**Random Sampling Practice Problems**

**Practice 1**

Here's a little theoretical problem that you might find fun. Think of it as a "brain buster." Suppose a sample set is infinite. Consider the set of all rational numbers between 0 and 1 inclusive. One-by-one sampling can be done forever from such a set, whether the sampled elements are replaced or not. If sampling is done at random and without replacement, a given number will never turn up twice during the course of the experiment, no matter how long the experiment continues. But suppose sampling is done with replacement. Can any given number turn up twice in that case?

**Solution 1**

If you calculate the probability of any given number turning up after it has been identified, you find that this probability is equal to "one divided by infinity." Presumably this means that the probability is 0, because it is reasonable to suppose that 1/∞ = 0. (The sideways 8 represents infinity.) This implies that any specific number can never turn up again once it has been selected from an infinite set, even if you replace it after it has been picked.

But wait! The above argument also implies that any specific number between 0 and 1 can never turn up *even once* in a random selection process! Suppose you pick a number between 0 and 1 out of the "clear blue." The probability of this number coming up in your experiment is not equal to 0, because you picked it. This isn't a random number, because your mind is, of course, biased. But it is a specific number. Now imagine a machine that generates truly random numbers between 0 and 1. You push the button, and out comes a number. What is the probability that it's the specific number you have picked? The probability is 0, because the number you happened to choose is only one of an infinite number of numbers between 0 and 1. The machine might choose any one of an infinitude of numbers, and the chance that it's the one you've thought of is therefore 1/∞. Therefore, the random-number machine cannot output any specific number.

Can you resolve this apparent paradox?

**Practice 2**

Interpolate the reading of the ammeter in Fig. 5-3 to the nearest tenth of an ampere, and to the nearest thousandth of an ampere.

**Fig. 5-3. **Visual interpolation of an analog meter reading is always subject to error. Illustration for Practice 2 through 4.

**Solution 2**

Note that the divisions get closer together as the value increases (toward the right). Based on this, and using some common sense, a good estimate of the reading is 3.6 amperes, accurate to the nearest tenth of an ampere.

It is impossible to visually interpolate this ammeter reading to the nearest thousandth of an ampere. We would need a better ammeter in order to do that.

**Practice 3**

How accurately can readings of the ammeter shown in Fig. 5-3 be visually interpolated?

**Solution 3**

This is a matter of opinion, but it's a good bet that experienced meter-readers would say that interpolation to the nearest 0.1 ampere is the limit near the upper end of the scale, and interpolation to the nearest 0.05 ampere is the limit near the lower end. Therefore, we can say that the visual interpolation error is about ±0.05 ampere (plus-or-minus 1/20 of an ampere) near the upper end of the scale, and ±0.025 ampere (plus-or-minus 1/40 of an ampere) near the lower end.

**Practice 4**

Suppose that, in addition to the visual interpolation error, the manufacturer of the ammeter shown in Fig. 5-3 tells us that we can expect a hardware error of up to ±10% of full scale. If that is the case, what is the best accuracy we can expect at the upper end of the scale?

**Solution 4**

A full-scale reading on this ammeter is 10 amperes. So at the upper end of the scale, this instrument could be off by as much as ± (10% × 10), or ±1 ampere. Adding this to the visual interpolation error of 0.05 ampere, we get a total measurement error of up to ±1.05 ampere at the upper end of the scale.

Practice problems for these concepts can be found at:

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