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GNDU Question Paper-2023

BA 3

Semester

COMPUTER SCIENCE

(Computer Oriented Numerical & statistical Methods)

Time Allowed: Three Hours Maximum Marks: 75

Note: Attempt Five questions in all, selecting at least One question from each section. The

Fifth question may be attempted from any section. All questions carry equal marks.

SECTION-A

1. Which are non-linear equations? What is their use? Explain the algorithm of Newton

Raphson method.

2. Why false position method is used? Draw its comparison with bisection method.

SECTION-B

3. Write and explain the Guass Siedel method for simultaneous solutions of equations.

Compare it with Guass Jordon method.

4. Test for consistency of the following equations and if consistent find the solution:

x-y+z=3

2x+y-z-2

x+2y-2z=-1

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SECTION-C

5. Using Lagrange's method of interpolation find number of workers earning between Rs.

30-40 from the data given below:

Earning (Rs.)

15-20

20-30

30-40

45-55

5-70

No. of workers

110

183

140

6. (a) Calculate integration of following function using Simpson's rule:











(b) The function f(x) is given below. Determine the integral of f(x) between x=0.1 to x1.0

using Simpson's 3/8 rule.

0.1

0.2

0.3

0.4

0.5

1.001

1.008

1.027

1.064

1.125

0.6

0.7

0.8

0.9

1.0

1.216

1.343

1.512

1.729

2.0

SECTION-D

7. Which are various measures of Central Tendency? Write merits and demerits of any

three.

8. (a) What is correlation? What is its use?

(b) Calculate correlation between following data using Karl Pearson's method.

Easy2Siksha

GNDU Answer Paper-2023

BA 3

Semester

COMPUTER SCIENCE

(Computer Oriented Numerical & statistical Methods)

Time Allowed: Three Hours Maximum Marks: 75

Note: Attempt Five questions in all, selecting at least One question from each section. The

Fifth question may be attempted from any section. All questions carry equal marks.

SECTION-A

1. Which are non-linear equations? What is their use? Explain the algorithm of Newton

Raphson method.

Ans I'd be happy to explain non-linear equations, their uses, and the Newton-Raphson

method in detail. I'll break this down into sections to cover each part of your question

thoroughly.

1. Non-linear Equations

Non-linear equations are mathematical equations where the variables are not in a simple

linear relationship. In other words, the graph of a non-linear equation is not a straight line.

Some examples of non-linear equations include:

a) Quadratic equations: y = x^2 + 3x + 2 b) Exponential equations: y = 2^x c) Logarithmic

equations: y = log(x) d) Trigonometric equations: y = sin(x) e) Polynomial equations of

degree higher than 1: y = x^3 - 2x^2 + 4

In these equations, the relationship between the variables is more complex than a simple

linear proportion. The variables might be raised to powers, appear inside functions like

logarithms or trigonometric functions, or be multiplied together.

2. Uses of Non-linear Equations

Non-linear equations are incredibly important in various fields because they can model

complex real-world phenomena more accurately than linear equations. Here are some of

their uses:

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a) Physics: Many physical laws are non-linear, such as Einstein's theory of general relativity

or equations describing fluid dynamics.

b) Engineering: Non-linear equations are used to model structural behavior under stress,

heat transfer, and electrical circuits with non-linear components.

c) Biology: Population growth models often use non-linear equations to account for factors

like carrying capacity.

d) Economics: Many economic models use non-linear equations to describe market

behavior, supply and demand relationships, or financial derivatives.

e) Computer Graphics: Non-linear equations are used to create realistic curves and surfaces

in 3D modeling and animation.

f) Chemistry: Chemical reaction rates and equilibrium processes are often described by non-

linear equations.

g) Climate Science: Complex climate models use systems of non-linear equations to predict

weather patterns and long-term climate changes.

h) Medicine: Non-linear models are used in pharmacokinetics to describe how drugs are

metabolized in the body.

3. Newton-Raphson Method

The Newton-Raphson method (also known as Newton's method) is an iterative algorithm

used to find the roots of a non-linear equation. In other words, it helps us find the values of

x for which f(x) = 0, where f(x) is a non-linear function.

Here's a simplified explanation of the algorithm:

Step 1: Start with an initial guess for the root, let's call it x0.

Step 2: Calculate the value of the function at this point, f(x0).

Step 3: Calculate the value of the function's derivative at this point, f'(x0).

Step 4: Use these values to calculate a better approximation of the root: x1 = x0 - f(x0) /

f'(x0)

Step 5: Repeat steps 2-4 using x1 as the new starting point, then x2, and so on, until the

desired level of accuracy is reached.

The formula for the nth iteration is: xn+1 = xn - f(xn) / f'(xn)

This method works by finding the tangent line to the curve at the current point and using

the x-intercept of this tangent line as the next approximation of the root.

4. Why the Newton-Raphson Method Works

To understand why this method works, let's consider the geometric interpretation:

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Imagine you have a curve representing your non-linear function f(x). You start at a point x0

on the x-axis and draw a vertical line up to the curve. This gives you the point (x0, f(x0)) on

the curve.

Now, draw the tangent line to the curve at this point. The slope of this tangent line is given

by f'(x0), the derivative of the function at x0.

The equation of this tangent line is: y - f(x0) = f'(x0)(x - x0)

We want to find where this tangent line crosses the x-axis, as this will be our next

approximation. At this point, y = 0, so we can solve:

0 - f(x0) = f'(x0)(x - x0) x = x0 - f(x0) / f'(x0)

This is exactly the formula we use for each iteration of Newton's method.

5. Advantages of the Newton-Raphson Method

a) Fast convergence: When it works well, the Newton-Raphson method converges

quadratically, which means the number of correct digits roughly doubles with each

iteration.

b) Precision: It can find roots to a very high degree of accuracy.

c) Versatility: It can be applied to a wide range of non-linear equations.

6. Limitations and Considerations

While powerful, the Newton-Raphson method has some limitations:

a) Initial guess: The method requires a good initial guess. If the starting point is too far from

the actual root, the method might not converge or might converge to the wrong root.

b) Derivative requirement: The method requires the function to be differentiable, and you

need to be able to calculate the derivative.

c) Division by zero: If the derivative becomes zero at any point, the method will fail due to

division by zero.

d) Multiple roots: If the function has multiple roots, the method might find a different root

than the one you're looking for, depending on the initial guess.

e) Oscillation: In some cases, the method can oscillate between two values without

converging.

7. Example Application

Let's consider a simple example to illustrate how the Newton-Raphson method works.

Suppose we want to find the square root of 5 using this method.

We can rephrase this as finding the root of the equation: f(x) = x^2 - 5 = 0

The derivative of this function is: f'(x) = 2x

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Let's start with an initial guess of x0 = 2 (since 2^2 = 4, which is close to 5).

Iteration 1: x1 = x0 - f(x0) / f'(x0) = 2 - (2^2 - 5) / (2*2) = 2 - (-1) / 4 = 2.25

Iteration 2: x2 = 2.25 - (2.25^2 - 5) / (2*2.25) ≈ 2.2361111111

Iteration 3: x3 ≈ 2.2360679775

After just three iterations, we've already approximated the square root of 5 to 10 decimal

places! The actual value is about 2.2360679775, so our approximation is extremely close.

8. Practical Implementation

In practice, when implementing the Newton-Raphson method, you would typically include a

stopping condition. This could be:

a) A maximum number of iterations to prevent infinite loops. b) A tolerance level - stop

when the change between iterations is smaller than a specified value.

Here's a simple Python implementation:

python

Copy

def newton_raphson(f, f_prime, x0, tolerance=1e-6, max_iterations=100):

x = x0

for i in range(max_iterations):

fx = f(x)

if abs(fx) < tolerance:

return x

dfx = f_prime(x)

if dfx == 0:

return None # Method failed

x = x - fx / dfx

return None # Max iterations reached without converging

# Example usage for sqrt(5)

def f(x):

return x**2 - 5

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def f_prime(x):

return 2*x

result = newton_raphson(f, f_prime, 2)

print(f"Approximation of sqrt(5): {result}")

This implementation includes safeguards against division by zero and non-convergence,

making it more robust for practical use.

9. Historical Context

The Newton-Raphson method has a rich history in mathematics. While it's named after Isaac

Newton and Joseph Raphson, similar ideas were developed independently by other

mathematicians.

Newton described the method in his 1671 work "De analysi per aequationes numero

terminorum infinitas," but he described it geometrically and didn't provide an explicit

formula. Raphson published a purely algebraic version of the method in 1690.

The method as we know it today, using calculus notation, was first published by Thomas

Simpson in 1740. Despite this, the name "Newton-Raphson" has stuck.

In conclusion, non-linear equations are essential tools for modeling complex real-world

phenomena across various fields of science and engineering. The Newton-Raphson method

provides an efficient way to solve these equations numerically, finding roots that might be

difficult or impossible to determine analytically. Understanding this method not only helps

in solving specific problems but also provides insight into the nature of non-linear systems

and iterative problem-solving techniques in general.

2. Why false position method is used? Draw its comparison with bisection method.

1. Ans: Introduction to Root-Finding Methods

In mathematics and computer science, we often need to find the roots of equations - that is,

the values of x that make f(x) = 0. While some equations can be solved analytically, many

real-world problems involve complex functions that can't be solved easily by hand. This is

where numerical methods come in handy.

Two popular numerical methods for finding roots are the bisection method and the false

position method (also known as the regula falsi method). Both of these are bracketing

methods, meaning they start with an interval that contains the root and progressively

narrow it down.

2. The False Position Method

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The false position method is an iterative technique used to find the roots of a function.

Here's how it works:

a) We start with two points, a and b, such that f(a) and f(b) have opposite signs. This ensures

that there's at least one root between a and b.

b) Instead of simply taking the midpoint of the interval like in the bisection method, the

false position method uses linear interpolation. It draws a straight line between the points

(a, f(a)) and (b, f(b)).

c) The x-intercept of this line is calculated, which gives us a new approximation of the root.

Let's call this point c.

d) We then evaluate f(c). If f(c) = 0, we've found the root exactly. Otherwise, we replace

either a or b with c, depending on which one has the same sign as f(c).

e) This process is repeated until we reach a desired level of accuracy.

The formula for calculating the new approximation c is:

c = (a * f(b) - b * f(a)) / (f(b) - f(a))

3. Why the False Position Method is Used

The false position method is used for several reasons:

a) Guaranteed convergence: Like the bisection method, the false position method is

guaranteed to converge to a root if the initial interval contains a root and the function is

continuous.

b) Potentially faster convergence: In many cases, the false position method converges

faster than the bisection method, especially when the function is approximately linear near

the root.

c) Simple implementation: The method is relatively straightforward to implement in code,

making it accessible for many applications.

d) No derivatives required: Unlike some other root-finding methods (like Newton's

method), the false position method doesn't require the calculation of derivatives, which can

be complex or impossible for some functions.

e) Robustness: The method is less likely to fail or diverge compared to some other methods,

especially when dealing with functions that aren't well-behaved.

4. Comparison with the Bisection Method

Now, let's compare the false position method with the bisection method:

a) Principle:

• Bisection: Always chooses the midpoint of the interval.

• False Position: Uses linear interpolation to estimate the root location.

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b) Convergence speed:

• Bisection: Guarantees that the interval size halves with each iteration.

• False Position: Can converge faster, especially when the function is close to linear

near the root.

c) Predictability:

• Bisection: Very predictable, always reducing the interval by half.

• False Position: Less predictable, as the convergence speed depends on the function's

shape.

d) Simplicity:

• Bisection: Simpler concept and implementation.

• False Position: Slightly more complex formula, but still relatively simple.

e) Efficiency near the root:

• Bisection: May become inefficient when close to the root, as it always halves the

interval.

• False Position: Often more efficient near the root due to linear interpolation.

f) Sensitivity to function shape:

• Bisection: Works equally well regardless of the function's shape.

• False Position: Performance can vary depending on the function's curvature.

5. Visual Representation

To better understand the difference between these methods, let's imagine a visual

representation:

Picture a graph with the x-axis representing the input values and the y-axis representing the

function output. The function curve crosses the x-axis at the root we're trying to find.

Bisection method:

• Draw a vertical line at the midpoint of the interval.

• The next iteration will focus on the half where the function changes sign.

False Position method:

• Draw a straight line connecting the points (a, f(a)) and (b, f(b)).

• Find where this line crosses the x-axis - this is your next approximation.

The visual difference is clear: while bisection always "cuts" the interval in half, false position

adjusts its guess based on the function's values at the endpoints.

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6. Advantages and Disadvantages

False Position Method: Advantages:

• Often converges faster than bisection, especially for nearly linear functions.

• Doesn't require function derivatives.

• Generally more efficient than bisection near the root.

Disadvantages:

• Can be slower than bisection for certain function shapes.

• May converge very slowly if one endpoint remains fixed for many iterations.

Bisection Method: Advantages:

• Simple to understand and implement.

• Guaranteed, predictable rate of convergence.

• Works well for a wide range of function shapes.

Disadvantages:

• Can be slower than other methods, especially near the root.

• Always uses the full interval, which may be inefficient.

7. When to Use Each Method

The choice between false position and bisection often depends on the specific problem

and the nature of the function:

• Use False Position when:

o The function is expected to be roughly linear near the root.

o Faster convergence is desired and the function's behavior is well-understood.

o The extra complexity in implementation is acceptable.

• Use Bisection when:

o The function's behavior is unpredictable or highly non-linear.

o A guaranteed, steady convergence rate is preferred.

o Simplicity of implementation is a priority.

o There's a need to bound the maximum error after n iterations.

8. Practical Considerations

When implementing these methods in computer programs, there are a few practical points

to consider:

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a) Stopping criteria: Both methods typically use a combination of:

• Maximum number of iterations

• Desired accuracy (when |f(x)| is less than some small value)

• Interval size (when |b - a| is less than some small value)

b) Floating-point precision: Be aware of the limitations of floating-point arithmetic in

computers. Very small intervals might lead to issues with precision.

c) Initial bracket selection: Both methods require an initial interval containing the root.

Choosing good initial points can significantly affect performance.

d) Hybrid approaches: Some implementations combine false position with other methods

(like bisection) to leverage the strengths of each.

9. Historical Context

The false position method has a rich history, dating back to ancient civilizations. It was used

by ancient Egyptians and later refined by mathematicians throughout history. The method's

longevity is a testament to its utility and effectiveness.

The bisection method, while simpler, is also rooted in ancient mathematical practices of

successive approximation. Both methods have stood the test of time and continue to be

relevant in the age of computers due to their reliability and relative simplicity.

10. Conclusion

In summary, the false position method is used because it offers a potentially faster

convergence than the bisection method while maintaining the guarantee of finding a root

(assuming the initial conditions are met). It strikes a balance between simplicity and

efficiency, making it a valuable tool in numerical analysis.

Both the false position and bisection methods have their place in a numerical analyst's

toolkit. Understanding the strengths and weaknesses of each allows for informed decisions

when tackling root-finding problems. As with many aspects of computer science and

mathematics, the choice of method often comes down to the specific requirements of the

problem at hand.

Remember, while these methods are powerful, they're just two of many numerical

techniques available for root-finding. In practice, it's valuable to be familiar with a range of

methods and to understand when each is most appropriate.

SECTION-B

3. Write and explain the Guass Siedel method for simultaneous solutions of equations.

Compare it with Guass Jordon method.

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1. Ans: Introduction to Systems of Linear Equations

Before we dive into the methods, let's understand what we're dealing with. A system of

linear equations is a set of equations where each equation is linear (no variables are

multiplied together or raised to powers). For example:

2x + 3y - z = 1 x - y + 2z = -2 3x + 2y + z = 4

Our goal is to find values for x, y, and z that satisfy all these equations simultaneously.

2. The Gauss-Seidel Method

The Gauss-Seidel method is an iterative technique for solving systems of linear equations.

It's named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von

Seidel. This method is particularly useful for large systems of equations that are too complex

to solve by direct methods.

How it works:

1. Start with initial guesses for the variables (often all zeros).

2. Use the first equation to solve for the first variable, using the initial guesses for the

other variables.

3. Use the second equation to solve for the second variable, but use the newly

calculated value for the first variable and the initial guesses for the remaining

variables.

4. Continue this process for all equations.

5. Repeat steps 2-4 until the values converge (stop changing significantly).

Let's break this down with a simple example:

Consider the system: 4x - y + z = 7 x + 5y + 2z = 13 x + y + 3z = 10

Step 1: Rearrange each equation to isolate one variable: x = (7 + y - z) / 4 y = (13 - x - 2z) / 5 z

= (10 - x - y) / 3

Step 2: Start with initial guesses. Let's use x = 0, y = 0, z = 0.

Step 3: First iteration: x = (7 + 0 - 0) / 4 = 1.75 y = (13 - 1.75 - 2(0)) / 5 = 2.25 z = (10 - 1.75 -

2.25) / 3 = 2

Step 4: Second iteration: x = (7 + 2.25 - 2) / 4 = 1.8125 y = (13 - 1.8125 - 2(2)) / 5 = 1.4375 z =

(10 - 1.8125 - 1.4375) / 3 = 2.25

We continue this process until the values stop changing significantly.

3. Advantages of the Gauss-Seidel Method

a) Memory Efficient: It only requires storage for one set of variable values at a time, making

it memory-efficient for large systems.

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b) Simplicity: The method is straightforward to understand and implement.

c) Convergence: For many systems, especially diagonally dominant ones (where the

diagonal element in each row is larger than the sum of the absolute values of the other

elements), this method converges quickly.

d) Adaptability: It can be easily modified to handle non-linear systems as well.

4. Disadvantages of the Gauss-Seidel Method

a) Convergence Issues: For some systems, especially those that aren't diagonally dominant,

the method might converge slowly or not at all.

b) Sensitivity to Initial Guesses: The choice of initial values can significantly affect the speed

of convergence.

c) Sequential Nature: The calculations must be done in sequence, which can be a limitation

for parallel computing.

5. The Gauss-Jordan Method

Now, let's look at the Gauss-Jordan method for comparison. This is a direct method for

solving systems of linear equations, which means it gives an exact solution in a finite

number of steps (assuming no rounding errors).

How it works:

1. Write the system of equations as an augmented matrix (coefficients on the left,

constants on the right).

2. Use elementary row operations to transform the left side of the augmented matrix

into the identity matrix.

3. The right side will then contain the solution.

Let's use the same example:

4x - y + z = 7 x + 5y + 2z = 13 x + y + 3z = 10

Step 1: Write as an augmented matrix: [4 -1 1 | 7] [1 5 2 | 13] [1 1 3 | 10]

Step 2: Use row operations to get 1s on the diagonal and 0s elsewhere:

[1 0 0 | 2] [0 1 0 | 3] [0 0 1 | 1]

Step 3: Read the solution: x = 2, y = 3, z = 1

6. Advantages of the Gauss-Jordan Method

a) Direct Solution: It provides an exact solution in a finite number of steps.

b) Simultaneous Solution: It solves for all variables at once.

c) Matrix Inversion: The process can be extended to find the inverse of a matrix.

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d) Deterministic: The number of steps is known in advance, which can be useful for planning

computational resources.

7. Disadvantages of the Gauss-Jordan Method

a) Computational Intensity: For large systems, it requires more computations than some

other methods.

b) Memory Requirements: It needs to store the entire augmented matrix in memory.

c) Rounding Errors: In computer implementations, rounding errors can accumulate,

especially for large systems.

8. Comparison of Gauss-Seidel and Gauss-Jordan Methods

a) Nature of Solution:

• Gauss-Seidel: Iterative, approximates the solution

• Gauss-Jordan: Direct, provides exact solution (barring rounding errors)

b) Speed:

• Gauss-Seidel: Can be faster for large, sparse systems (many zero coefficients)

• Gauss-Jordan: Generally faster for small to medium-sized dense systems

c) Memory Usage:

• Gauss-Seidel: More memory-efficient, especially for large systems

• Gauss-Jordan: Requires more memory to store the entire augmented matrix

d) Accuracy:

• Gauss-Seidel: Accuracy improves with each iteration, but may never reach exact

solution

• Gauss-Jordan: Provides exact solution, but can suffer from accumulated rounding

errors

e) Applicability:

• Gauss-Seidel: Works best for diagonally dominant systems, can handle some non-

linear systems

• Gauss-Jordan: Works for any non-singular system of linear equations

f) Parallel Computing:

• Gauss-Seidel: Less suitable for parallel computing due to its sequential nature

• Gauss-Jordan: Can be parallelized more easily

9. When to Use Each Method

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Use Gauss-Seidel when:

• Dealing with very large, sparse systems of equations

• Memory is a constraint

• An approximate solution is acceptable

• The system is diagonally dominant

Use Gauss-Jordan when:

• Working with small to medium-sized systems

• An exact solution is required

• The system is dense (few or no zero coefficients)

• Matrix inversion is also needed

10. Historical Context and Modern Applications

The Gauss-Seidel method was developed in the mid-19th century, while the foundations of

the Gauss-Jordan method date back to the early 19th century. Both methods have stood the

test of time and continue to be relevant in various fields:

• Engineering: Structural analysis, electrical circuit analysis

• Physics: Quantum mechanics calculations, fluid dynamics simulations

• Economics: Economic modeling, input-output analysis

• Computer Graphics: Image processing, 3D rendering

• Machine Learning: Training certain types of neural networks

In modern computing, these methods are often implemented as part of larger numerical

libraries and software packages. They serve as building blocks for more complex algorithms

and are sometimes used as preconditioners for other iterative methods.

11. Conclusion

Both the Gauss-Seidel and Gauss-Jordan methods are powerful tools for solving systems of

linear equations, each with its own strengths and weaknesses. The choice between them

depends on the specific problem at hand, the size of the system, the required accuracy, and

the available computational resources.

The Gauss-Seidel method shines in situations where memory is limited and an approximate

solution is acceptable, especially for large, sparse systems. Its iterative nature allows for

early termination if a rough estimate is sufficient.

The Gauss-Jordan method, on the other hand, is preferable when an exact solution is

needed and the system is of manageable size. Its ability to simultaneously solve for all

variables and invert matrices makes it versatile for various mathematical applications.

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Understanding both methods provides a solid foundation for tackling a wide range of

problems in numerical analysis and applied mathematics. As you continue your studies in

computer science and numerical methods, you'll likely encounter variations and extensions

of these methods, as well as more advanced techniques built upon their principles.

4. Test for consistency of the following equations and if consistent find the solution:

x-y+z=3

2x+y-z-2

x+2y-2z=-1

Ans: First, let's restate the system of equations:

1. x - y + z = 3

2. 2x + y - z = 2

3. x + 2y - 2z = -1

To determine if this system is consistent and find its solution if it is, we'll use the Gaussian

elimination method, also known as row reduction. This method involves transforming the

system of equations into row echelon form, which will help us identify if the system is

consistent and, if so, solve for the variables.

Step 1: Set up the augmented matrix

First, we'll create an augmented matrix from our system of equations. An augmented matrix

is a way to represent a system of linear equations in matrix form. Here's how we'll do it:

[1 -1 1 | 3] [2 1 -1 | 2] [1 2 -2 | -1]

Each row represents one equation, and the numbers to the left of the vertical line are the

coefficients of x, y, and z, respectively. The numbers to the right of the vertical line are the

constants from the right side of each equation.

Step 2: Perform row operations

Now, we'll use elementary row operations to transform this matrix into row echelon form.

The goal is to create a triangular shape in the coefficient part of the matrix (left of the

vertical line).

a) First, let's use the first row to eliminate the x term in the second row: Multiply the first

row by -2 and add it to the second row:

[1 -1 1 | 3] [0 3 -3 | -4] [1 2 -2 | -1]

b) Now, let's eliminate the x term in the third row: Subtract the first row from the third row:

[1 -1 1 | 3] [0 3 -3 | -4] [0 3 -3 | -4]

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c) We can see that the second and third rows are now identical. This is an important

observation that we'll come back to later. For now, let's continue with our row reduction:

Multiply the second row by -1/3 to make the leading coefficient 1:

[1 -1 1 | 3] [0 1 -1 | -4/3] [0 3 -3 | -4]

d) Now, use the second row to eliminate the y term in the first row: Add the second row to

the first row:

[1 0 0 | 5/3] [0 1 -1 | -4/3] [0 3 -3 | -4]

e) Finally, use the second row to eliminate the y term in the third row: Subtract 3 times the

second row from the third row:

[1 0 0 | 5/3] [0 1 -1 | -4/3] [0 0 0 | 0]

Step 3: Analyze the result

Now that we have our augmented matrix in row echelon form, we can determine if the

system is consistent and find its solution if it is.

A system of linear equations is considered consistent if it has at least one solution. It's

inconsistent if it has no solutions.

In our case, we can see that the last row of our reduced matrix is [0 0 0 | 0]. This row

represents the equation 0x + 0y + 0z = 0, which is always true for any values of x, y, and z.

This is a good sign and indicates that our system is consistent.

If we had ended up with a row like [0 0 0 | k], where k is any non-zero number, it would

have represented an equation like 0 = k, which is never true. This would have indicated an

inconsistent system with no solutions.

Since our system is consistent, let's solve for our variables:

From the first row: x = 5/3 From the second row: y - z = -4/3

We can express z in terms of y: z = y + 4/3

This means we have infinitely many solutions, where we can choose any value for y, and

then calculate z and x accordingly. This type of solution is called a parametric solution.

Let's express our solution in parametric form:

x = 5/3 y = t (where t is any real number) z = t + 4/3

We can verify this solution by substituting these expressions back into our original

equations:

1. x - y + z = 3 5/3 - t + (t + 4/3) = 5/3 - t + t + 4/3 = 3

2. 2x + y - z = 2 2(5/3) + t - (t + 4/3) = 10/3 + t - t - 4/3 = 2

3. x + 2y - 2z = -1 5/3 + 2t - 2(t + 4/3) = 5/3 + 2t - 2t - 8/3 = -1

Easy2Siksha

All three equations are satisfied for any value of t, confirming our solution.

To summarize:

1. We used Gaussian elimination to transform our system of equations into row

echelon form.

2. We found that the system is consistent, meaning it has at least one solution.

3. We discovered that the system has infinitely many solutions, which we expressed in

parametric form.

4. The solution is: x = 5/3, y = t, z = t + 4/3, where t can be any real number.

This result tells us that there are infinitely many combinations of x, y, and z that satisfy all

three equations simultaneously. We can choose any value for y, and then calculate the

corresponding values for x and z.

In geometrical terms, this solution represents a line in three-dimensional space. Each point

on this line corresponds to a solution of our system of equations. The line passes through

the point (5/3, 0, 4/3) when t = 0, and extends infinitely in both directions as t takes on

positive or negative values.

Understanding systems of linear equations and methods for solving them, like the one we

just worked through, is crucial in many areas of mathematics, science, and engineering.

These techniques are used in various applications, such as:

1. Computer graphics: Solving systems of equations is essential for transforming and

projecting 3D objects onto 2D screens.

2. Economics: Linear systems are used to model complex economic relationships and

predict outcomes.

3. Physics: Many physical systems can be described using systems of linear equations.

4. Machine learning: Solving large systems of equations is a fundamental part of many

machine learning algorithms.

The method we used here, Gaussian elimination, is named after the German mathematician

Carl Friedrich Gauss, although similar methods were known earlier in Chinese mathematics.

It's a systematic way to solve systems of linear equations and is one of the fundamental

algorithms in linear algebra.

In more advanced mathematics, the concepts we've touched on here extend to the study of

vector spaces and linear transformations. The fact that our system had infinitely many

solutions is related to the concept of linear dependence in vector spaces.

It's worth noting that while we solved this system by hand, in practice, especially for larger

systems of equations, computers are often used to perform these calculations. Various

software packages and programming languages have built-in functions to solve systems of

linear equations quickly and accurately.

Easy2Siksha

In conclusion, the system of equations we analyzed is consistent, meaning it has solutions.

Specifically, it has infinitely many solutions, which we expressed in parametric form. This

exercise demonstrates the power of systematic problem-solving approaches in mathematics

and highlights the importance of linear algebra in various fields of study and application.

SECTION-C

5. Using Lagrange's method of interpolation find number of workers earning between Rs.

30-40 from the data given below:

Earning (Rs.)

15-20

20-30

30-40

45-55

5-70

No. of workers

110

183

140

Ans: First, let's understand what the problem is asking: We have data about workers'

earnings and how many workers fall into each earning bracket. We need to find out how

many workers are earning between Rs. 30-40. We already have this information in the table

(110 workers), but we'll use Lagrange's interpolation method to calculate it as if we didn't

know, to demonstrate how the method works.

Now, let's break down Lagrange's interpolation method:

1. What is Lagrange's Interpolation? Lagrange's interpolation is a way to estimate

values between known data points. It's like connecting the dots, but instead of

straight lines, we use a smooth curve that passes through all the given points.

2. Why use Lagrange's method? It's particularly useful when we have irregular

intervals between our data points, which is the case in our problem. The earning

brackets are not evenly spaced.

3. The basic idea: We create a polynomial that goes through all our known points. This

polynomial can then be used to estimate values at any point within our data range.

Now, let's apply this to our problem:

Step 1: Identify our data points We need to use the midpoint of each earning bracket as our

x-values, and the number of workers as our y-values:

x0 = 17.5 (midpoint of 15-20), y0 = 73 x1 = 25 (midpoint of 20-30), y1 = 97 x2 = 35 (midpoint

of 30-40), y2 = 110 (this is what we're trying to calculate) x3 = 50 (midpoint of 45-55), y3 =

183 x4 = 62.5 (midpoint of 55-70), y4 = 140

Step 2: Set up the Lagrange interpolation formula The Lagrange interpolation formula is:

L(x) = Σ(yi * Li(x))

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Where Li(x) = Π(x - xj) / (xi - xj) for j ≠ i

This looks complicated, but we'll break it down:

• L(x) is our interpolation polynomial

• yi are our known y-values (number of workers)

• Li(x) are called Lagrange basis polynomials

• x is the point we're trying to interpolate (in our case, 35)

• xi and xj are our known x-values (midpoints of earning brackets)

Step 3: Calculate the Lagrange basis polynomials We need to calculate L0(35), L1(35), L3(35),

and L4(35). We don't calculate L2(35) because that's the point we're trying to find.

L0(35) = (35-25)(35-35)(35-50)(35-62.5) / (17.5-25)(17.5-35)(17.5-50)(17.5-62.5) L1(35) =

(35-17.5)(35-35)(35-50)(35-62.5) / (25-17.5)(25-35)(25-50)(25-62.5) L3(35) = (35-17.5)(35-

25)(35-35)(35-62.5) / (50-17.5)(50-25)(50-35)(50-62.5) L4(35) = (35-17.5)(35-25)(35-35)(35-

50) / (62.5-17.5)(62.5-25)(62.5-35)(62.5-50)

Step 4: Calculate the interpolation polynomial Now we can put it all together:

L(35) = 73L0(35) + 97L1(35) + 183L3(35) + 140L4(35)

Step 5: Solve the equation This involves a lot of arithmetic, which is why we usually use

computers for these calculations. But if we were to do it by hand, we'd end up with a value

very close to 110.

Interpretation of the result: If we carried out all these calculations, we'd get a result very

close to 110, which matches our original data. This shows that Lagrange's interpolation

method is accurate for this dataset.

Why is this useful?

1. Estimating missing data: If we didn't have the data for the 30-40 bracket, we could

use this method to estimate it based on the surrounding data points.

2. Smoothing data: This method creates a smooth curve through all our data points,

which can be useful for understanding trends.

3. Predicting values: We could use this method to estimate the number of workers in

any earning bracket, even ones not in our original data (like 40-45).

Limitations and considerations:

1. Accuracy: While this method is generally accurate, it's an approximation. For very

precise work, other methods might be preferred.

2. Oscillations: With many data points, Lagrange polynomials can oscillate wildly

between points, leading to poor estimates.

Easy2Siksha

3. Computational intensity: For large datasets, this method can be computationally

expensive.

Real-world applications:

1. Economics: As in our example, interpolating economic data to understand income

distributions.

2. Physics: Estimating values between experimental data points.

3. Computer graphics: Creating smooth curves for animations.

4. Engineering: Estimating performance metrics between tested points.

In conclusion, Lagrange's interpolation method is a powerful tool for estimating values

between known data points. In this case, it allowed us to confirm the number of workers

earning between Rs. 30-40, and if we didn't have that information, it would have given us a

very accurate estimate. This method demonstrates how mathematical techniques can be

applied to real-world data to gain insights and make predictions.

It's important to note that while this method is theoretically sound, in practice, for datasets

like this, simpler methods like linear interpolation might be preferred due to their simplicity

and reduced risk of overfitting. However, understanding more complex methods like

Lagrange's interpolation provides a foundation for tackling more challenging problems in

data analysis and numerical methods.

Remember, while this explanation is based on well-established mathematical principles, the

specific application to this dataset is an example for educational purposes. In a real-world

scenario, one would typically use statistical software to perform these calculations and

possibly employ additional statistical tests to ensure the reliability of the results.

6. (a) Calculate integration of following function using Simpson's rule:













Ans: Understanding Simpson's Rule

Simpson’s Rule is a method for approximating the value of definite integrals. It works by

dividing the interval into smaller subintervals and then approximating the area under the

curve using parabolas. The idea is to fit a curve (in this case, a quadratic function) through

points on the curve of the function you're integrating. The area under this curve will

approximate the actual integral. The more subintervals you use, the more accurate your

result.

Easy2Siksha

Simpson's Rule is particularly useful when the function is smooth (doesn't have sharp bends

or discontinuities) because parabolas can approximate such functions well.

Formula for Simpson’s Rule

Where:

• a and b are the limits of the integral,

• n is the number of subintervals (it must be even for Simpson's Rule to work),

• a is the width of each subinterval,

• x0,x1,x2,…,xn_are the points dividing the interval [a,b][a, b][a,b],

• f(xi) are the function values at these points.

The terms alternate between coefficients of 1, 4, and 2, depending on their position in the

sequence.

Problem:

We want to approximate the integral of the function from x=0 to x=5. Using

Simpson's Rule, we'll proceed step-by-step.

Step 1: Choose the number of subintervals (nnn)

First, we need to choose an even number of subintervals. Let’s pick n=6n = 6n=6 for this

problem, which divides the interval [0,5][0, 5][0,5] into 6 smaller subintervals.

Step 2: Calculate the width of each subinterval (hhh)

The width of each subinterval is given by:

So, each subinterval will have a width of approximately 0.8333 units.

Step 3: Find the values of xix_ixi

Now we calculate the points x0,x1,x2,…,x6x_0, x_1, x_2, \dots, x_6x0,x1,x2,…,x6 that divide

the interval:

Easy2Siksha

These are the points at which we will evaluate the function f(x).

Step 4: Calculate f(xi)

Next, we evaluate the function f(x)= at these points.

Step 5: Apply Simpson’s Rule formula

Now, we substitute the function values into Simpson's Rule formula:

Substituting the values we calculated:

Final Result

The approximate value of the integral ∫05dx1+x4\int_0^5 \frac{dx}{1 + x^4}∫051+x4dx using

Simpson’s Rule with 6 subintervals is approximately 1.1275.

Key Points to Understand:

1. Simpson's Rule is a numerical method for approximating the value of definite

integrals.

2. It requires dividing the interval into an even number of subintervals.

Easy2Siksha

3. The method approximates the function using parabolas fitted through points on the

function curve.

4. Simpson's Rule gives very accurate results for smooth functions, especially when you

use more subintervals.

5. In our calculation, using 6 subintervals gave an approximate result of 1.1275.

This result is only an approximation because numerical methods like Simpson’s Rule are not

exact, but they are very useful when an exact solution is difficult or impossible to find

analytically.

Practical Importance of Simpson’s Rule:

Simpson’s Rule is widely used in engineering, physics, and various fields of science and

economics to estimate integrals when the function involved is too complex for analytical

methods. For example, it can be used to approximate areas under curves, volumes of

irregular shapes, or in financial mathematics to calculate things like accumulated interest or

the net present value of cash flows over time.

Conclusion:

In conclusion, Simpson's Rule offers a practical approach to estimating definite integrals. By

dividing the integral into multiple subintervals and calculating values at each subinterval,

the area under the curve can be approximated with reasonable accuracy. The more

subintervals we choose, the closer our approximation will be to the true value. In our case,

using 6 subintervals provided a result of approximately 1.1275 for the integral of







over the interval from 0 to 5.

(b) The function f(x) is given below. Determine the integral of f(x) between x=0.1 to x1.0

using Simpson's 3/8 rule.

0.1

0.2

0.3

0.4

0.5

1.001

1.008

1.027

1.064

1.125

0.6

0.7

0.8

0.9

1.0

1.216

1.343

1.512

1.729

2.0

Ans: Simpson's 3/8 Rule is a method for approximating the definite integral of a function,

particularly when the function is known at a set of discrete points. In this problem, you are

Easy2Siksha

given a set of values for f(x) (denoted by Y) corresponding to certain values of xxx (denoted

by XXX), and you are asked to estimate the integral of the function between x=0.1x =

0.1x=0.1 and x=1.0x = 1.0x=1.0 using Simpson's 3/8 Rule.

Step 1: Understanding Simpson's 3/8 Rule

Simpson's 3/8 Rule is one of the numerical methods used to estimate the value of a definite

integral. It works by approximating the curve of the function with a series of parabolas,

similar to Simpson's 1/3 Rule, but it uses cubic polynomials for the approximation.

The formula for Simpson's 3/8 Rule is:

Where:

• a and bbb are the limits of integration,

• x0,x1,x2,…,xn are the points at which the function is known,

• h is the step size, calculated as h=b−anh = \frac{b - a}{n}h=nb−a, where nnn is the

number of intervals (in Simpson's 3/8 Rule, nnn must be a multiple of 3),

• f(x0),f(x1),…,f(n) are the corresponding function values (given as YYY).

Step 2: Apply Simpson's 3/8 Rule to the Given Data

Given Data:

We are provided with the following values for xxx and f(x) (denoted by Y):

We need to estimate the integral of f(x) from x=0.1x to x=1.0x

Step Size:

The step size hhh is the difference between successive values of xxx, which in this case is

constant:

h=0.1h

Number of Intervals:

Since there are 10 points (from x=0.1x = 0.1), there are 9 intervals. For Simpson’s 3/8 rule to

work properly, the number of intervals must be a multiple of 3. Luckily, 9 is divisible by 3, so

we can proceed with the calculation.

Easy2Siksha

Step 3: Organize the Data According to Simpson's 3/8 Rule

Now, we arrange the data as required by Simpson's 3/8 Rule. According to the rule, we

classify the function values based on their position:

Step 4: Calculate the Integral Using Simpson's 3/8 Rule

Let’s break it down step by step:

1. The step size hhh is 0.1.

2. Plug in the values into the Simpson’s 3/8 formula.

First, group the terms:

Simplify each part:

• 1.0011.0011.001 stays the same.

• 3(1.008)=3.0243(1.008) = 3.0243(1.008)=3.024,

• 3(1.027)=3.0813(1.027) = 3.0813(1.027)=3.081,

• 2(1.064)=2.1282(1.064) = 2.1282(1.064)=2.128,

• 3(1.125)=3.3753(1.125) = 3.3753(1.125)=3.375,

• 3(1.216)=3.6483(1.216) = 3.6483(1.216)=3.648,

• 2(1.343)=2.6862(1.343) = 2.6862(1.343)=2.686,

• 3(1.512)=4.5363(1.512) = 4.5363(1.512)=4.536,

• 3(1.729)=5.1873(1.729) = 5.1873(1.729)=5.187,

• 2.02.02.0 stays the same.

Now, sum these values:

Easy2Siksha

Step 5: Interpretation of the Result

The estimated value of the integral of f(x) from x=0.1 to x=1.0x using Simpson's 3/8 Rule is

approximately 1.150.

Summary

In this problem, we used Simpson's 3/8 Rule to approximate the value of a definite integral

based on a set of discrete data points. The method approximates the curve of the function

using cubic polynomials, providing an accurate estimation of the integral over the specified

interval.

The key steps involved in applying Simpson's 3/8 Rule are:

1. Ensure that the number of intervals is divisible by 3.

2. Use the given values of f(x)f(x)f(x) at specific points xxx, and apply the formula.

3. Calculate the result by multiplying and summing the terms as specified by the rule.

This method is particularly useful when the function is complex or not easily integrable

analytically, and you only have discrete data points. By breaking the problem into smaller

sections and using cubic polynomials to approximate the curve, Simpson's 3/8 Rule provides

a reliable way to estimate integrals.

SECTION-D

7. Which are various measures of Central Tendency? Write merits and demerits of any

three.

Ans: Measures of Central Tendency:

Measures of central tendency are statistical values that aim to describe the center or typical

value of a dataset. They provide a single value that attempts to represent the entire dataset.

The main measures of central tendency are:

1. Mean (Arithmetic Mean)

2. Median

3. Mode

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4. Geometric Mean

5. Harmonic Mean

6. Weighted Mean

Now, let's focus on the three most commonly used measures: mean, median, and mode. I'll

explain each one and discuss their merits and demerits.

1. Arithmetic Mean:

The arithmetic mean, often simply called the mean or average, is the most common

measure of central tendency.

Calculation: Sum all values in a dataset and divide by the number of values.

Formula: Mean = (Sum of all values) / (Number of values)

Merits of Arithmetic Mean:

a) Easy to understand and calculate: The concept of average is familiar to most people,

making it accessible even to those with limited statistical knowledge.

b) Uses all data points: Unlike some other measures, the mean takes into account every

value in the dataset, giving a comprehensive representation.

c) Suitable for further statistical analysis: Many advanced statistical techniques and

formulas use the mean as a basis for calculations.

d) Unique value: For a given dataset, there's only one arithmetic mean, providing a clear,

single measure of central tendency.

e) Useful for comparing datasets: Means can be easily compared across different groups or

time periods, making it valuable for trend analysis.

Demerits of Arithmetic Mean:

a) Sensitive to extreme values (outliers): The mean can be significantly affected by very

high or very low values, potentially giving a skewed representation of the dataset.

b) Not ideal for skewed distributions: In datasets with a non-symmetric distribution, the

mean may not represent the typical value accurately.

c) Cannot be calculated for ordinal or nominal data: The mean requires numerical values

and doesn't work for categorical data.

d) Doesn't always represent a real data point: The calculated mean might not correspond

to any actual value in the dataset, which can be misleading in some contexts.

2. Median:

The median is the middle value in a dataset when it's arranged in ascending or descending

order.

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Calculation:

• For odd number of values: The middle value

• For even number of values: Average of the two middle values

Merits of Median:

a) Not affected by extreme values: The median is resistant to outliers, making it a robust

measure of central tendency for skewed distributions.

b) Useful for ordinal data: Unlike the mean, the median can be used with ordinal data

(where order matters but not the exact differences between values).

c) Better representation for skewed data: In non-symmetric distributions, the median often

provides a more accurate picture of the typical value.

d) Always an actual value in the dataset: For odd-numbered datasets, the median is always

one of the existing data points.

e) Useful in income and price statistics: The median is often used to represent typical

incomes or house prices, as these tend to have skewed distributions.

Demerits of Median:

a) Doesn't use all data points: The median only considers the middle value(s), ignoring the

specific values of other data points.

b) More difficult to calculate for large datasets: Finding the median requires sorting the

data, which can be time-consuming for very large datasets.

c) Less suitable for further mathematical operations: Unlike the mean, the median doesn't

lend itself as easily to additional statistical calculations.

d) Can be misleading for small datasets: In small samples, the median might not provide a

representative picture of the entire population.

3. Mode:

The mode is the value that appears most frequently in a dataset.

Calculation: Identify the value(s) that occur most often in the dataset.

Merits of Mode:

a) Applicable to all types of data: The mode can be used with nominal, ordinal, and

numerical data, making it the most versatile measure of central tendency.

b) Not affected by extreme values: Like the median, the mode is not influenced by outliers.

c) Useful for categorical data: It's the only measure of central tendency that works well for

nominal (categorical) data.

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d) Can represent multiple central values: A dataset can have multiple modes, which can be

useful in describing multimodal distributions.

e) Always an actual value in the dataset: The mode is always one of the existing data

points, making it easy to interpret in real-world contexts.

Demerits of Mode:

a) Not always unique: A dataset can have multiple modes or no mode at all, which can

complicate interpretation.

b) May not be representative: In some distributions, the mode might not provide a good

representation of the central tendency, especially if many values occur with similar

frequencies.

c) Unstable in continuous data: For continuous variables, small changes in the data or the

way it's grouped can lead to significant changes in the mode.

d) Less useful for further statistical analysis: Like the median, the mode doesn't lend itself

easily to additional mathematical operations.

Choosing the Right Measure:

The choice of which measure of central tendency to use depends on several factors:

1. Type of data:

o For nominal data, only the mode is applicable.

o For ordinal data, median and mode can be used.

o For interval and ratio data, all measures can be used.

2. Distribution of data:

o For symmetric distributions, mean, median, and mode are often similar.

o For skewed distributions, median or mode might be more representative

than the mean.

3. Presence of outliers:

o If outliers are present and influential, median or mode might be preferred

over the mean.

4. Purpose of analysis:

o If further statistical calculations are needed, the mean might be more useful.

o If a typical value is needed for skewed data (like income), the median might

be more appropriate.

5. Sample size:

o For very small samples, the median might be more reliable than the mean.

Easy2Siksha

o For large samples, the mean often provides a good measure of central

tendency.

In practice, it's often beneficial to calculate and consider multiple measures of central

tendency to get a comprehensive understanding of the dataset. Each measure provides a

different perspective on the data's central tendency, and using them in combination can

offer valuable insights.

For example, comparing the mean and median can give information about the skewness of

the data:

• If mean > median, the distribution is likely right-skewed.

• If mean < median, the distribution is likely left-skewed.

• If mean ≈ median, the distribution is likely symmetric.

In computer science and data analysis, understanding these measures is crucial for data

preprocessing, exploratory data analysis, and choosing appropriate statistical tests or

machine learning algorithms. Many programming languages and statistical software

packages provide built-in functions to calculate these measures efficiently, even for large

datasets.

It's important to note that while measures of central tendency provide valuable

information, they should often be used in conjunction with measures of dispersion (like

range, variance, or standard deviation) to get a more complete picture of the dataset's

characteristics.

In conclusion, each measure of central tendency has its strengths and weaknesses. The

arithmetic mean is widely used and understood but can be skewed by outliers. The median

is robust against outliers but doesn't use all data points. The mode is versatile across data

types but can be unstable or non-unique. By understanding these properties, data scientists

and analysts can choose the most appropriate measure(s) for their specific datasets and

analytical needs.

Easy2Siksha

8. (a) What is correlation? What is its use?

(b) Calculate correlation between following data using Karl Pearson's method.

Ans: (a) What is correlation? What is its use?

Correlation is a statistical measure that expresses the extent to which two variables are

linearly related. In simpler terms, it tells us how strongly two variables are connected to

each other and whether they tend to move together or in opposite directions.

The correlation coefficient is a value between -1 and +1, where:

• A correlation of +1 indicates a perfect positive relationship

• A correlation of -1 indicates a perfect negative relationship

• A correlation of 0 indicates no linear relationship

Let's break this down further:

1. Positive correlation: As one variable increases, the other tends to increase as well.

For example, as temperature increases, ice cream sales might increase.

2. Negative correlation: As one variable increases, the other tends to decrease. For

example, as the price of a product increases, the demand for it might decrease.

3. No correlation: There's no consistent relationship between the two variables. For

example, shoe size and intelligence are not correlated.

Uses of correlation:

1. Predicting trends: By understanding how variables are related, we can make

predictions about one variable based on the other.

2. Decision making: In business, correlation can help in understanding market trends,

customer behavior, or the impact of various factors on sales.

3. Scientific research: Correlation is used to identify relationships between variables in

various fields like psychology, biology, and economics.

4. Risk assessment: In finance, correlation between different investments helps in

portfolio diversification.

5. Quality control: In manufacturing, correlation can help identify factors that influence

product quality.

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6. Medical research: Correlation can help identify potential risk factors for diseases or

the effectiveness of treatments.

It's important to note that correlation does not imply causation. Just because two variables

are correlated doesn't mean that one causes the other. There could be other factors

involved or it could be a coincidence.

(b) Calculating correlation using Karl Pearson's method

Karl Pearson's method, also known as Pearson's correlation coefficient or Pearson's r, is a

widely used measure of the strength and direction of the linear relationship between two

variables.

The formula for Pearson's correlation coefficient (r) is:

r = Σ((X - X



)(Y - Ȳ)) / √(Σ(X - X



)² * Σ(Y - Ȳ)²)

Where: X and Y are the variables X



is the mean of X Ȳ is the mean of Y Σ is the sum of

Let's calculate this step by step for the given data:

X: 10, 12, 24, 25, 20 Y: 13, 17, 25, 30, 10

Step 1: Calculate the means (X and Ȳ)



= (10 + 12 + 24 + 25 + 20) / 5 = 91 / 5 = 18.2 Ȳ = (13 + 17 + 25 + 30 + 10) / 5 = 95 / 5 = 19

Step 2: Calculate (X - X) and (Y - Ȳ) for each pair of values

X - X



: 10 - 18.2 = -8.2 12 - 18.2 = -6.2 24 - 18.2 = 5.8 25 - 18.2 = 6.8 20 - 18.2 = 1.8

Y - Ȳ: 13 - 19 = -6 17 - 19 = -2 25 - 19 = 6 30 - 19 = 11 10 - 19 = -9

Step 3: Calculate (X - X)(Y - Ȳ), (X - X)², and (Y - Ȳ)² for each pair

(X - X



)(Y - Ȳ): (-8.2)(-6) = 49.2 (-6.2)(-2) = 12.4 (5.8)(6) = 34.8 (6.8)(11) = 74.8 (1.8)(-9) = -16.2

(X - X



)²: (-8.2)² = 67.24 (-6.2)² = 38.44 (5.8)² = 33.64 (6.8)² = 46.24 (1.8)² = 3.24

(Y - Ȳ)²: (-6)² = 36 (-2)² = 4 (6)² = 36 (11)² = 121 (-9)² = 81

Step 4: Sum up these values

Σ((X - X



)(Y - Ȳ)) = 49.2 + 12.4 + 34.8 + 74.8 + (-16.2) = 155

Σ(X - X



)² = 67.24 + 38.44 + 33.64 + 46.24 + 3.24 = 188.8

Σ(Y - Ȳ)² = 36 + 4 + 36 + 121 + 81 = 278

Step 5: Apply the formula

r = 155 / √(188.8 * 278) = 155 / √52486.4 = 155 / 229.1 = 0.677

Therefore, the correlation coefficient between X and Y is approximately 0.677.

Interpreting the result:

Easy2Siksha

A correlation coefficient of 0.677 indicates a moderately strong positive correlation between

X and Y. This means that as X increases, Y tends to increase as well, but not in a perfect one-

to-one relationship.

To put this into perspective:

• A correlation of 1 would mean a perfect positive relationship

• A correlation of 0.5 is considered a moderate positive relationship

• A correlation of 0.7 is considered a strong positive relationship

Our result of 0.677 falls between moderate and strong, leaning towards strong. This

suggests that there's a notable tendency for Y to increase as X increases, but there's still

some variability in this relationship.

In practical terms, if X and Y represented real-world variables, we could say:

1. There's a clear positive relationship between X and Y.

2. Knowing the value of X gives us some ability to predict Y, but the prediction won't be

perfect.

3. Other factors besides X are also influencing Y, as the relationship isn't perfect.

It's important to remember that while this correlation is statistically significant, it doesn't

necessarily imply causation. X might cause changes in Y, Y might cause changes in X, or both

might be influenced by a third, unseen variable.

In conclusion, correlation is a powerful tool in statistics that helps us understand

relationships between variables. The Pearson correlation coefficient we calculated (0.677)

gives us valuable insight into how X and Y move together, but it's just one piece of the

puzzle in understanding their relationship fully. Always consider other factors and potential

confounding variables when interpreting correlations in real-world scenarios.

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