## Rounding Error

#### Problem

A number is rounded to two decimal places and this answer is rounded to one decimal place. If the final number is 0.2 then what is the chance that the original number lies between 0.15 and 0.25?

#### Solution

When rounding to $k$ decimal places (d.p.) we truncate the number at the $k^{th}$ decimal position, but if the next digit is 5 or more then we increase the last digit by 1; this may have a knock-on effect. Consider the following examples of rounding to 2 d.p.:

3.61483 3.61

7.7853301 7.79

0.5982 0.60

Suppose we were told that 0.2 were the result of rounding to one decimal place. The smallest number that would be rounded up to 0.2 is 0.15. However, there is no actual value we can assign to the greatest value that would round down to 0.2. For example, 0.25 would round up to 0.3 and although 0.249 would round down to 0.2, what about 0.2499, or 0.24999, or ... ?

Under these circumstances we define bounds (or limits) in the following way:

- The lower bound is the least value that rounds up to the given number.
- The upper bound is the least value that fails to round down to the given number.

So if 0.2 is the result of rounding to 1 d.p. then the lower and upper bounds are 0.15 and 0.25 respectively.

Working backwards, the lower bound of 0.2 is 0.15 and the lower bound of 0.15 is 0.145. We can easily check this: 0.145 = 0.15 (2 d.p.) and then 0.15 = 0.2 (1 d.p.).

However, we need to tread very carefully when working backwards with the upper bound. Although it is true to say that the upper bound of 0.2 is 0.25, if the original number were rounded to 0.25 (to 2 d.p.), then next process of rounding to 1 d.p. would make it 0.3, not 0.2.

In fact, it is necessary that the result of rounding to 2 d.p. takes it to 0.24; and the upper bound of 0.24 is 0.245.

Hence we are certain that the original number, $x$, lies somewhere in the interval $0.145 \le x \lt 0.245$.

If we are considering if the number lies between 0.15 and 0.25 then it would have to lie in the interval $0.15 \le x \lt 0.245$.

$$\begin{align}\therefore P(0.15 \le x \lt 0.245 \mid 0.145 \le x \lt 0.245) &= \dfrac{0.245 - 0.15}{0.245 - 0.145}\\\\&= \dfrac{0.095}{0.1}\\\\&= 95\%\end{align}$$If instead of rounding to 2 d.p. then to 1 d.p. a number is just rounded to 1 d.p., what is the chance that both methods give the same answer?