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Limits of Accuracy

- Calculus
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Suppose we measure the length of a string using a ruler and say that it is 8.2 cm. Well, it may not be exactly 8.2 cm… it may be 8.23 cm, but since our ruler is only capable of measuring things to the nearest **millimetre**, 8.2 cm is as good as we are going to get. In this case, the **nearest** **millimetre**, or 0.1 cm, is known as the **limit of accurac**y. In this article, we will be learning all about what limits of accuracy are, and how we can learn to deal with them in the examples. Let's start by defining what we actually mean by a limit of accuracy.

The limit of accuracy of a measurement is the range of possible values a measurement can actually be when a measurement is given.

- The length of a room is 10 metres to the nearest metre. In this case, the degree of accuracy is
**one****metre**. - Connie weighs 65 kg to the nearest 5 kg. In this case, the degree of accuracy is
**5 kg.** - The length of a piece of string is 10.2 cm to the nearest 0.1 cm. In this case, the degree of accuracy is
**0.1 cm.** - Sebastian earned £110 yesterday to the nearest £10. In this case, the degree of accuracy is
**£10**

We could say that we are going to meet up with our friends at 10 am. But, do we actually mean precisely 10 am? What we really mean could be 10 am plus or minus ten minutes or so. In this case, the lower limit is 9:50 am, and the upper limit is 10:10 am. Any times outside that range are considered early or late. In the below section, we will consider upper and lower bounds in more detail.

To determine limits of accuracy, we need to work out **upper** and **lower** **bounds**. These can be found by using a formula. Let's first define what we mean by upper and lower bounds.

The **lower bound** of measurement is the smallest possible value it could be when a rounded measurement is given. Similarly, the **upper bound** is the largest possible value it could be.

To work out the upper bound, we simply **half** the degree of accuracy and **add** it to the stated measurement. To work out the lower bound, we **half** the degree of accuracy and **subtract** it from the stated measurement.

We will now look at some examples involving finding upper and lower bounds.

**A bag of oranges weighs 3 kg to the nearest kilogram. What are the upper and lower bounds for the weights of the oranges? **

**Solution:**

Suppose the bag of oranges weighed 2.8 kg. Well, if we rounded the weight to the nearest kilogram, we would say that the oranges weighed 3 kg. Thus, the actual weight of the oranges could be 2.8 kg. Similarly, the weight could be 2.6 kg or 2.55 kg.

The question is, what is the smallest weight the oranges could be that would get rounded up to 3 kg? Here, the degree of accuracy is 1 kg, so we half this to get 0.5 kg and subtract this from 3 kg to get 2.5 kg. Thus, **the lower bound is 2.5 kg. **

The upper bound is a little more confusing. If we take half of the degree of accuracy and add it to 3 kg, we get 3.5 kg, but surely that would round up to 4kg...

What is the largest possible value that would round down to 3 kg? The oranges could weigh 3.4 kg because that would round to 3 kg. Similarly, the could oranges weigh 3.49 kg, because that also would round down to 3 kg. If we said that the oranges weighed 3.4999 kg, that would round down to 3 kg, but if we said the oranges weighed 3.5 kg, that would round up to 4 kg.

The answer is that there is no largest value. The largest value would be 3.49 recurring but it is not possible for a bag of oranges to weigh 3.49 recurring kilograms. Therefore, despite the fact that 3.5 kg rounds up to 4 kg, we would say **3.5 kg is the upper bound** because there is no single largest value less than 3.5 kg.

**Therefore, in this case, the lower bound is 2.5 kg and the upper bound is 3.5 kg. **

Here we will look at a typical GCSE exam style question for limits of accuracy.

**The mass of the average apple is 175 g to the nearest gram. What are the least and greatest possible masses for the apple?**

**Solution:**

Here, the degree of accuracy is 1 gram. If we half 1 gram, we get 0.5 grams. To get the lower bound, we subtract 0.5g from 175 g to get 174.5 g. To get the upper bound we add 0.5 g to 175 g to get 175.5 g.

If we defined the mass of the apple to be m, we could say. This is called the **error interval**. We use the < inequality for the upper bound to say that the mass of the apple can't actually equal 175.5 g, but it can be anything less than 175.5 g.

An **error interval** is the range of possible values that a measurement can be. It can be formed by expressing the upper and lower bound of measurement in an inequality.

**A car is travelling at a speed of 70 mph, to the nearest 10 mph. What are the maximum and minimum possible speeds of the car?**

**Solution:**

Here, the degree of accuracy is 10 mph. If we half 10 mph we get 5 mph. Adding 5 mph onto 70 mph, we get the upper bound of 75 mph. Subtracting 5 mph from 70 mph, we get the lower bound of 65 mph.

If we defined the speed of the car to be x mph, we could say that the range of possible speeds of the car could be expressed as.

These are GCSE-style questions where limits of accuracy are used but they involve other topics.

**Matthew measured the length of the height and base of a triangle. **

**He measured the height to be 16 cm to the nearest centimetre and the base to be 20 cm to the nearest 10 centimetres. Work out the maximum possible area for the triangle. **

**Solution:**

Recall that the area of a triangle is given by.

The maximum possible area is obtained by using the maximum possible values for both the base and the height.

The upper bound for the height is 16.5 cm, and the upper bound for the base is 25 cm. Thus, the maximum possible area is .

**Tony went on a 40-mile bike ride to the nearest 10 miles. He travelled at a constant speed of 15 mph to the nearest mph. What is the minimum time he completed his bike ride in? **

**Solution:**

Recall that. Rearranging this, we obtain that,. In order to get the smallest possible time, we need to divide the smallest possible distance by the largest possible speed. The minimum distance is 35 miles since 40 has been rounded to the nearest 10. Similarly, the maximum speed is 15.5 mph. Thus, the minimum time ishours.

**Below is triangle ABC. **

**Angle ABC is a right angle. **

**BC = 10 cm to the nearest centimetre. Angle ACB is 34 degrees to the nearest degree. **

**Work out the upper and lower bounds for the length of AB**

**Solution:**

Here we can use trigonometry. If you are unsure about trigonometry, maybe review this before answering this question. However, if you feel good about trigonometry, keep reading!

We have the adjacent side, and we are seeking to find the opposite side. Thus, we use the trigonometric ratio tan.

Recall thatwhere O is the opposite side, A is the adjacent side andis the angle. We are trying to find the opposite side, so we can say that.

For the lower bound, we need to find the value of A andthat give the smallest possible value for the opposite side. In this case, it is the lower bound for A andthat do this. The lower bound for A is 9.5 cm and the lower bound foris. Thus, the lower bound for the opposite side is.

Similarly, for the upper bound, we need to find the value of A andthat give the largest possible value. The upper bound for A is 10.5 cm and the upper bound for the angle is. Thus, the upper bound for the opposite side is.

**Thus, the lower bound for the side AB is 6.29 cm and the upper bound is 7.22 cm. **

- The
**limit of accuracy**of a measurement is the range of possible values a measurement can actually be when a measurement is given. - The
**lower bound**of measurement is the smallest possible value it could be when a rounded measurement is given. Similarly, the**upper bound**is the largest possible value it could be. - The
**error interval**is the range of possible values possible when a measurement is started. It is defined using inequalities.

They tell us that a value has been rounded and so the true value can be a range of possible values.

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