampling are

• simple random sampling
• Stratified random sampling and
• Systematic sampling

(i) In simple random sampling, every item of the population has an equal chance for being selected. The sampling can be done with replacement (or) without replacement. A random sampling from a finite population with replacement is equivalent to sampling from an infinite population without replacement. This technique will give useful results only if the population is homogeneous. The following are some of the methods of selecting a random sample.

(a) Use of an unbiased die or coin: If we have to choose between two alternatives, a coin is tossed, and depending on the head or tail course of action is taken. A die can be employed if there are six different alternatives.

(b) Lottery sampling: Here a random sample is selected by identifying each element of the population by means of a card of a pack of uniform cards or (by writing the number on pieces of paper) and to select a required number of cards after thorough mixing of the cards.

(c) Random numbers: Random numbers are formed of ‘random digits’ and arranged in the form of a table having a number of rows and columns. Tippett’s numbers form one such table wherein 40,000 digits were selected at random from census reports and combined by groups of four into 10,000 numbers.

(ii) In stratified random sampling, a population of units is divided into L sub-populations of N₁, N₂, …… N_L. The sub-populations being non-overlapping and mutually exhaustive so that N = N₁ + N₂ + …….. + N_L. Each subpopulation is known as a stratum. If we select n₁, n₂, ……. n_l items, respectively, from these strata, we get a stratified sample. If a simple random sample is taken from each stratum, the whole procedure is referred to as stratified random sampling.

(iii) Systematic sampling is a form of restricted random selection which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in serial order and every ith element, starting from any of the first items is chosen. For example, suppose we require a 5% sample of students from a college where there are 2000 students, we select a random number from 1 to 20. If it is 12, then our sample consists of students with numbers 12, 32, 52, 72, …… 1992.

Question 2.
Write a short note on sampling distribution and standard error.
Solution:
sampling distribution:
The sampling distribution of a statistic is the frequency distribution which is formed with various values of a statistic computed from different samples of the same size drawn from the same population.

For instance, if we draw a sample a size n from a given finite population of N, then the total number of possible samples is Nc_n = \(\frac { N! }{n!(N-n)!}\) = k(say)

Standard Error:
The standard deviation of the sampling distribution of a statistic is known as its standard Error abbreviated as S.E. The standard error (S.E) of some of the well-known statistics, for large samples, are given below, where n is the sample size, σ² is the population variance.

Question 3.
Explain the procedures of testing of hypothesis
Solution:
The following are the steps involved in hypothesis testing problems:
1. Null hypothesis: Set up the null hypothesis H₀

2. Alternative hypothesis: Set up the alternative hypothesis. This will enable us to decide whether we have to use two-tailed test or single-tailed test (right or left tailed)

3. Level of significance: Choose the appropriate level of significance (a) depending on the reliability of the estimates and permissible risk. This is to be fixed before the sample is drawn, i.e., a is fixed in advance.

4. Test statistic: Compute the test statistic
Z = \(\frac { t-E(t) }{\sqrt{var(t)}}\) = \(\frac { t-E(t) }{S.E(t)}\) N(0, 1) as n → ∞

5. Conclusion: We compare the computed value of Z in step 4 with the significant value or critical value or table value Zα at the given level of significance.
(i) If |Z | < Zα i.e., if the calculated value is less than the critical value we say it is not significant. This may due to fluctuations of sampling and sample data do not provide us sufficient evidence against the null hypothesis which may therefore be accepted.

(ii) If |Z |> Zα i.e., if the calculated value of Z is greater than critical value Zα then we say it is significant and the null hypothesis is rejected at the level of significance α.

Question 4.
Explain in detail the test of significance for a single mean.
Solution:
Let xi, (i = 1, 2, 3, …, n) is a random sample of size from a normal population with mean µ and variance σ² then the sample mean is distributed normally with mean and variance
\(\frac { σ^2 }{n}\), i.e \(\bar { x }\) N(µ, \(\frac { σ^2 }{n}\))

Thus for large samples, the standard normal variate corresponding to \(\bar { x }\) is
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\) N (0, 1)

Under the null hypothesis that the sample has been drawn from a population with mean and variance σ², i.e., there is no significant difference between the sample mean (\(\bar { x }\)) and the population mean (α), the test statistic (for large samples) is:
Z = \(\frac {\bar { x } -µ }{\frac{σ}{√n}}\)

Question 5.
Determine the standard error of proportion for a random sample of 500 pineapples was taken from a large consignment and 65 were found to be bad?
Solution:
sample size n = 500
No. of bad pine apples = 65
sample proportion = P = \(\frac { 65 }{500}\) = 0.13
Q = 1 – p ⇒ Q = 1 – 0.13
∴ Q = 0.87
The S.E for sample proportion is given by
S.E = \(\sqrt{\frac { PQ }{N}}\) = \(\sqrt{\frac { (0.13)(0.87) }{500}}\)
= \(\sqrt{\frac { 0.1131 }{500}}\) = \(\sqrt{0.0002262}\)
= 0.01504
∴ S.E = 0.015
Hence the standard error for sample proportion is S.E = 0.015

Question 6.
A sample of 100 students are drawn from a school The mean weight and variance of the sample are 67.45 kg and 9 kg respectively find (a) 95% and (b) 66% confidence intervals for estimating the mean weight of the students.
Solution:
sample size n = 100
The sample mean = \(\bar { x }\) = 67.45
The sample variance S² = 9
The sample standard deviation S = 3
S.E = \(\frac { S }{√n}\) = \(\frac { 3 }{\sqrt{100}}\) = \(\frac { 3 }{10}\) = 0.3

(a) The 95% confidence limits for µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E < µ < \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (1.96 × 0.3) ≤ µ ≤ 67.45 + (1.96 × 0.3) 67.45 – 0.588 ≤ µ ≤ 67.45 + 0.588
66.862 ≤ µ ≤ 68.038
The confidential limits is (66.86, 68.04)

(b) The 99% confidence limits for estimating µ are given by
\(\bar { x }\) – z_{\(\frac { α }{2}\)} S.E ≤ µ ≤ \(\bar { x }\) + z_{\(\frac { α }{2}\)} S.E
67.45 – (2.58 × 0.3) ≤ µ ≤ 67.45 + (2.58 × 0.3)
67.45 – 0.774 ≤ µ ≤ 67.45 + 0.774
66.676 ≤ µ ≤ 68.224
∴ The 99% confidence limits is (66.68, 68.22)

Question 7.
The mean I.Q of a sample of 1600 children was 99. it is likely that this was a random sample from a population with a mean I.Q 100 and standard deviation of 15? (Test at 5% level of significance)
Solution:
sample size n = 1600
\(\bar { x }\) = 99
sample mean
Population mean µ = 100
population S.D σ = 15
under the Null hypothesis H₀ : µ = 100
Alternative hypothesis H₁ : µ = 100 (two tails)
Level of significance µ = 0.05

z = -2.666
z = -2.67
Calculated value |z| = 2.67
critical value at 5% level of significance is
z_{\(\frac { α }{2}\)} = 1.96
Inference:
Since the calculated value is greater than table value i.e z ⇒ z_{\(\frac { α }{2}\)} at 5% level of significance, the null hypothesis is rejected. Therefore we conclude that the sample mean differs, significantly from the population mean.