Step 8: Discrete Random Variables

What a random variable is

Think about a factory where about 3% of the units coming off the line are faulty. Nobody can say which unit will be the bad one. But the manager can still plan: she can say how many faulty units to expect in a box of 50, and how likely it is that a box has none at all. She is not predicting single units. She is modelling the pattern.

To do that, we attach numbers to chance outcomes. “The number of faulty units in a box” is a number that depends on luck. That is a random variable. We write it with a capital letter, such as \( X \), and its particular values with a small letter, such as \( x \).

There are two kinds, and telling them apart is the first skill of this step.

• A discrete random variable takes values you can list, usually counts: emails received today, customers arriving, heads in three coin flips. Between 2 and 3 there is nothing.
• A continuous random variable can take any value in a range, usually measurements: the time until your next email, a person’s height, the weight of a bag of flour. Between 2 and 3 there are endless values.

The quick test is simple: if you count it, it is discrete; if you measure it, it is continuous. This step covers discrete variables. Step 8 deals with the continuous case.

Probability distributions and expected value

A probability distribution for a discrete variable is a table of every possible value next to its probability. It must pass two checks: every probability is between 0 and 1, and all the probabilities add up to exactly 1. Always check the total first.

Suppose a shop sells a luxury item. On any day it sells 0, 1, 2, 3 or 4 of them, with probabilities 0.10, 0.25, 0.30, 0.25 and 0.10. These add to 1.00, so the distribution is valid. The chance of selling at least 2 is \( 0.30 + 0.25 + 0.10 = 0.65 \).

How many does the shop sell on an average day? Not a plain average of 0 to 4, because some values are more likely than others. Instead you weight each value by its probability.

The variance follows the same idea as the variance of data in Step 4: take each squared distance from the mean, but weight it by probability instead of dividing by \( n \). The standard deviation brings you back to the original units.

Expected value is also how you judge a bet or an investment. Enter losses as negative numbers. A carnival game costs \( 3 \) dollars and pays \( 10 \) dollars with probability 0.2. Your net result is \( +7 \) with probability 0.2 and \( -3 \) with probability 0.8, so \( E(\text{profit}) = 0.2(7) + 0.8(-3) = 1.4 – 2.4 = -1 \). You lose a dollar per game on average. That is how carnivals stay in business.

Transforming and combining random variables

Real questions often build on a basic variable. The shop earns 150 dollars per sale and pays 200 dollars a day in fixed costs, so its profit is \( P = 150X – 200 \). Multiplying stretches both the centre and the spread. Adding or subtracting a constant slides every outcome by the same amount, so it moves the centre but leaves the spread alone.

So expected profit is \( 150(2.0) – 200 = 100 \) dollars a day, and its standard deviation is \( 150 \times 1.14 \approx 171 \) dollars. The spread is bigger than the average profit, which the mean alone would hide.

When you combine two variables, means always add. Variances add too, but only if the variables are independent, and they add even when you subtract one variable from the other.

For example, assembly takes a mean of 12 minutes with \( \sigma = 3 \) and inspection a mean of 5 minutes with \( \sigma = 4 \), independently. The total has mean \( 12 + 5 = 17 \) minutes and variance \( 9 + 16 = 25 \), so its standard deviation is \( \sqrt{25} = 5 \) minutes, not 7.

The binomial distribution

Some patterns turn up so often that they have names. The most important discrete one is the binomial. It counts successes in a set number of tries: faulty units in a box of 50, heads in ten flips, correct guesses on a quiz.

A situation is binomial only when all four conditions hold:

Read the formula in pieces. \( p^k \) is the chance of the \( k \) successes, \( (1-p)^{n-k} \) is the chance of the failures, and \( C(n,k) \) counts how many different orders give the same total.

Watch the wording. “At most 2” means 0, 1 or 2, so you add \( P(0) + P(1) + P(2) = 0.16807 + 0.36015 + 0.3087 \approx 0.837 \). “Fewer than 2” stops at 1.

Now the factory question. With \( n = 50 \) and \( p = 0.03 \), the expected number of faulty units is \( 50 \times 0.03 = 1.5 \), with \( \sigma = \sqrt{50 \times 0.03 \times 0.97} \approx 1.21 \). A perfect box means \( X = 0 \): \( 0.97^{50} \approx 0.218 \). Only about 22% of boxes arrive flawless.

The shape of a binomial depends on \( p \). It is symmetric when \( p = 0.5 \), right-skewed when \( p \) is small, and left-skewed when \( p \) is large. As \( n \) grows, the bars start to look like a bell, which leads into Step 8.

The Poisson distribution

The binomial needs a fixed number of trials. But some counts have no natural \( n \): calls to a help desk in an hour, typos on a page, accidents at a junction in a month. There is only a rate over an interval. For these you use the Poisson distribution, which has one parameter, \( \lambda \) (lambda), the average number of events per interval.

How do you choose between the two? Ask what is being counted. Successes out of a fixed number of trials means binomial. Events per hour, page or month with no fixed \( n \) means Poisson.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

Check yourself

The short version

• A random variable gives a number to each chance outcome. Counted means discrete; measured means continuous.
• A probability distribution lists every value with its probability, and the probabilities must sum to 1. \( E(X) = \sum x P(x) \) is the long-run average, and \( \text{Var}(X) = \sum (x-\mu)^2 P(x) \).
• For \( aX + b \), the mean follows the whole change but the spread ignores \( b \). For independent variables, variances add (even for a difference); standard deviations never add.
• The binomial counts successes in \( n \) independent trials with constant \( p \): \( \mu = np \), \( \sigma = \sqrt{np(1-p)} \). Use \( 1 – P(X = 0) \) for “at least one”. The Poisson counts events per interval, with mean and variance both equal to \( \lambda \).
• The full chapter in Statistics Made Simple has many more worked examples and a full set of tiered practice questions with a complete answer key.