ICS Statistics: Probability Distributions Made Simple

A probability distribution is simply a complete list of the values a random quantity can take, together with how likely each value is. Once you see a distribution this way, topics like expected value, the binomial distribution and the normal curve stop looking like separate chapters and start looking like variations on one idea. This guide explains the distributions you meet in ICS Statistics and A Level, and works through each with a clear, original example.

Random Variables

A random variable \( X \) assigns a number to each outcome of a random experiment. For example, if you toss three coins, \( X \) could be the number of heads, taking the values 0, 1, 2 or 3.

• A discrete random variable takes separate, countable values (number of heads, number of defective bulbs, number of customers).
• A continuous random variable can take any value in an interval (height, weight, time, marks measured precisely).

For a discrete variable, the probability distribution lists each value \( x \) with its probability \( P(X = x) \). Two conditions must hold:

$$ 0 \le P(X = x) \le 1 \quad \text{and} \quad \sum P(X = x) = 1 $$

Expected Value and Variance

The expected value (mean) is the long-run average of \( X \):

$$ E(X) = \mu = \sum x\,P(x) $$

The variance measures spread:

$$ \text{Var}(X) = E(X^2) - [E(X)]^2, \qquad \text{where } E(X^2) = \sum x^2 P(x) $$

The standard deviation is the square root of the variance.

Worked Example 1: A Discrete Distribution

A mobile repair shop in Lahore records the number of phones brought in for screen repair each hour. The distribution is:

Step 1. Check it is a valid distribution. \( 0.1 + 0.3 + 0.4 + 0.2 = 1.0 \). Valid.

Step 2. Mean. \( E(X) = 0 + 0.3 + 0.8 + 0.6 = 1.7 \) phones per hour.

Step 3. \( E(X^2) \). \( 0 + 0.3 + 1.6 + 1.8 = 3.7 \).

Step 4. Variance.

$$ \text{Var}(X) = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81 $$

Step 5. Standard deviation. \( \sigma = \sqrt{0.81} = 0.9 \) phones.

Interpretation. On average the shop gets 1.7 screen repairs per hour, typically varying by about 0.9 from this figure. Over an 8-hour day, it can expect about \( 8 \times 1.7 = 13.6 \), so roughly 14 repairs.

The Binomial Distribution

The binomial distribution applies when:

1. There is a fixed number of trials, \( n \).
2. Each trial has only two outcomes, "success" or "failure".
3. The probability of success, \( p \), is the same on every trial.
4. The trials are independent.

Then the probability of exactly \( x \) successes is

$$ P(X = x) = \binom{n}{x} p^x q^{n-x}, \qquad q = 1 - p $$

with mean \( \mu = np \) and variance \( \sigma^2 = npq \).

Worked Example 2: Binomial Probability

A cricketer hits a boundary on 30% of the deliveries he faces in a certain phase of play. Assuming independence, find the probability he hits exactly 2 boundaries in 5 deliveries, and the mean and variance of the number of boundaries.

Step 1. Identify the parameters. \( n = 5 \), \( p = 0.3 \), \( q = 0.7 \), \( x = 2 \).

Step 2. Substitute.

$$ P(X = 2) = \binom{5}{2}(0.3)^2(0.7)^3 $$

Step 3. Calculate each part.

• \( \binom{5}{2} = 10 \)
• \( (0.3)^2 = 0.09 \)
• \( (0.7)^3 = 0.343 \)

Step 4. Multiply. \( P(X = 2) = 10 \times 0.09 \times 0.343 = 0.3087 \).

Step 5. Mean and variance. \( \mu = np = 5 \times 0.3 = 1.5 \) boundaries; \( \sigma^2 = npq = 5 \times 0.3 \times 0.7 = 1.05 \).

Sanity check. The most likely outcomes should be near the mean of 1.5. Indeed, \( P(X = 1) = 5(0.3)(0.7)^4 \approx 0.3602 \) and \( P(X = 2) \approx 0.3087 \) are the two largest probabilities.

The Hypergeometric Distribution

When you sample without replacement from a small population, the probability of success changes after each draw, so the binomial model no longer applies. Instead use the hypergeometric distribution. If a population of \( N \) items contains \( k \) successes and you draw \( n \) items,

$$ P(X = x) = \frac{\binom{k}{x}\binom{N - k}{n - x}}{\binom{N}{n}} $$

Worked Example 3: Defective Bulbs

A box contains 10 LED bulbs, of which 3 are defective. A shopkeeper picks 4 at random without replacement. Find the probability that exactly 1 is defective.

Step 1. Parameters. \( N = 10 \), \( k = 3 \), \( n = 4 \), \( x = 1 \).

Step 2. Count the favourable selections. Choose 1 defective from 3 and 3 good from 7:

$$ \binom{3}{1}\binom{7}{3} = 3 \times 35 = 105 $$

Step 3. Count all selections. \( \binom{10}{4} = 210 \).

Step 4. Divide. \( P(X = 1) = \dfrac{105}{210} = 0.5 \).

So there is an even chance that exactly one of the four bulbs is defective.

The Normal Distribution

The normal distribution is the famous bell-shaped curve for continuous data. It is symmetric about its mean \( \mu \), and its spread is set by the standard deviation \( \sigma \). The empirical rule says that about 68% of values lie within one standard deviation of the mean, about 95% within two, and about 99.7% within three.

To find probabilities, convert to the standard normal variable:

$$ Z = \frac{X - \mu}{\sigma} $$

and use standard normal tables.

Worked Example 4: Exam Marks

Marks in a statistics test are normally distributed with mean 60 and standard deviation 8. Find the proportion of students scoring above 72, and between 52 and 68.

Above 72.

$$ Z = \frac{72 - 60}{8} = 1.5 $$

From tables, \( P(Z < 1.5) = 0.9332 \), so \( P(Z > 1.5) = 1 - 0.9332 = 0.0668 \). About 6.7% of students score above 72.

Between 52 and 68. The \( Z \)-values are \( \tfrac{52 - 60}{8} = -1 \) and \( \tfrac{68 - 60}{8} = 1 \). From tables, \( P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 \). About 68% of students fall in this range, exactly as the empirical rule predicts.

Choosing the Right Distribution

• Fixed number of independent trials with constant \( p \): binomial.
• Sampling without replacement from a small, finite population: hypergeometric.
• Continuous measurements that cluster symmetrically around a mean: normal.
• A custom table of values and probabilities: use the general discrete formulas for \( E(X) \) and \( \text{Var}(X) \).

Did You Know?

• Jacob Bernoulli studied repeated success-failure trials in his book Ars Conjectandi, published in 1713, eight years after his death.
• Abraham de Moivre found the normal curve in 1733 as an approximation to binomial probabilities with many trials.
• When the population is large compared with the sample, the hypergeometric and binomial distributions give almost identical answers.

Key Takeaways

• A probability distribution lists every value of a random variable with its probability, and the probabilities sum to 1.
• \( E(X) = \sum xP(x) \) and \( \text{Var}(X) = E(X^2) - [E(X)]^2 \).
• Use the binomial model for independent trials with constant success probability; mean \( np \), variance \( npq \).
• Use the hypergeometric model when sampling without replacement from a small population.
• For normal probabilities, standardise with \( Z = \tfrac{X - \mu}{\sigma} \) and read the tables carefully.

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