Track 5 · Probability Distributions · Lesson 1 of 10

Random variables and their distributions

Describe a random variable with its pmf or pdf and cdf, and calculate its mean, variance and standard deviation.

30 minIntermediate8 graded exercisesFree

An insurer cannot know how many claims a particular policy will produce next year, but it must still price the policy. Treating the claim count as a random variable with a distribution turns that uncertainty into numbers the insurer can plan with: the probability of no claim, the expected number of claims, and how much that number varies.

A random variable is a numerical outcome of a random process: the number of claims on a policy, the number of heads in ten tosses, the time until the next call arrives. We write it with a capital letter, $X$ , and a particular value it can take with a small letter, $x$ .

The distribution of $X$ lists which values are possible and how probable they are. Every model in this track — binomial, Poisson, normal, t and the rest — is a distribution with a name and a few parameters.

	Discrete	Continuous
Values	Separate, countable values (0, 1, 2, …)	Any value in an interval
Typical source	Counting	Measuring (time, weight, distance)
Described by	Probability mass function (pmf)	Probability density function (pdf)
Probability of one exact value	Can be positive	Always 0
Examples	Claims per policy, defective items in a batch	Waiting time, finish time of a race

Two kinds of random variable

p(x) = P(X = x), \qquad \sum_{x} p(x) = 1, \qquad F(x) = P(X \le x)

Probability mass function and cumulative distribution function of a discrete random variable

$p(x)$: the probability that $X$ takes exactly the value $x$ ; each $p(x)$ is between 0 and 1
$\sum_x p(x) = 1$: the probabilities of all possible values add up to 1
$F(x)$: the cumulative distribution function (cdf): the probability that $X$ is at most $x$ . Convention: $P(X \le x)$ includes $x$ itself

For a discrete variable the cdf is a staircase: it jumps by $p(x)$ at each possible value and is flat in between. Because the jump at $k$ belongs to $P(X \le k)$ , the two neighbouring questions have different answers:

P(X < k) = P(X \le k-1), \qquad P(X \ge k) = 1 - P(X \le k-1).

Getting $<$ versus $\le$ right is the most common source of wrong answers with discrete distributions.

\mu = \E[X] = \sum_x x\,p(x), \qquad \sigma^2 = \Var(X) = \sum_x (x-\mu)^2 p(x) = \E[X^2] - \mu^2, \qquad \sigma = \sqrt{\sigma^2}

Mean (expected value), variance and standard deviation of a discrete random variable

$\mu = \E[X]$: the expected value: the probability-weighted average of the possible values, i.e. the long-run average over many repetitions
$\sigma^2$: the variance: the probability-weighted average squared distance from $\mu$ (in squared units)
$\E[X^2]$: $\sum_x x^2 p(x)$ , the expected value of the square of $X$
$\sigma$: the standard deviation: the square root of the variance, in the same units as $X$

Worked example

Complaints per day

A shop's number of customer complaints in a day, $X$ , has this distribution: $P(X=0) = 0.5$ , $P(X=1) = 0.3$ , $P(X=2) = 0.15$ , $P(X=3) = 0.05$ . Find $P(X \le 1)$ , $P(X < 1)$ , the mean, the variance and the standard deviation.

Check it is a distribution. Every probability is between 0 and 1 and they sum to 1.00.
Cumulative probabilities. $P(X \le 1) = 0.5 + 0.3 = 0.80$ , but $P(X < 1) = P(X = 0) = 0.5$ — the value 1 itself is excluded.
Mean. $\mu = 0(0.5) + 1(0.3) + 2(0.15) + 3(0.05) = 0.75$ complaints per day.
Variance. $\E[X^2] = 0(0.5) + 1(0.3) + 4(0.15) + 9(0.05) = 1.35$ , so $\sigma^2 = 1.35 - 0.75^2 = 0.7875$ complaints².
Standard deviation. $\sigma = \sqrt{0.7875} = 0.8874$ complaints.

$P(X \le 1) = 0.80$ , $P(X < 1) = 0.5$ , $\mu = 0.75$ , $\sigma^2 = 0.7875$ , $\sigma ≈ 0.89$ . Over many days the shop averages 0.75 complaints a day, even though no single day can have exactly 0.75 complaints.

A continuous random variable, such as a waiting time, can take any value in an interval. It is described by a probability density function $f(x)$ : a curve that is never negative and has total area 1 underneath it. Probabilities are areas under the curve:

P(a \le X \le b) = \int_a^b f(x)\,dx.

A single point has no width, so it has no area: for continuous $X$ , $P(X = x) = 0$ for every $x$ . That is why $P(X \le b)$ and $P(X < b)$ are equal for continuous variables, unlike discrete ones. The mean and variance are defined as for discrete variables with the sum replaced by an integral.

A discrete distribution: probability bars at 0, 1, …, 10. The shaded bars make up $P(X \le 3) = 0.6496$ — the bar at 3 is included.

Show the data as a table

x (successes in 10 trials)	P(X = x)
0	0.0282475
1	0.121061
2	0.233474
3	0.266828
4	0.200121
5	0.102919
6	0.0367569
7	0.00900169
8	0.0014467
9	0.000137781
10	0.0000059049

Explore a discrete distribution, then a continuous one. The explorer shades the probability you ask for and reports the mean and variance.

Quick check

The time $T$ a customer spends on hold is continuous. Which statement is true?

Dataset

Motor insurance claims

1,500 rows · one row = one motor policy (one policy year) · synthetic data

About this dataset Open in the lab

Policy	Driver age band	Exposure (years)	Claims (claims)
POL0001	36-50	1	0
POL0002	51-65	1	0
POL0003	26-35	1	0
POL0004	26-35	0.94	0
POL0005	26-35	1	2
POL0006	51-65	0.98	0

One row per motor policy in a (synthetic) insurer's portfolio. claim_count is a discrete random variable observed 1 500 times.

Exercises

The lesson is complete when every exercise is answered correctly. Hints are free; viewing a worked solution before solving means the exercise won't count towards mastery.

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Exercise 1 · Practice, difficulty 1 of 3

Discrete or continuous?

Not started

Classify each random variable as discrete or continuous.

Exercise 2 · Guided, difficulty 1 of 3

Expected number of items

Not started

$X$ is the number of items a visitor to an online shop puts in the basket. Its distribution is

$x$	0	1	2	3	4
$P(X = x)$	0.30	0.25	0.05	0.20	0.20

Find the expected value $\E[X]$ . Give your answer to 2 decimal places.

Exercise 3 · Practice, difficulty 2 of 3

Variance and standard deviation

Not started

The number of support tickets $X$ a technician closes in an hour has this distribution:

$x$	0	1	2	3	4
$P(X = x)$	0.25	0.20	0.20	0.20	0.15

The mean is $\mu = 1.80$ . Find the variance and the standard deviation of $X$ , each to 4 decimal places.

Exercise 4 · Practice, difficulty 2 of 3

Reading the cdf: ≤ versus <

Not started

$X$ is the number of parcels a courier fails to deliver on a shift:

$x$	0	1	2	3	4	5
$P(X = x)$	0.15	0.10	0.25	0.20	0.15	0.15

Find $P(X \le 2)$ and $P(X < 2)$ . Give both to 2 decimal places.

Exercise 5 · Practice, difficulty 2 of 3

The missing probability

Not started

A game awards $X$ points per round with the distribution below, but one probability is missing.

$x$	0	1	2	3	4
$P(X = x)$	?	0.15	0.15	0.30	0.30

Find the missing probability $P(X = 0)$ , and then $P(X \ge 2)$ . Give both to 2 decimal places.

Exercise 6 · Challenge, difficulty 3 of 3

Probability as area under a density

Not started

The time $T$ (minutes) a printing job waits in a queue has density $f(t) = t/50$ for $0 \le t \le 10$ (and 0 elsewhere): a triangle with total area 1.

Find $P(2 < T \le 9)$ to 4 decimal places. (Hint: the area under $f$ from 0 to $t$ is $t^2/100$ .)

Exercise 7 · Application, difficulty 2 of 3

The empirical distribution of claim counts

Not started

Treat claim_count in the insurance_claims dataset (1 500 policies) as observations of a random variable $X$ . Using the relative frequencies as probabilities, find the mean number of claims per policy (3 decimal places) and the proportion of policies with at least one claim, $P(X \ge 1)$ (4 decimal places).

Data: insurance_claims · Open in the lab to explore and calculate.

Exercise 8 · Interpretation, difficulty 2 of 3

What an expected value means

Not started

A raffle ticket's net gain $X$ (in SAR) has expected value $\E[X] = -2.5$ and standard deviation $\sigma = 40$ . Which statement is correct?

What you can now do

Classify a random variable as discrete or continuous
Read probabilities from a pmf and a cdf, distinguishing P(X ≤ k) from P(X < k)
Explain why P(X = x) = 0 for a continuous variable and find probabilities as areas
Calculate the mean, variance and standard deviation of a discrete random variable and interpret the mean as a long-run average