Expected value, binomial and normal distribution in Mathematik (CBK) at WU

The fifth of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna is probability (Wahrscheinlichkeitsrechnung); the syllabus does not list its subtopics.¹ In the textbook by Birgit Rudloff, who leads the course, and Achim Zeileis, the fifth chapter has the same title. Its second half is about random variables (Zufallsgrößen), numbers that depend on chance: a weekly income, a day's takings, the hits among your guesses. The questions are what comes out on average, how much the result varies, and how likely a given range is.

Events and conditional probability have an article of their own. The points here are lost mostly to a variance without the square, a missing binomial coefficient and the wrong side of the normal curve. According to the syllabus, help is available in the forum on Canvas and at the walk-in tutorial, where you work through exercises with a tutor on hand.¹ German terms are given in brackets.

Random variables and their distribution: Mathematik (CBK)

A random variable assigns a number to each outcome of a random experiment, a process whose outcome is not known in advance. A student gives private lessons (Nachhilfe), and she does not know in advance how many hours will be booked in a week. The number of hours $X$ is a discrete random variable, because it only takes separate values. Its distribution lists each value together with its probability:

Hours $x$	0	1	2	3
$P(X=x)$	0.1	0.3	0.4	0.2

No probability is negative, and together they add up to 1. For a range you add them: at least two hours are booked with probability $0.4+0.2=0.6$.

A continuous random variable, such as a travel time, can take any value in an interval. The table is replaced by a density function, and the probability of a range is the area under it; the whole area is 1. A single value therefore has probability 0, and it makes no difference whether a question writes $<$ or $\le$.

Expected value and variance: Mathematik (CBK)

The expected value (Erwartungswert) is the average of the values, weighted by their probabilities. For the lessons:

$$E(X)=0\cdot 0.1+1\cdot 0.3+2\cdot 0.4+3\cdot 0.2=1.7$$

Over many weeks that is 1.7 hours on average. The plain average of the four values, 1.5, ignores the probabilities.

The variance (Varianz) measures how far the values lie from the expected value, as an average of squared distances. The quick way to compute it is the shortcut formula (Verschiebungssatz):

$$\mathrm{Var}(X)=E(X^2)-E(X{)}^2$$

Here $E(X^2)=1\cdot 0.3+4\cdot 0.4+9\cdot 0.2=3.7$, so $\mathrm{Var}(X)=3.7-1.7^2=0.81$. The standard deviation is its square root, $\sigma =0.9$ hours, in the same unit as $X$. Forget to square $E(X)$ and you get $3.7-1.7=2$.

The student earns 25 GE (a generic currency unit used in German-language exam tasks) per hour and pays 12 GE a week for travel. Her weekly result $Y=25X-12$ is a linear transformation of $X$:

$$E(aX+b)=a\cdot E(X)+b,\mathrm{Var}(aX+b)=a^2\cdot \mathrm{Var}(X)$$

So $E(Y)=25\cdot 1.7-12=30.5$ GE, $\mathrm{Var}(Y)=625\cdot 0.81=506.25$ and ${\sigma }_Y=22.5$ GE. The constant shifts every value by the same amount and leaves the spread alone; the factor enters squared, so $25\cdot 0.81=20.25$ is wrong as a variance.

Expected values always add up; variances add up only for independent variables. Two independent weeks of lessons have an expected value of 3.4 and a variance of 1.62, so the standard deviation is $\sqrt{1.62}\approx 1.27$ hours, less than $2\cdot 0.9$. The same two rules give the spread of a portfolio return or of a mean: the average of $n$ independent variables has one $n$-th of their common variance. For dependent variables, that sum also picks up their covariance, not covered here.

Binomial distribution: Mathematik (CBK)

A Bernoulli trial has two outcomes, success with probability $p$ and failure. Repeat it $n$ times independently, and the number of successes $X$ follows a binomial distribution (Binomialverteilung):

$$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)\cdot p^k\cdot (1-p{)}^{n-k},E(X)=np,\mathrm{Var}(X)=np(1-p)$$

An example from the exam itself, by our own calculation from the rule in the syllabus: you have 11 answers you are sure of and guess blind on the other 9 questions, each guess right with $p=0.2$, how many of those guesses land. No hit at all has probability $0.8^9\approx 0.134$. One particular sequence with exactly one hit, say on the first question, has probability $0.2\cdot 0.8^8\approx 0.034$. But the hit can be on any of the 9 questions; that number of sequences is the binomial coefficient $\left(\genfrac{}{}{0ex}{}{9}{1}\right)=9$, and exactly one hit has probability $9\cdot 0.2\cdot 0.8^8\approx 0.302$. Without the coefficient you count a single sequence.

Under the syllabus rule (first three wrong free, then 0.25 per answer), no hit or one hit leaves you with 9.5 or 10.75 points.¹ So you fail with probability $0.134+0.302=0.436$ and pass with 56.4%, as the negative-marking calculator shows for this situation. On average you hit $np=1.8$ times, with variance $9\cdot 0.2\cdot 0.8=1.44$.

Normal distribution and standardising: Mathematik (CBK)

Many continuous quantities spread around a mean in a bell shape: they are normally distributed with expected value $\mu$ and standard deviation $\sigma$. Our example is the daily takings $X$ of a sausage stand (Würstelstand), with $\mu =460$ MU and $\sigma =30$ MU. To read off a probability, you standardise:

$$z=\frac{x-\mu }{\sigma },P(X\le x)=\Phi (z)$$

$\Phi$ is the distribution function of the standard normal distribution, the one with $\mu =0$ and $\sigma =1$; its values are in tables. Takings of up to 490 MU have $z=(490-460)/30=1$ and probability $\Phi (1)\approx 0.841$.

1. Daily takings in MU, μ = 460, σ = 30; wanted: the area F up to 490.
2. Minus 460: the centre slides to 0.
3. Divided by 30: 30 MU become 1 wide, the area F stays the same.
4. F = P(X ≤ 490) = Φ(1) ≈ 0.841.

Our calculation

Numbers from the article’s example (our own numbers): μ = 460 MU, σ = 30 MU, wanted is P(X ≤ 490).

z = (490 − 460) / 30 = 1. Shifting every value by 460 and dividing it by 30 makes the bell 30 times narrower and 30 times taller; every area under it stays the same. So P(X ≤ 490) = P(Z ≤ 1) = Φ(1) ≈ 0.841, read off the table of the standard normal distribution.

Tables usually list only positive $z$. Because the bell is symmetric, $\Phi (-z)=1-\Phi (z)$: takings below 430 MU, at $z=-1$, have probability $1-0.841=0.159$. Takings above 490 MU are just as likely, since $P(X>x)=1-\Phi (z)$. So about 68% of days lie between $\mu -\sigma$ and $\mu +\sigma$.

A quantile asks the other way round, for the value that belongs to a probability, for instance the takings the stand beats on 95% of days. From $\Phi (1.645)\approx 0.95$, 5% of days lie below $z=-1.645$, that is below $\mu +z\cdot \sigma =460-1.645\cdot 30\approx 410.65$ MU.

By the central limit theorem (zentraler Grenzwertsatz), a sum of many independent variables is approximately normal. That is why a binomial distribution with large $n$ can be approximated by the normal distribution with $\mu =np$ and $\sigma =\sqrt{np(1-p)}$.

1. n = 9: μ = 1.8, σ = 1.2; the bell is cut off at 0.
2. n = 16: the bars move right and get flatter.
3. n = 25: μ = 5, σ = 2.
4. n = 50: hardly any gaps between bell and bars.
5. n = 100: μ = 20, σ = 4; the bell lies almost exactly on the bars.

Our calculation

p = 0.2 and n = 9 come from the article’s guessing example (9 questions guessed blind). We chose n = 16, 25, 50 and 100.

n = 9: μ = 1.8, σ = 1.2. n = 16: μ = 3.2, σ = 1.6. n = 25: μ = 5, σ = 2. n = 50: μ = 10, σ ≈ 2.83. n = 100: μ = 20, σ = 4.

Each bar is P(X = k) tall, the bell is the density of the normal distribution with the same μ and σ; both on the same scale.

A worked normal-distribution question: Mathematik (CBK)

Find the probability that the sausage stand takes between 430 and 505 MU on a given day, with $\Phi (1)\approx 0.841$ and $\Phi (1.5)\approx 0.933$. Standardised, the limits are $z_1=(430-460)/30=-1$ and $z_2=(505-460)/30=1.5$, and the probability of a range is the difference of the distribution function at its two ends: $P(430<X<505)=\Phi (1.5)-\Phi (-1)=0.933-(1-0.841)=0.774$.

Put in $\Phi (1)$ where $\Phi (-1)$ belongs and you get $0.933-0.841=0.092$. That cannot be right, because the stretch from $\mu -\sigma$ to $\mu$ alone holds about 34% of days. Leave out the lower limit and you are left with $\Phi (1.5)\approx 0.933$; divide by the variance 900 instead of by $\sigma$ and both $z$ values end up near 0, giving about 3%.

Practice questions on expected value and distributions: Mathematik (CBK)

The exam has 20 multiple-choice questions with 5 options each, exactly one of them right. A right answer earns 1 point. Your first 3 wrong answers cost nothing, every further one costs 0.25 points, and you pass with 11 points.¹

The three questions below are our own, written in that format; none comes from a WU exam or from the textbook. Every wrong option is a typical mistake, and each solution shows the working and how to rule options out.

Practice question 1Probability

An ice-cream stand on the Donauinsel makes, on a Saturday, a profit of −30 MU (monetary units) if it rains, 40 MU if it is cloudy and 90 MU if the sun shines. Rain has probability 0.25, cloud 0.45 and sun 0.30. What is the standard deviation of the profit (rounded)?

1. a)24.87 MU
2. b)37.50 MU
3. c)44.37 MU
4. d)58.09 MU
5. e)1,968.75 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Expected value: E(X) = −30 · 0.25 + 40 · 0.45 + 90 · 0.30 = −7.5 + 18 + 27 = 37.5 MU.
2. E(X²) = 900 · 0.25 + 1,600 · 0.45 + 8,100 · 0.30 = 225 + 720 + 2,430 = 3,375.
3. Variance: Var(X) = 3,375 − 37.5² = 3,375 − 1,406.25 = 1,968.75.
4. Standard deviation: σ = √1,968.75 ≈ 44.37 MU.

a)

Wrong. The loss in the rain entered as +30 instead of −30: E(X) = 52.5 instead of 37.5, Var(X) = 3,375 − 52.5² = 618.75, its root ≈ 24.87.

0 or −0.25

b)

Wrong. The expected value E(X) = −30 · 0.25 + 40 · 0.45 + 90 · 0.30 = 37.5 instead of the standard deviation.

0 or −0.25

c)

Right. σ = √(E(X²) − E(X)²) = √(3,375 − 1,406.25) = √1,968.75 ≈ 44.37.

+1

d)

Wrong. E(X)² not subtracted: √E(X²) = √3,375 ≈ 58.09.

0 or −0.25

e)

Wrong. The variance instead of the standard deviation, the square root forgotten; its unit would be MU².

0 or −0.25

–

Left blank.

0

Typical mistake. Forgetting the square root. The variance 1,968.75 is in MU² and is not a spread in MU; only its root, 44.37 MU, is the standard deviation. Losing the minus on the loss costs just as much: with +30 instead of −30 the expected value goes up, the variance goes down, and you get 24.87.

Ruling out before you finish. The standard deviation is never larger than half the range of the values, (90 − (−30)) / 2 = 60 MU; that rules out e. The expected value takes three products, 37.5, and it is option b; it says nothing about the spread, so b goes too. Three are left.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 2Probability

In her part-time job a student advises customers on mobile phone plans. Each conversation ends in a contract with probability 0.35, independently of the others. One afternoon she has 6 conversations. What is the probability that she closes at most 2 contracts (rounded)?

1. a)0.1379
2. b)0.3191
3. c)0.3280
4. d)0.6471
5. e)0.6809

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Binomial distribution with n = 6 and p = 0.35; at most 2 means X = 0, 1 or 2.
2. P(X = 0) = 0.65⁶ ≈ 0.0754 and P(X = 1) = 6 · 0.35 · 0.65⁵ ≈ 0.2437.
3. P(X = 2) = 15 · 0.35² · 0.65⁴ ≈ 0.3280, with the binomial coefficient 6 · 5 / 2 = 15.
4. Sum: P(X ≤ 2) ≈ 0.0754 + 0.2437 + 0.3280 = 0.6471.

a)

Wrong. The binomial coefficients left out: 0.65⁶ + 0.35 · 0.65⁵ + 0.35² · 0.65⁴ ≈ 0.1379, a single sequence counted for each number of successes.

0 or −0.25

b)

Wrong. The case “exactly 2” forgotten, so fewer than 2 worked out: P(X = 0) + P(X = 1) ≈ 0.0754 + 0.2437.

0 or −0.25

c)

Wrong. Exactly 2 only: P(X = 2) = 15 · 0.35² · 0.65⁴ ≈ 0.3280.

0 or −0.25

d)

Right. P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) ≈ 0.0754 + 0.2437 + 0.3280 = 0.6471.

+1

e)

Wrong. “At most” mixed up with “at least”: P(X ≥ 2) = 1 − P(X = 0) − P(X = 1) ≈ 0.6809.

0 or −0.25

–

Left blank.

0

Typical mistake. The forgotten binomial coefficient. 0.35² · 0.65⁴ is the probability of one particular sequence, say contracts in the first two conversations and none in the next four. But two contracts can fall on the 6 conversations in 15 ways. Without the coefficients 1, 6 and 15 you get 0.1379, less than a quarter of the right value.

Ruling out before you finish. Work out only P(X = 2) = 15 · 0.1225 · 0.1785 ≈ 0.3280. “At most 2” contains this case plus P(X = 0) and P(X = 1), both positive, so the answer is above 0.3280. That rules out a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 3Probability

Suppose the monthly rent of rooms in shared flats in a city is normally distributed with expected value 412 MU (monetary units) and variance 576 MU². What share of the rooms costs less than 382 MU (rounded)? Use Φ(0.05) = 0.5199 and Φ(1.25) = 0.8944.

1. a)0.1056
2. b)0.2112
3. c)0.3944
4. d)0.4801
5. e)0.8944

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Standard deviation: σ = √576 = 24 MU.
2. Standardise: z = (382 − 412) / 24 = −30 / 24 = −1.25.
3. Symmetry: Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056.

a)

Right. σ = √576 = 24, z = (382 − 412) / 24 = −1.25 and Φ(−1.25) = 1 − Φ(1.25) = 1 − 0.8944 = 0.1056.

+1

b)

Wrong. Both tails counted, the rooms above 442 MU as well: 2 · 0.1056.

0 or −0.25

c)

Wrong. Only the area between 382 MU and the expected value: Φ(1.25) − 0.5 = 0.3944.

0 or −0.25

d)

Wrong. Divided by the variance instead of σ: z = −30 / 576 ≈ −0.05 and Φ(−0.05) = 1 − 0.5199 = 0.4801.

0 or −0.25

e)

Wrong. The “1 −” forgotten: Φ(1.25) instead of Φ(−1.25), the share of rooms above 382 MU.

0 or −0.25

–

Left blank.

0

Typical mistake. Dividing by the variance. The question gives the variance, 576, but you standardise with σ = 24. Divide by 576 and you get z ≈ −0.05 and so 0.4801, almost half, although 382 MU is well below the mean.

Ruling out before you finish. The standard deviation is √576 = 24 MU, and 382 is more than one standard deviation below the expected value (412 − 24 = 388). Well above that share, c, d and e are certainly wrong; only the calculation decides between a and its doubled-tail cousin b.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Tables and calculator for probability in Mathematik (CBK)

WU does not publish which of the 20 questions covers which chapter; this article follows the textbook. There, "Zufallsgrößen" (random variables) and "Diskrete Verteilungen" (discrete distributions) still belong to "Grundbegriffe" (basic concepts). Then come "Erwartungswert und Varianz" with "Die Rendite eines Portfolios" (portfolio return), "Die Normalverteilung" with "Quantile", and "Die Binomialverteilung" with "Approximation durch die Normalverteilung".²

A calculator is allowed if it has no functions for differential calculus, integration or matrices, cannot solve linear systems and has no text memory.¹ Whether it may have statistics functions, the syllabus does not say.

Under WU's guideline for conducting exams, formula sheets and watches of any kind are not allowed unless the syllabus says otherwise.³ So you cannot bring your own normal table, and WU does not say whether one is handed out in the exam. That is why our questions give the values of $\Phi$ they need. Keep in your head the shortcut formula for the variance, $np$ and $np(1-p)$, standardising, and $\Phi (-z)=1-\Phi (z)$.

The exam plan sets aside two hours for the whole exam; the syllabus does not give the writing time.⁴ With 20 questions, that leaves at most six minutes for each.

Next step after expected value and distributions: Mathematik (CBK)

1. Do the three questions above before opening the solutions, and for each one write down first what is asked: an expected value, a standard deviation or the probability of a range.
2. If you got one wrong, find the mistake in the solution that matches your answer. For more practice, see the fifth chapter of the free textbook by Birgit Rudloff and Achim Zeileis.⁵
3. If events and conditional probability are not yet solid, start with the article on probability and conditional probability. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 5: Wahrscheinlichkeitsrechnungmathe4wiwi.org · undated page · checked on 27 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Prüfungsplan Semestermitte WiSe 2026/27, Prüfungswoche November 2026 (PDF)WU Wien, Prüfungsorganisation · document dated 30 March 2026 · checked on 27 September 2026
5. 5Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026