02.Time Series Analysis

Basic definition related to Bounds

Definition

Lower Bounds

The lower bound (LB) refers to the lowest number that can be rounded to get an estimated value.

Upper Bounds

The upper bound (UB) refers to the highest number that can be rounded to get an estimated value.

Error interval

Error intervals show the range of numbers that are within the limits of accuracy. They are written in the form of inequalities.

• Notes

  Info

  The lower and upper bounds can also be called the limits of accuracy.

Degree of accuracy (DA)

The degree of accuracy is the measure to which a value is rounded.

Info

You may come across questions involving formulas, and you will have to work with multiplication, division, addition, and subtraction.

$LowerBounds = value - \frac{DA}{2}$
$UpperBounds = value + \frac{DA}{2}$

How to get Bounds

For addition

This usually happens when we have a value that undergoes an increase. We then have an original value and its range of increase.

When you have a question involving addition, do the following:

1. Find the upper and lower bounds of the original value, $UB_{value}$, and of its range of increase, $UB_{range}$.
2. Use the following formulas to find the upper and lower bounds of the answer.
   $UB_{new} = UB_{value} + UB_{range}$
   $LB_{new} = LB_{value} + LB_{range}$
3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

For Subtraction

This usually happens when we have a value that undergoes a decrease. We then have an original value and its range of decrease.
When you have a question involving subtraction, do the following.

1. Find the upper and lower bounds of the original value, $UB_{value}$, and of its range of increase, $UB_{range}$.
2. Use the following formulas to find the upper and lower bounds of the answer.
   $UB_{new} = UB_{value} - UB_{range}$
   $LB_{new} = LB_{value} - LB_{range}$
3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

For Multiplication.

This usually happens when we have quantities that involve the multiplication of other quantities, such as areas, volumes, and forces.
When you have a question involving multiplication, do the following.

1. Find the upper and lower bounds of the numbers involved. Let them be quantity 1, q1, and quantity 2, q2.
2. Use the following formulas to find the upper and lower bounds of the answer.
   $UB_{new} = UB_{q1}*UB_{q2}$
   $LB_{new} = LB_{q1}*LB_{q2}$
3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

For Division

Similarly to the multiplication, this usually happens when we have a quantity that involves the division of other quantities, such as velocity, and density.

When you have a question involving division, do the following.

1. Find the upper and lower bounds of the numbers involved. Let's denote them quantity 1, q1, and quantity 2, q2.
2. Use the following formulas to find the upper and lower bounds of the answer.
   $UB_{new} = \frac{UB_{q1}}{LB_{q2}}$
   $LB_{new} = \frac{LB_{q1}}{UB_{q2}}$
3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

Examples

E1

Find the upper and lower bound of the number 40 rounded to the nearest 10.

Solution
There are lots of values that could be rounded to 40 to the nearest 10. It can be 37, 39, 2.5, 43, 44.9, 44.9999, and so on.

But the lowest number which will be the lower bound is 35 and the highest number is 44.444, so we will say the upper bound is 44.

Let's call the number that we start with, 40, x. The error interval will be:

$$35\leq x \lt 45$$

This means x can be equal to or more than 35, but less than 44.

E2

The length of an object y is 250 cm long, rounded to the nearest 10 cm. What is the error interval for y?

Solution

To know the error interval, you have to first find the upper and lower bound. Let's use the steps we mentioned earlier to get this.

Step 1: First, we have to know the degree of accuracy, DA. From the question, the degree of accuracy is DA = 10 cm.
Step 2: The next step is to divide it by 2.

$$\frac{DA}{2}=\frac{10}{2} = 5$$

Step 3: We will now subtract and add 5 to 250 to get the lower and upper bound.

$$UpperBound = value + \frac{D_a}{2}=250 + 5 = 255 \\ LowerBound = value + \frac{D_a}{2}=250 - 5 = 245$$

The error interval will be:

$$245 \leq y \lt 255$$

This means that the length of the object can be equal to or more than 245 cm, but less than 255 cm.

E3: addition

The length of a rope x is 33.7 cm. The length is to be increased by 15.5 cm. Considering the bounds, what will be the new length of the rope?

Solution.
This is a case of addition. So, following the steps for addition above, the first thing is to find the upper and lower bounds for the values involved.

Step 1: Let's start with the original length of the rope.

The lowest number that can be rounded to 33.7 is 33.65, meaning that 33.65 is the lower bound, $LB_{value}$.

The highest number is 33.74, but we will use 33.75 which can be rounded down to 33.7, $UB_{value}$.
So, we can write the error interval as: $33.65 \leq x \lt 33.75$

We will do the same for 15.5 cm, let's denote it $y$.

The lowest number that can be rounded to 15.5 is 15.45 meaning that 15.45 is the lower bound, $LB_{range}$.

The highest number is 15.54, but we will use 15.55 which can be rounded down to 15.5, $UB_{range}$.
So, we can write the error interval as: $15.45 \leq y \lt 15.55$

Step 2: We will use the formulas for finding upper and lower bounds for addition.

$$UB_{new} = UB_{value} + UB_{range} = 33.75 + 15.55 = 49.3cm$$

The lower bound is:

$$LB_{new} = LB_{value} + LB_{range} = 33.65 + 15.45 = 49.1cm$$

Step 3: We now have to decide what the new length will be using the upper and lower bound we just calculated.

The question we should be asking ourselves is to what degree of accuracy does the upper and lower bound round to the same number? That will be the new length.

Well, we have 49.3 and 49.1 and they both round to 49 at 1 decimal place. Therefore, the new length is 49 cm.

E4: multiplication

The length L of a rectangle is 5.74 cm and the breadth B is 3.3 cm. What is the upper bound of the area of the rectangle to 2 decimal places?

Solution

Step 1: First thing is to get the error interval for the length and breadth of the rectangle.

The lowest number that can be rounded to the length of 5.74 is 5.735 meaning that 5.735 is the lower bound, $LB_{value}$.

The highest number is 5.744, but we will use 5.745 which can be rounded down to 5.74, $UB_{value}$.
So, we can write the error interval as: $5.735 \leq y \lt 5.745$

The lowest number that can be rounded to the breadth of 3.3 is 3.25 meaning that 3.25 is the lower bound.

The highest number is 3.34, but we will use 3.35, so we can write the error interval as:
$3.25 \leq y \lt 3.35$

The area of a rectangle is: $Length * Breadth$

Step 2: So to get the upper bound, we will use the upper bound formula for multiplication.
$UB_{new} = UB_{value}*UB_{range} = 5.745*3.35 = 19.24575$

Step 3: The question says to get the answer in 2 decimal places. Therefore, the upper bound is: $UB_{new} = 19.25cm$

Example 5: multiplication

A man runs 14.8 km in 4.25 hrs. Find the upper and lower bounds of the man's speed. Give your answer in 2 decimal places.

Solution
We are asked to find the speed, and the formula for finding speed is:

$$Speed = \frac{Distance}{Time} = \frac{d}{t}$$

Step 1: We will first find the upper and lower bounds of the numbers involved.

The distance is 14.8 and the lowest number that can be rounded to 14.8 is 14.75 meaning that 14.75 is the lower bound, LBd.

The highest number is 14.84, but we will use 14.85 which can be rounded down to 14.8, UBd.

So, we can write the error interval as: $14.75 \leq d \lt 14.85$

The speed is 4.25 and the lowest number that can be rounded to 4.25 is 4.245 meaning that 4.245 is the lower bound, $LB_t$.

The highest number is 4.254, but we will use 4.255 (which can be rounded down to 4.25), UBt, so we can write the error interval as: $4.245 \leq d \lt 4.255$

Step 2: We are dealing with division here. So, we will use the division formula for calculating the upper and lower bound.

$$UB_{new} = \frac{UB_{d}}{LB_t} = \frac{14.85}{4.245} = 3.4982 \approx 3.50$$

The lower bound of the man's speed is:

$$LB_{new} = \frac{LB_{d}}{UB_t} = \frac{14.75}{4.255} = 0.4665 \approx 4.7$$

Step 3: The answers for the upper and lower bound are approximated because we are to give our answer in 2 decimal places.

Therefore, the upper and lower bound for the man's speed are 3.50 km/hr and 0.47 km/hr respectively.

E6

The height of a door is 93 cm to the nearest centimetre. Find the upper and lower bounds of the height.

Solution

The first step is to determine the degree of accuracy. The degree of accuracy is to the nearest 1 cm.

Knowing that the next step is to divide by 2. $\frac{1}{2} = 0.5$

To find the upper and lower bound, we will add and subtract 0,5 from 93 cm.

The Upper bound is: $UB = 93 + 0.5 = 93.5cm$
The Lower bound is: $LB = 93 - 0.5 = 92.5cm$

References

[1] Lower and Upper Bounds
[2] Lower and Upper Bound Theory
[3] Upper And Lower Bounds
[4] Upper & Lower Bounds | Number | Maths | FuseSchool
[5] What is a z-score? What is a p-value?
[6] Lower and Upper Bound Theory