FSP Training 07

Preparation for Mock Exam 08

Instructions

This training page covers all of Section 07 and prepares you for Mock Exam 08. Set A trains setting up: you translate a text into mathematics and check your translation. Set B is a practice mock in the format of Mock Exam 08: 180 minutes plus 5 minutes reading time, 100 points, and all 3 problems must be completed. Write down every step as you would in the exam. The points show what each part is worth.

Round probabilities to four decimal places (or give them in percent with two decimal places) and keep intermediate results unrounded. Round descriptive statistics such as the mean or the standard deviation to two decimal places. A minimum number \(n\) is always rounded up: check \(n - 1\) and \(n\).

In the real FSP, decimals are written with a comma (1,25 means 1.25).

Set A: Set Up & Check (40 minutes)

From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. A salad bar prepares 60 salad bowls every morning. On 8 randomly chosen days, the numbers of bowls left unsold at closing time were 10, 4, 13, 7, 14, 10, 3 and 11. Compute the mean and the sample standard deviation of the daily number of unsold bowls. Write the probability that a bowl stays unsold as a relative frequency and compute it.

    Check: A classmate writes \(P(\text{unsold}) = \frac{72}{60} = 1.2\). Explain why this number cannot be a probability and give the correct base.

  2. In a public library, 70 % of the visitors borrow a book (\(B\)), 20 % use a study place (\(L\)), and 14 % do both. Decide with \(P(B \cap L) = P(B) \cdot P(L)\) whether \(B\) and \(L\) are independent. Write “a visitor does neither” as a set and compute its probability.

    Check: A classmate computes the share of the visitors who do neither as \(1 - 0.70 - 0.20 = 0.10\). Draw a Venn diagram with the probabilities of its four regions and explain the mistake.

  3. A school receives 13 new laptops. 4 of them have a faulty keyboard, but nobody knows which. The IT teacher tests 3 of the laptops, chosen at random. Write the probability that exactly one tested laptop has a faulty keyboard as a quotient of counts and compute it. Compute the probability that at least one tested laptop has a faulty keyboard.

    Check: A classmate draws a tree and computes \(\frac{4}{13} \cdot \frac{9}{12} \cdot \frac{8}{11}\) for “exactly one”. Compute this product and explain why it is smaller than your result.

  4. At a car-sharing company, 45 % of the rentals are made by new customers (\(N\)), the rest by regular customers. 2 % of the rentals by regular customers end with a parking fine (\(F\)). In total, 5.6 % of all rentals end with a parking fine. Set up the law of total probability with the unknown \(x = P(F \mid N)\) and show that \(x = 0.1\).

    Check: Put \(x = 0.1\) into the tree and add the two paths that end in \(F\). A classmate computes \(x = 0.056 - 0.02 = 0.036\). Explain the mistake.

  5. A heating company services gas boilers at a fixed price of 149 EUR per service. From experience, a service costs the company 100 EUR with probability 0.7, 180 EUR with probability 0.25 (a small spare part is needed) and 400 EUR with probability 0.05 (a larger repair that the fixed price includes). \(X\) is the company’s cost of one service. Write the expected cost as a formula and compute it. Decide whether the fixed price covers the costs on average.

    Check: A classmate takes the mean of the three costs, \(\frac{100 + 180 + 400}{3} \approx 226.67\) EUR, and says that the company loses money on every service on average. Explain the mistake.

  6. A sawmill scans every log for hidden metal (nails, old fence wire) before cutting it, because metal destroys the saw blades. 1 % of the logs contain metal (\(M\)). The scanner raises an alarm (\(+\)) for 97 % of the logs with metal and for 4 % of the logs without metal. Write the three rates in probability notation and name them. Compute the probability that a log with an alarm does indeed contain metal, and the probability that a log contains metal but is incorrectly identified as free of metal.

    Check: Complete a table of counts for 10,000 logs and compute both probabilities again from the table. The scanner’s supplier says: “Our scanner is right in 97 % of its alarms.” Decide whether this is true.

  7. A high ropes course (a climbing park in the trees) had 1,200 visitors last month. 480 of them came with a school class (\(S\)). Of the visitors with a school class, 15 % booked the lunch package (\(F\)). In total, 300 visitors booked the lunch package. Complete the contingency table of counts. Write the expected count of the field \(S \cap F\) if booking the lunch package were independent of coming with a school class, compute it and compare it with the observed count.

    Check: A classmate computes the expected count as \(0.4 \cdot 0.15 \cdot 1200 = 72\) and says: “Observed and expected count agree, so the two properties are independent.” Explain the mistake.

  8. A removal company carries out 15 moves per week. In each move, some furniture gets damaged with probability 15 %, independently of the other moves. \(X\) is the number of moves with damage in one week. Write “more than 2 but fewer than 6 moves with damage” as a probability with the correct endpoints and as a sum of binomial terms, and compute it. Write “at least 3 moves with damage” with the complement and compute it.

    Check: A classmate writes the first event as \(P(2 \leq X \leq 6)\). List the values of \(X\) in the classmate’s event and in the event from the text.

  1. Mean: the 8 values add up to 72, so \(\bar{x} = \frac{72}{8} = 9\) unsold bowls per day.

    Sample standard deviation: the 8 days are a sample of all days, so divide by \(n - 1 = 7\). Deviations \(x_i - 9\): \(1, -5, 4, -2, 5, 1, -6, 2\).

    \[\sum (x_i - \bar{x})^2 = 1 + 25 + 16 + 4 + 25 + 1 + 36 + 4 = 112, \qquad s^2 = \frac{112}{7} = 16, \qquad s = 4\]

    Relative frequency ≈ probability: on the 8 days, the salad bar prepared \(8 \cdot 60 = 480\) bowls, and 72 stayed unsold:

    \[P(\text{unsold}) \approx \frac{72}{480} = \frac{9}{60} = 0.15 = 15.00\,\%\]

    Check: by the axioms, a probability lies between 0 and 1. The classmate divides the unsold bowls of 8 days by the bowls of one day. The base is all bowls prepared on the 8 days (480), or, the same, the mean per day divided by the 60 bowls of one day.

    Typical mistake: dividing by \(n = 8\) in the standard deviation. The 8 days are a sample, so the sample standard deviation divides by \(n - 1\).

  2. Independence test: \(P(B) \cdot P(L) = 0.70 \cdot 0.20 = 0.14 = P(B \cap L)\), so \(B\) and \(L\) are independent.

    “Neither” is the complement of the union, \(\overline{B \cup L}\):

    \[P(B \cup L) = 0.70 + 0.20 - 0.14 = 0.76, \qquad P(\overline{B \cup L}) = 1 - 0.76 = 0.24\]

    Check: the four regions of the Venn diagram are only \(B\): \(0.70 - 0.14 = 0.56\), both: \(0.14\), only \(L\): \(0.20 - 0.14 = 0.06\), outside both circles: \(0.24\). They add up to 1 ✓. The classmate removes the 14 % who do both twice, once with \(B\) and once with \(L\).

    Typical mistake: “14 % do both, so \(B\) and \(L\) are not independent.” Events that can happen together can still be independent. Independence only asks whether \(P(B \cap L) = P(B) \cdot P(L)\).

  3. All \(\binom{13}{3} = 286\) selections of 3 laptops are equally likely. Exactly one faulty laptop: 1 of the 4 faulty and 2 of the 9 good laptops.

    \[P(\text{exactly one}) = \frac{\binom{4}{1} \cdot \binom{9}{2}}{\binom{13}{3}} = \frac{4 \cdot 36}{286} = \frac{144}{286} = \frac{72}{143} \approx 0.5035\]

    At least one, with the complement “all 3 from the 9 good laptops”:

    \[P(\text{at least one}) = 1 - \frac{\binom{9}{3}}{\binom{13}{3}} = 1 - \frac{84}{286} = \frac{101}{143} \approx 0.7063\]

    Check: \(\frac{4}{13} \cdot \frac{9}{12} \cdot \frac{8}{11} = \frac{288}{1716} = \frac{24}{143} \approx 0.1678\). This is only the path “faulty, good, good”. The faulty laptop can also be the second or the third one tested, and each of the three paths has the same probability: \(3 \cdot \frac{24}{143} = \frac{72}{143}\) ✓.

    Typical mistake: \(\frac{\binom{4}{1}}{\binom{13}{3}}\). The other 2 tested laptops must be chosen too, from the 9 good ones.

  4. Tree: \(P(N) = 0.45\) with \(P(F \mid N) = x\), and \(P(\bar{N}) = 0.55\) with \(P(F \mid \bar{N}) = 0.02\). Law of total probability:

    \[P(F) = 0.45 \cdot x + 0.55 \cdot 0.02 = 0.45x + 0.011 = 0.056 \iff 0.45x = 0.045 \iff x = 0.1 \;\checkmark\]

    Check: the paths are \(0.45 \cdot 0.1 = 0.045\) and \(0.55 \cdot 0.02 = 0.011\), and \(0.045 + 0.011 = 0.056\) ✓. The classmate subtracts two rates with different bases: 5.6 % refers to all rentals, 2 % only to the rentals by regular customers. With \(x = 0.036\), the tree gives \(0.0162 + 0.011 = 0.0272 \neq 0.056\).

    Typical mistake: stopping at \(0.45x = 0.045\) and answering 4.5 %. \(0.045\) is the path probability \(P(N \cap F)\). The share among the rentals by new customers needs the division by \(0.45\).

  5. Expected value: each cost is weighted with its probability.

    \[E[X] = 100 \cdot 0.7 + 180 \cdot 0.25 + 400 \cdot 0.05 = 70 + 45 + 20 = 135 \text{ EUR}\]

    The fixed price of 149 EUR covers the costs on average: the company earns \(149 - 135 = 14\) EUR per service on average.

    Check: the probabilities add up to \(0.7 + 0.25 + 0.05 = 1\) ✓. The plain mean gives the rare large repair (5 %) the same weight as the normal service (70 %). The expected value weights each cost with how often it occurs.

    Typical mistake: the classmate’s mean. It would only be right if all three costs had the same probability \(\frac{1}{3}\).

  6. Given: base rate \(P(M) = 0.01\), true positive rate \(P(+ \mid M) = 0.97\), false positive rate \(P(+ \mid \bar{M}) = 0.04\) (so the true negative rate is \(P(- \mid \bar{M}) = 0.96\)).

    Positive predictive value with Bayes’ theorem, \(P(+) = 0.01 \cdot 0.97 + 0.99 \cdot 0.04 = 0.0097 + 0.0396 = 0.0493\):

    \[P(M \mid +) = \frac{0.0097}{0.0493} = \frac{97}{493} \approx 0.1968\]

    Metal, but incorrectly identified as free of metal (one path, base: all logs):

    \[P(M \cap -) = P(M) \cdot P(- \mid M) = 0.01 \cdot 0.03 = 0.0003\]

    Check with 10,000 logs:

    \(M\) \(\bar{M}\) sum
    \(+\) 97 396 493
    \(-\) 3 9,504 9,507
    sum 100 9,900 10,000

    \(P(M \mid +) = \frac{97}{493} \approx 0.1968\) ✓ and \(P(M \cap -) = \frac{3}{10000} = 0.0003\) ✓. The supplier is wrong: 97 % is the true positive rate. Only about 19.68 % of the alarms are right, because the many logs without metal produce 396 false alarms.

    Typical mistake: computing \(P(M \mid -) = \frac{3}{9507}\) for “contains metal but is incorrectly identified as free of metal”. The sentence speaks about all logs, so it is the joint probability with the base 10,000.

  7. Table: \(S \cap F = 0.15 \cdot 480 = 72\). The rest follows from the totals:

    \(F\) \(\bar{F}\) sum
    \(S\) 72 408 480
    \(\bar{S}\) 228 492 720
    sum 300 900 1,200

    Expected count under independence:

    \[E_{S,F} = \frac{\text{row total} \cdot \text{column total}}{\text{grand total}} = \frac{480 \cdot 300}{1200} = 120\]

    Observed: 72, far fewer than 120. So the two properties are not independent: visitors with a school class book the lunch package less often (15 %) than the other visitors (\(\frac{228}{720} \approx 31.67\,\%\)).

    Check: the classmate’s 15 % is \(P(F \mid S)\), and it comes from the observed field. So \(0.4 \cdot 0.15 \cdot 1200 = 72\) is the observed count again, and the comparison proves nothing. The expected count uses only the totals: \(P(S) \cdot P(F) \cdot 1200 = 0.4 \cdot 0.25 \cdot 1200 = 120\).

    Typical mistake: the classmate’s. Under independence, the field only depends on its row and column totals.

  8. \(X \sim B(15;\ 0.15)\). “More than 2” leaves out 2, “fewer than 6” leaves out 6:

    \[P(3 \leq X \leq 5) = \sum_{k=3}^{5} \binom{15}{k} \cdot 0.15^k \cdot 0.85^{15 - k} \approx 0.2184 + 0.1156 + 0.0449 \approx 0.3790\]

    (sum of the unrounded terms). At least 3 with the complement “at most 2”:

    \[P(X \geq 3) = 1 - P(X \leq 2) = 1 - \left(0.85^{15} + 15 \cdot 0.15 \cdot 0.85^{14} + 105 \cdot 0.15^2 \cdot 0.85^{13}\right) \approx 1 - (0.0874 + 0.2312 + 0.2856) \approx 0.3958\]

    Check: the classmate’s event contains \(X = 2, 3, 4, 5, 6\), the text’s event only \(X = 3, 4, 5\). “More than” and “fewer than” leave out the endpoint, “at least” and “not more than” include it.

    Typical mistake: \(P(X \geq 3) = 1 - P(X \leq 3)\). The complement of “at least 3” is “at most 2”.

One Paragraph, All Conditions (xx)

A company maintains the lifts in office buildings. Once a day, a remote sensor checks every lift and reports a possible fault (\(+\)). On a given day, 6 % of the lifts have a fault (\(F\)). The sensor reports 80 % of the lifts with a fault, and it also reports 3 % of the lifts without a fault. After every report, a technician drives to the lift. The manager wants to know the probability that a technician who drives out finds a real fault, the probability that a lift has a fault but the sensor does not report it, and the number of trips to expect on one day for the 250 lifts of the company. The manager also asks for the probability that exactly 2 of the 12 lifts in one office tower are reported on the same day, and for the smallest number of lifts that must be checked so that, with a likelihood (probability) of at least 99 %, at least one lift with a fault is reported.

List every given value and every question in this paragraph as a mathematical statement, with the course name of each rate. Name the rule or the distribution you would use. Do not solve anything.

Check: An assistant writes the last question as \(1 - 0.2^n \geq 0.99\). Explain what went wrong.

  1. Random experiment: one randomly chosen lift on one day, with the events \(F\) = “the lift has a fault” and \(+\) = “the sensor reports the lift”
  2. Base rate: \(P(F) = 0.06\), so \(P(\bar{F}) = 1 - 0.06\)
  3. True positive rate: \(P(+ \mid F) = 0.8\), so the false negative rate is \(P(- \mid F) = 1 - 0.8\)
  4. False positive rate: \(P(+ \mid \bar{F}) = 0.03\), so the true negative rate is \(P(- \mid \bar{F}) = 1 - 0.03\)
  5. A technician who drives out finds a real fault: positive predictive value \(P(F \mid +) = \frac{P(F) \cdot P(+ \mid F)}{P(+)}\) with \(P(+) = P(F) \cdot P(+ \mid F) + P(\bar{F}) \cdot P(+ \mid \bar{F})\), Bayes’ theorem with the law of total probability
  6. A fault that the sensor does not report: \(P(F \cap -) = P(F) \cdot P(- \mid F)\), multiplication rule (one path, a joint probability with all lifts as the base)
  7. Expected number of trips for 250 lifts: \(250 \cdot P(+)\) (expected value \(n \cdot p\))
  8. Exactly 2 of the 12 lifts in the tower are reported: \(X\) = number of reported lifts, \(X \sim B(12;\ P(+))\) if the lifts are reported independently of each other, \(P(X = 2) = \binom{12}{2} \cdot P(+)^2 \cdot \left(1 - P(+)\right)^{10}\)
  9. Smallest number of lifts: for one checked lift, the “success” is “the lift has a fault and is reported”, with \(p = P(F \cap +) = P(F) \cdot P(+ \mid F)\). Condition \(1 - \left(1 - p\right)^n \geq 0.99\), solved with logarithms (the sign flips when dividing by \(\ln(1 - p) < 0\)), the smallest whole \(n\) (round up)

Check: \(0.2^n\) is the probability that \(n\) lifts with a fault are all missed. But the \(n\) checked lifts are random lifts, and most of them have no fault. For one checked lift, the “success” is a fault that is reported, \(F \cap +\), with \(p = 0.06 \cdot 0.8\), so the “none” probability is \((1 - p)^n\).

Typical mistake: taking \(p = P(+)\) in question 9. The reports also contain the false alarms, and a false alarm is not a lift with a fault.

FSP Words So Far

The FSP problems use these words from Sections 01 to 07. Learn what each one means and how to write it in mathematics.

Word Meaning Mathematics
quantity unit (QU) a fixed number of items, e.g. 1 QU = 100 items the output \(x\) is counted in QU
monetary unit (MU) a fixed amount of money, e.g. 1 MU = 100 EUR results in MU, convert to EUR at the end
fixed costs costs that arise even if nothing is produced the constant term of \(C(x)\)
variable costs costs that grow with the output \(C(x) - C(0)\) (for linear costs: output times costs per QU)
total costs fixed costs + variable costs \(C(x)\)
revenue quantity times price \(R(x) = x \cdot \text{price}\)
profit revenue − total costs \(P(x) = R(x) - C(x)\)
break-even point the output at which the profit changes from negative (a loss) to positive \(P(x) = 0\), the smaller zero
profit limit (upper break-even point) the larger output with zero profit: above it, the profit turns negative again \(P(x) = 0\), the larger zero
profit zone all outputs with a positive profit \(P(x) > 0\): from the break-even point to the profit limit (or the capacity)
capacity the maximum possible output an upper bound: \(x \leq\) capacity
growth factor the factor by which a quantity is multiplied each period \(q\) in (starting value) \(\cdot\, q^t\)
growth rate the change per period in percent \(q = 1 + r\) with \(r\) as a decimal, a loss of 4 % gives \(q = 0.96\)
doubling time the time after which a quantity has doubled \(q^T = 2\), continuous: \(e^{kT} = 2\), so \(T = \frac{\ln 2}{k}\)
half-life (halving time) the time after which a quantity has halved \(q^T = 0.5\), continuous: \(e^{-kT} = 0.5\), so \(T = \frac{\ln 2}{k}\)
“deviates from \(a\) by at most \(d\)” (tolerance) the distance from \(a\) is at most \(d\) \(\lvert x - a \rvert \leq d \iff a - d \leq x \leq a + d\)
“share of \(A\) among \(B\)” the part of \(B\) that also belongs to \(A\): the base is \(B\) \(\dfrac{\text{number in } A \cap B}{\text{number in } B}\)
“neither \(A\) nor \(B\)” not in \(A\) and not in \(B\), i.e. outside both circles \(\overline{A \cup B}\)
contingency table a table of counts for two properties: the inner fields are the regions of the Venn diagram, the margins are the totals inner fields \(A \cap B\), \(A \cap \overline{B}\), \(\overline{A} \cap B\), \(\overline{A} \cap \overline{B}\). Joint probability: field divided by the grand total, conditional probability: field divided by the total of the condition’s row or column
gross price, net price gross price = net price + VAT with 19 % VAT: gross \(= 1.19 \cdot\) net
system of linear equations several linear equations that must hold at the same time solve by substitution or elimination, check the solution in every equation
no solution, infinitely many solutions the conditions contradict each other, or one condition only repeats another elimination leaves \(0 = d\) with \(d \neq 0\), or \(0 = 0\)
demand, supply the quantity buyers want to buy and the quantity sellers offer at a given price as terms in the price \(p\), e.g. \(Q_d(p)\), \(Q_s(p)\), or as prices \(p_D(x)\), \(p_S(x)\) for the quantity \(x\)
market equilibrium the quantity demanded equals the quantity supplied demand = supply: \(Q_d(p) = Q_s(p)\), or \(p_D(x) = p_S(x)\). With demand and supply as prices: \(D(q) = S(q)\) gives \(q^*\), and \(p^* = D(q^*)\)
price-demand function gives the price at which \(x\) QU can be sold \(p(x)\), revenue \(R(x) = x \cdot p(x)\)
average cost total costs divided by the number of units \(\dfrac{C(x)}{x}\) with \(x \neq 0\)
zero form all terms on one side, 0 on the other side \(ax^2 + bx + c = 0\), then factor or use the quadratic formula
discriminant decides how many real solutions a quadratic equation has \(\Delta = b^2 - 4ac\): \(> 0\) two, \(= 0\) one, \(< 0\) none
biquadratic equation an equation with only \(x^4\), \(x^2\) and a constant substitute \(u = x^2\) in \(ax^4 + bx^2 + c\), solve for \(u\), then \(x = \pm\sqrt{u}\) for \(u \geq 0\)
domain restrictions (of an equation) the conditions the unknown must meet so that every term is defined denominators \(\neq 0\), under a root \(\geq 0\), inside a logarithm \(> 0\)
extraneous solution a candidate that solves the squared equation but not the original one check every candidate in the original equation
function assigns to every input exactly one output \(f(x)\), its graph passes the vertical line test
domain (of a function) all inputs for which the function is defined, in a context only the sensible ones e.g. \([0; \text{capacity}]\)
range all outputs that the function takes on its domain if the function grows or falls: between the values at the two ends, for a parabola also check the vertex
maximum price the price at which nobody buys any more \(Q_d(p) = 0\), or \(c\) in \(p = c - dQ\)
market saturation the quantity demanded at a price of 0 \(a\) in \(Q_d = a - bp\), or the zero of \(p(x)\)
shortage, surplus at a price below the equilibrium price, demand exceeds supply (shortage), above it, supply exceeds demand (surplus) \(Q_d(p) > Q_s(p)\), or \(Q_d(p) < Q_s(p)\)
contribution margin price minus variable costs per unit: the part of the price that covers the fixed costs break-even quantity \(= \dfrac{\text{fixed costs}}{\text{contribution margin}}\)
vertex the highest or lowest point of a parabola \(x_v = -\dfrac{b}{2a}\), or \((h, k)\) in \(a(x - h)^2 + k\)
profit-maximizing output the output with the highest profit \(x_v\) of \(P(x)\), a maximum since \(a < 0\), and the price \(p(x_v)\). If the capacity lies below \(x_v\): the capacity and its price
shift, stretch, compression moving or scaling a graph \(a \cdot f(b(x - h)) + k\): first \(b\), then \(h\), then \(a\), then \(k\)
mirror image at the \(y\)-axis the graph flipped from right to left \(f(-x)\)
mirror image at the \(x\)-axis the graph flipped upside down \(-f(x)\)
composite function first the inner function, then the outer function \((f \circ g)(x) = f(g(x))\)
inverse function reverses a function: gives the input that belongs to an output solve \(y = f(x)\) for \(x\), e.g. \(x(p)\) from \(p(x)\)
polynomial of third degree a function with powers of \(x\) up to \(x^3\) \(ax^3 + bx^2 + cx + d\) with \(a \neq 0\): four unknowns, so four conditions
degree, leading coefficient the highest power of \(x\), and the number in front of it together they decide the end behavior
zero of multiplicity \(m\) the factor \((x - c)\) appears \(m\) times in the factored form odd \(m\): the graph crosses the \(x\)-axis at \(c\), even \(m\): it touches it
polynomial long division divides a polynomial by \((x - c)\) once \(c\) is a known zero \(f(x) = (x - c) \cdot q(x)\), the remainder must be 0
asymptotic behavior (end behavior) what happens to \(f(x)\) for \(x \to \infty\) and \(x \to -\infty\) polynomial: degree and sign of the leading coefficient. Rational function: its asymptotes
intercepts the points where the graph meets the axes \(y\)-intercept \((0, f(0))\), \(x\)-intercepts from \(f(x) = 0\)
rational function a quotient of two polynomials \(\dfrac{p(x)}{q(x)}\) with the domain \(q(x) \neq 0\)
vertical asymptote a vertical line near which the values grow without bound a zero of the denominator that does not cancel
hole a single missing point of the graph a factor that cancels, its \(y\)-value comes from the simplified term
horizontal asymptote a horizontal line that the graph approaches for \(x \to \pm\infty\) \(y = c\) with \(f(x) \to c\)
oblique asymptote a slanted line that the graph approaches for \(x \to \pm\infty\) the linear part of the long division, if the numerator’s degree is one higher
grows less than proportionally grows, but twice the input gives less than twice the output \(a \cdot x^b\) with \(0 < b < 1\)
continuous growth rate the rate \(k\) of a model with the base \(e\) \(N_0 \cdot e^{kt}\), growth factor per period \(e^k\)
saturation level a level that a quantity approaches but never exceeds a horizontal asymptote \(y = L\), e.g. of \(L - a \cdot e^{-kt}\) or \(\dfrac{L}{1 + A e^{-kt}}\)
behavior for \(t \to \infty\) the value a quantity approaches in the long run \(\lim\limits_{t \to \infty} N(t)\), often with \(e^{-kt} \to 0\)
radian an angle measure in which a full turn is \(2\pi\) \(180° = \pi\), arc length \(s = r\theta\)
amplitude half the distance between the highest and the lowest value \(\lvert A \rvert = \dfrac{\max - \min}{2}\)
midline the horizontal line halfway between the highest and the lowest value \(y = D = \dfrac{\max + \min}{2}\)
period the length of one full cycle \(B = \dfrac{2\pi}{\text{period}}\)
phase shift the horizontal shift of a sine or cosine curve \(C\) in \(A\sin(B(t - C)) + D\) or \(A\cos(B(t - C)) + D\)
mirror image at the midline the graph flipped upside down about its midline \(y = D\) \(2D - f(t)\)
limit the value that \(f(x)\) approaches when \(x\) approaches a number \(a\), or \(\pm\infty\) \(\lim\limits_{x \to a} f(x)\), one-sided: \(x \to a^-\) (from the left), \(x \to a^+\) (from the right)
continuous (at \(a\)) the graph has no jump and no gap at \(a\) \(f(a)\) is defined, \(\lim\limits_{x \to a} f(x)\) exists, and both are equal
removable discontinuity a gap that a single value could close, e.g. a hole the limit exists, but \(f(a)\) is not defined or differs from it
derivative the instantaneous rate of change: the slope of the tangent \(f'(x) = \lim\limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\)
secant, slope of the secant the line through two points of the graph, and its slope (the average rate of change) \(\dfrac{f(b) - f(a)}{b - a}\) on \([a; b]\)
tangent, tangential gradient the line that touches the graph at a point, and its slope \(y = f'(a)(x - a) + f(a)\) with the slope \(f'(a)\)
angle of intersection (with the \(x\)-axis) the angle at which a line with slope \(m\) crosses the \(x\)-axis \(\alpha = \arctan(m)\), with the indicator D for degrees
marginal cost, marginal revenue, marginal profit about the change of the costs, the revenue or the profit caused by one more unit \(C'(x)\), \(R'(x)\), \(P'(x) = R'(x) - C'(x)\)
stationary point (critical point) a point of the graph with a horizontal tangent \(f'(x_0) = 0\)
nature of a stationary point whether it is a local maximum, a local minimum or a saddle point sign change of \(f'\) at \(x_0\), or the sign of \(f''(x_0)\)
local maximum, local minimum a point that is higher (lower) than all points nearby \(f'(x_0) = 0\) and \(f''(x_0) < 0\) (maximum) or \(f''(x_0) > 0\) (minimum)
saddle point a stationary point at which \(f'\) does not change its sign: the graph levels off and keeps its direction \(f'(x_0) = 0\) without a sign change, e.g. \(f''(x_0) = 0\) and \(f'''(x_0) \neq 0\)
necessary condition, sufficient condition a condition every candidate must meet, and a condition that proves the kind of point (“prove it really is a maximum”) necessary: \(f'(x_0) = 0\), sufficient: e.g. \(f''(x_0) < 0\) for a maximum
global maximum, global minimum (on an interval) the highest or lowest value on the whole interval compare the local extrema with the values at the ends of the interval
concave up, concave down the graph curves upward like a cup, or downward like a cap \(f''(x) > 0\), or \(f''(x) < 0\)
inflection point the point at which the curvature changes sign: the graph is steepest or flattest there \(f''(x_0) = 0\) and \(f'''(x_0) \neq 0\), or a sign change of \(f''\)
“the curvature changes sign” the graph changes from concave down to concave up, or back an inflection point: \(f''(x_0) = 0\). For a cubic cost function, the marginal costs are lowest there
inflectional tangent the tangent at an inflection point: it shows the fastest growth or decline \(y = f'(a)(x - a) + f(a)\) with \(f''(a) = 0\)
unit costs, lowest unit costs the total costs per unit, and the output at which they are lowest \(k(x) = \dfrac{C(x)}{x}\), minimum from \(k'(x) = 0\) and \(k''(x) > 0\)
long-term price floor the lowest unit costs: below this price, the company cannot cover its total costs in the long run the minimum value of \(k(x) = \dfrac{C(x)}{x}\)
variable costs per unit the variable costs divided by the output \(k_v(x) = \dfrac{C(x) - C(0)}{x}\)
short-term lower limit price the lowest variable costs per unit: below this price, the company cannot even cover its variable costs the minimum value of \(k_v(x)\)
function family (German: Funktionsschar) functions with the same term that differ only in a parameter \(f_a(x)\), e.g. \(f_a(x) = x^2 - ax + 1\)
parameter a letter that is fixed within one function of the family but changes from one function to the next solve a condition for it, e.g. \(f_a'(2) = 0\)
antiderivative (German: Stammfunktion) a function whose derivative is \(f\) \(F'(x) = f(x)\)
constant of integration the free constant: two antiderivatives of \(f\) differ only by a constant \(F(x) + C\), fixed by one condition, e.g. \(C(0) =\) fixed costs or \(P(0) = -\)fixed costs
indefinite integral the family of all antiderivatives \(\int f(x) \, dx = F(x) + C\)
definite integral a number: the signed area between the graph and the \(x\)-axis from \(a\) to \(b\) \(\int_a^b f(x) \, dx = F(b) - F(a)\) (Fundamental Theorem of Calculus)
lower limit, upper limit the start \(a\) and the end \(b\) of the interval in \(\int_a^b f(x) \, dx\) “in the second month”: from \(t = 1\) to \(t = 2\)
signed area area above the \(x\)-axis counts positive, area below it negative \(\int_a^b f(x) \, dx\), it can be negative or 0
area (between a graph and the \(x\)-axis) always a positive number split at the zeros and add the absolute values of the integrals of the pieces
net change the total change of a quantity, computed from its rate of change \(F(b) - F(a) = \int_a^b F'(x) \, dx\), e.g. extra costs \(\int_a^b C'(x) \, dx\)
“total … within the first … months” everything that has come together from the start up to month \(b\) \(\int_0^b (\text{rate}) \, dt\)
area enclosed by two graphs the region between the graphs from one intersection point to the next \(\int_a^b [f(x) - g(x)] \, dx\) with \(f\) on top, limits from \(f(x) = g(x)\), split where the graphs cross
“both regions have equal size” two regions have the same area (not: the integral over both is 0) compute both areas separately and compare them
consumer surplus what the buyers gain: they would have paid more than the market price \(CS = \int_0^{q^*} [D(q) - p^*] \, dq\)
producer surplus what the sellers gain: they would have sold for less than the market price \(PS = \int_0^{q^*} [p^* - S(q)] \, dq\)
total surplus consumer surplus + producer surplus \(\int_0^{q^*} [D(q) - S(q)] \, dq\), the area between demand and supply
minimum price a price set by the state: no unit may be sold for less above \(p^*\): only the quantity demanded at this price is traded, and the surpluses end there. Below \(p^*\): no effect
average value the constant height that gives the same area (the same total) over the interval \(f_{avg} = \frac{1}{b - a}\int_a^b f(x) \, dx\), “average monthly … within the first \(b\) months”: \(\frac{1}{b}\int_0^b f(t) \, dt\)
“prove that \(F\) is an antiderivative of \(f\)” show it by differentiating, no integrating needed compute \(F'(x)\), e.g. with the product and chain rule, and show \(F'(x) = f(x)\)
integration by parts the integration rule that reverses the product rule \(\int u \, dv = uv - \int v \, du\), choose \(u\) with LIATE
mean the sum of all values divided by their number \(\bar{x} = \dfrac{1}{n}\sum x_i\)
median the middle value of the sorted data odd \(n\): the middle value, even \(n\): the mean of the two middle values
quartiles \(Q_1\) and \(Q_3\): the medians of the lower and the upper half of the sorted data interquartile range \(\text{IQR} = Q_3 - Q_1\), the middle 50 % of the data. With an odd \(n\), leave the median out of both halves
box plot a picture of the five-number summary a box from \(Q_1\) to \(Q_3\) with a line at the median, whiskers to the smallest and the largest value that are not outliers
outlier a value far away from the rest of the data a value outside the fences, i.e. outside \(\left[Q_1 - 1.5 \cdot \text{IQR};\ Q_3 + 1.5 \cdot \text{IQR}\right]\)
sample standard deviation the typical distance of the values of a sample from their mean \(s = \sqrt{\dfrac{\sum (x_i - \bar{x})^2}{n - 1}}\), for a whole population divide by \(N\)
sample space the set of all possible outcomes of a random experiment \(S\), e.g. \(S = \{1, 2, \ldots, 6\}\) for one die
event a set of outcomes \(A \subseteq S\), with the probability \(P(A)\) between 0 and 1
complement (German: Gegenwahrscheinlichkeit) “not \(A\)”, everything outside \(A\) \(P(\bar{A}) = 1 - P(A)\), e.g. “at least one” \(= 1 -\) “none”
mutually exclusive the events cannot happen together \(A \cap B = \emptyset\), so \(P(A \cup B) = P(A) + P(B)\)
independent one event does not change the probability of the other \(P(A \cap B) = P(A) \cdot P(B)\), or \(P(A \mid B) = P(A)\), or observed count = expected count
“of the …”, “among the …”, “given that …” names the group that a percentage refers to: the condition behind the bar: “of the members, 30 % redeem” is \(P(R \mid M) = 0.3\)
tree diagram the stages of a random experiment, with a probability on every branch the branches that leave one point add up to 1, the second stage carries conditional probabilities
path probability the probability of one path through the tree multiply along the path (multiplication rule), all path probabilities add up to 1
law of total probability the probability of an event that several paths lead to add these paths: \(P(B) = P(A) \cdot P(B \mid A) + P(\bar{A}) \cdot P(B \mid \bar{A})\)
Bayes’ theorem reverses a conditional probability \(P(A \mid B) = \dfrac{P(A) \cdot P(B \mid A)}{P(B)}\), with \(P(B)\) from the law of total probability
base rate (prevalence) the share of all items (or people) that have the property, before any test \(P(D)\)
true positive rate (sensitivity) the share of the items with the property that test positive \(P(+ \mid D)\)
true negative rate (specificity) the share of the items without the property that test negative \(P(- \mid \bar{D})\)
false positive, false negative a positive result without the property, a negative result with the property \(\bar{D} \cap +\) and \(D \cap -\), with the rates \(P(+ \mid \bar{D}) = 1 -\) true negative rate and \(P(- \mid D) = 1 -\) true positive rate
“… with a positive test result does indeed …” among the positive results, the ones that are right the positive predictive value \(P(D \mid +)\), the base is all positive results
“… has … but is incorrectly identified as …” an item has the property, but the test says it has not, seen among all items the joint probability \(P(D \cap -) = P(D) \cdot P(- \mid D)\), one path of the tree
positive predictive value, negative predictive value the share of the positive results that are right, and the share of the negative results that are right \(P(D \mid +)\) and \(P(\bar{D} \mid -)\)
“Display the fractions in a contingency table. Fill in the blank fields.” write every given share into its field or total, then complete the table the fields of a row add up to the row total, the totals add up to the grand total (1 or 100 %)
“State the businesslike meaning of the quantities …” say in words what a set means for the company e.g. \(\bar{M} \cap R\): “a customer who is not a member and redeems the coupon”
expected count the count of a field if the two properties were independent \(E = \dfrac{\text{row total} \cdot \text{column total}}{\text{grand total}}\), compare it with the observed count
binomial distribution the number of successes in \(n\) trials \(P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}\). Requirements: fixed \(n\), two outcomes per trial, the same \(p\) in every trial, independent trials
“exactly / not more than / at least … but not more than …” which values of \(X\) an event contains \(P(X = k)\), \(P(X \leq k)\), \(P(a \leq X \leq b)\) with both endpoints. “Fewer than” and “more than” leave out the endpoint
expected value the long-run average of a random variable \(E[X] = \sum x_i \cdot P(X = x_i)\). Binomial: \(\mu = n \cdot p\) and \(\sigma = \sqrt{n \cdot p \cdot (1 - p)}\)
“with a likelihood of at least …” the probability must reach the given value \(P(\ldots) \geq\) the value, e.g. \(1 - \left(1 - p\right)^n \geq 0.95\) for “at least one”
minimum number (of trials) the smallest whole number \(n\) that meets a condition solve with logarithms, flip the sign when dividing by \(\ln(1 - p) < 0\), round up and check \(n - 1\) and \(n\)

In set-builder notation, \(\{x \in \mathbb{R} \mid \dots\}\) and \(\{x \in \mathbb{R} : \dots\}\) mean the same.

Set B: Practice Mock (180 minutes, 100 points)

Problem 1: Wrong Orders at a Drive-Through [30 pts. total]

A fast-food restaurant takes 200 orders at its drive-through on every lunch shift. Customers either order in advance with the restaurant’s app (\(A\)) or at the speaker in the drive-through lane (\(\bar{A}\)). An order is wrong (\(W\)) if an item is missing or is not what the customer ordered. On 14 randomly chosen lunch shifts, the numbers of wrong orders were

\[18,\ 13,\ 21,\ 16,\ 11,\ 19,\ 15,\ 24,\ 14,\ 17,\ 20,\ 12,\ 16,\ 8\]

25 % of all orders are app orders, and 2 % of the app orders are wrong.

  1. Determine the median, the quartiles \(Q_1\) and \(Q_3\) and the interquartile range of the number of wrong orders per shift. On a 15th shift, a new employee worked at the speaker, and 29 orders were wrong. Decide with the fences whether this shift is an outlier compared with the 14 shifts. [5 pts.]

  2. Compute the mean and the sample standard deviation of the number of wrong orders per shift. Use the mean to estimate the probability that an order is wrong.

(for verification: \(P(W) = 8\,\%\)) [4 pts.]

  1. The shift manager claims that one in ten lane orders is wrong. Draw a tree diagram with the type of order in the first stage and the unknown \(x = P(W \mid \bar{A})\) on the branch from \(\bar{A}\) to \(W\). Prove the manager’s claim with the law of total probability. [4 pts.]

  2. An assistant says: “App orders are a quarter of all orders, so they are also a quarter of the wrong orders.” Compute the probability that a wrong order is an app order and assess the assistant’s statement. Check with \(P(A \cap W) = P(A) \cdot P(W)\) whether the events \(A\) and \(W\) are independent, and explain what the result means for the restaurant. [5 pts.]

  3. On an evening shift, the drive-through takes 150 orders. In the evening, 60 % of the orders are app orders. As at lunch, 2 % of the app orders are wrong, but 15 % of the lane orders are wrong. The restaurant puts an extra employee at the pickup window on every shift on which more than 12 wrong orders are expected. Compute the probability that an order on an evening shift is wrong and the expected number of wrong orders per evening shift. Decide for the lunch shift and for the evening shift whether the extra employee is needed. [3 pts.]

  4. On one lunch shift, a trainer checks 22 lane orders. Each lane order is wrong with probability 10 %, independently of the other orders. Compute the probability that

(1) exactly 2 of the checked orders are wrong,

(2) fewer than 2 of the checked orders are wrong,

(3) more than 3 of the checked orders are wrong.

[5 pts.]

  1. A mystery shopper (a person who is paid to test the service like a normal customer) orders with the app, on different days. Determine how many app orders the mystery shopper must place to receive at least one wrong order with a likelihood of at least 90 %. [4 pts.]

a) Sorted data (\(n = 14\)): \(8,\ 11,\ 12,\ 13,\ 14,\ 15,\ 16,\ 16,\ 17,\ 18,\ 19,\ 20,\ 21,\ 24\)

Median: \(n\) is even, so the median is the mean of the 7th and the 8th value: \(\tilde{x} = \frac{16 + 16}{2} = 16\).

Quartiles: the lower half \(8, 11, 12, 13, 14, 15, 16\) has the median \(Q_1 = 13\), the upper half \(16, 17, 18, 19, 20, 21, 24\) has the median \(Q_3 = 19\). So \(\text{IQR} = 19 - 13 = 6\).

Fences: \(1.5 \cdot \text{IQR} = 9\), so

\[Q_1 - 9 = 4, \qquad Q_3 + 9 = 28\]

The 15th shift with 29 wrong orders lies above the upper fence 28: it is an outlier. (All 14 shifts lie between 4 and 28.)

Grading: 1 Pt. sorted data and median, 2 Pts. quartiles, 1 Pt. IQR and fences, 1 Pt. decision.

b) Mean: the 14 values add up to 224, so \(\bar{x} = \frac{224}{14} = 16\) wrong orders per shift.

Sample standard deviation (the 14 shifts are a sample, divide by \(n - 1 = 13\)). Deviations \(x_i - 16\): \(2, -3, 5, 0, -5, 3, -1, 8, -2, 1, 4, -4, 0, -8\).

\[\sum (x_i - \bar{x})^2 = 4 + 9 + 25 + 0 + 25 + 9 + 1 + 64 + 4 + 1 + 16 + 16 + 0 + 64 = 238\]

\[s^2 = \frac{238}{13} \approx 18.31, \qquad s = \sqrt{\tfrac{238}{13}} \approx 4.28\]

Probability: on average, 16 of the 200 orders of a shift are wrong (relative frequency ≈ probability):

\[P(W) \approx \frac{16}{200} = 0.08 = 8\,\%\]

Grading: 1 Pt. mean, 2 Pts. sample standard deviation (with \(n - 1\)), 1 Pt. \(P(W) = 0.08\).

c) Tree diagram (first stage: type of order, second stage: wrong or not):

  • \(A\) (\(0.25\)) \(\to\) \(W\) (\(0.02\)): \(0.25 \cdot 0.02 = 0.005\), \(\quad A \to \bar{W}\) (\(0.98\)): \(0.25 \cdot 0.98 = 0.245\)
  • \(\bar{A}\) (\(0.75\)) \(\to\) \(W\) (\(x\)): \(0.75x\), \(\quad \bar{A} \to \bar{W}\) (\(1 - x\)): \(0.75(1 - x)\)

Law of total probability with \(P(W) = 0.08\) from b):

\[P(W) = 0.25 \cdot 0.02 + 0.75 \cdot x = 0.005 + 0.75x = 0.08 \iff 0.75x = 0.075 \iff x = 0.1\]

So \(P(W \mid \bar{A}) = 0.1\): one in ten lane orders is wrong ✓. The paths of the lane orders are \(0.075\) and \(0.675\), and \(0.005 + 0.245 + 0.075 + 0.675 = 1\) ✓.

Grading: 1 Pt. tree with all branch probabilities (including the complements), 1 Pt. law of total probability, 1 Pt. solving for \(x\), 1 Pt. path probabilities with the check.

d) Conditional probability with the base “wrong orders”:

\[P(A \mid W) = \frac{P(A \cap W)}{P(W)} = \frac{0.005}{0.08} = 0.0625 = 6.25\,\%\]

The assistant is wrong: app orders are 25 % of all orders, but only 6.25 % of the wrong orders.

Independence test: \(P(A \cap W) = 0.005\), but \(P(A) \cdot P(W) = 0.25 \cdot 0.08 = 0.02\). Since \(0.005 \neq 0.02\), \(A\) and \(W\) are not independent. (The assistant’s statement would only be true if they were.)

Meaning: an app order is wrong much less often (2 %) than a lane order (10 %), probably because nobody has to understand the order over the speaker. Every customer who switches to the app lowers the number of wrong orders.

Grading: 1 Pt. \(P(A \mid W)\) with the base \(W\), 1 Pt. assessment, 1 Pt. independence test \(P(A \cap W)\) vs \(P(A) \cdot P(W)\), 1 Pt. decision, 1 Pt. meaning.

e) Let \(p_E\) be the probability that an order on an evening shift is wrong. Law of total probability:

\[p_E = 0.6 \cdot 0.02 + 0.4 \cdot 0.15 = 0.012 + 0.06 = 0.072\]

Expected number of wrong orders (expected value \(n \cdot p\)):

\[\text{evening: } 150 \cdot 0.072 = 10.8, \qquad \text{lunch: } 200 \cdot 0.08 = 16\]

Check: on an evening shift, \(90 \cdot 0.02 + 60 \cdot 0.15 = 1.8 + 9 = 10.8\) ✓.

Decision: the lunch shift needs the extra employee (\(16 > 12\)), the evening shift does not (\(10.8 \leq 12\)). A lane order is wrong more often in the evening, but more customers order with the app, so even the share of wrong orders is lower (7.2 % instead of 8 %).

Grading: 1 Pt. \(p_E = 0.072\) with the law of total probability, 1 Pt. 10.8 and 16 expected wrong orders, 1 Pt. decision for both shifts.

f) \(X\) = number of wrong orders among the 22 checked lane orders. Fixed \(n = 22\), wrong or not, the same \(p = 0.1\), independent orders, so \(X \sim B(22;\ 0.1)\):

\[P(X = k) = \binom{22}{k} \cdot 0.1^k \cdot 0.9^{22 - k}\]

(1) Exactly 2:

\[P(X = 2) = 231 \cdot 0.1^2 \cdot 0.9^{20} \approx 0.2808\]

(2) Fewer than 2 (0 or 1, “fewer than” leaves out 2):

\[P(X \leq 1) = 0.9^{22} + 22 \cdot 0.1 \cdot 0.9^{21} \approx 0.0985 + 0.2407 \approx 0.3392\]

(3) More than 3 (4 or more), with the complement:

\[P(X \geq 4) = 1 - P(X \leq 3) \approx 1 - (0.0985 + 0.2407 + 0.2808 + 0.2080) \approx 1 - 0.8281 = 0.1719\]

with \(P(X = 3) = 1540 \cdot 0.1^3 \cdot 0.9^{19} \approx 0.2080\) (sums of the unrounded terms).

Grading: 1 Pt. binomial model with \(n\) and \(p\), 1 Pt. exactly 2, 1 Pt. fewer than 2, 2 Pts. more than 3 with the complement.

g) An app order is wrong with \(p = 0.02\). Complement: “at least one wrong order” \(= 1 -\) “no wrong order”:

\[1 - 0.98^n \geq 0.9 \iff 0.98^n \leq 0.1\]

Logarithm (power law): \(n \cdot \ln(0.98) \leq \ln(0.1)\). Since \(\ln(0.98) < 0\), dividing by it reverses the inequality sign:

\[n \geq \frac{\ln(0.1)}{\ln(0.98)} \approx \frac{-2.3026}{-0.020203} \approx 113.97\]

Round up: the mystery shopper must place 114 app orders. Check: \(1 - 0.98^{113} \approx 0.8980 < 0.9\), but \(1 - 0.98^{114} \approx 0.9001 \geq 0.9\) ✓.

Grading: 1 Pt. complement inequality, 1 Pt. logarithm, 1 Pt. sign reversal, 1 Pt. rounding up with the check.

Problem 2: Testing Honey at an Importer [35 pts. total]

An importer buys honey in barrels from beekeepers’ cooperatives abroad and bottles it for supermarket chains. Some barrels are mixed with cheap sugar syrup (\(M\)). The importer tests every barrel with a quick test. Barrels with a positive result (\(+\)) go back to the supplier, barrels with a negative result (\(-\)) are treated as pure honey and are delivered to the supermarkets. The prevalence (base rate) of barrels mixed with syrup is 3 %, the test’s sensitivity (true positive rate) is 94 % and its specificity (true negative rate) is 96 %.

  1. Create a tree diagram for the testing and complete the following table for 20,000 barrels. [7 pts.]
positive (\(+\)) negative (\(-\)) sum
mixed with syrup (\(M\))
pure honey (\(\bar{M}\))
sum 20,000

(for verification: 1,340 of the 20,000 barrels test positive)

  1. Compute the probability that a barrel with a positive test result does indeed contain syrup. Explain why this probability is so much lower than the sensitivity of 94 %. [4 pts.]

  2. Compute the probability that a barrel contains syrup but is incorrectly identified as pure honey. Compute the negative predictive value, i.e. the share of the delivered barrels that really are pure honey. [4 pts.]

  3. Consider a random sample of 140 barrels from all barrels that arrive, before the test. Compute the probability for the following events:

(1) Exactly 3 barrels are mixed with syrup.

(2) Not more than 6 barrels are mixed with syrup.

(3) At least 4 but not more than 10 barrels are mixed with syrup.

[5 pts.]

  1. Before the quick test existed, the importer sent barrels to a laboratory one by one. Determine the number of barrels which need to be tested to find at least one barrel mixed with syrup with a likelihood of 97 %. [4 pts.]

  2. The importer works with 10 cooperatives. This season, 4 of them delivered at least one barrel mixed with syrup. A food inspector visits 3 of the 10 cooperatives, picked at random. Give the number of ways to pick the 3 cooperatives. Compute the probability that the inspector’s visits include exactly 2 of the cooperatives that delivered syrup, and the probability that the inspector visits at least one cooperative that delivered syrup. [5 pts.]

  3. For their own brand, the supermarket chains accept the honey only if at most 1 of 1,000 delivered barrels contains syrup. Only the sensitivity of the test can be changed. Find the smallest sensitivity the test needs and the largest false negative rate that goes with it.

Hint: The denominator of a probability is positive, so multiplying an inequality by it keeps the inequality sign.

[6 pts.]

a) Given: \(P(M) = 0.03\) (prevalence), \(P(+ \mid M) = 0.94\) (sensitivity), \(P(- \mid \bar{M}) = 0.96\) (specificity). Complements: \(P(\bar{M}) = 0.97\), \(P(- \mid M) = 0.06\) (false negative rate), \(P(+ \mid \bar{M}) = 0.04\) (false positive rate).

Tree diagram (first stage: syrup or not, second stage: test result, path probability = product along the path):

  • \(M\) (\(0.03\)) \(\to\) \(+\) (\(0.94\)): \(0.0282\), \(\quad M \to -\) (\(0.06\)): \(0.0018\)
  • \(\bar{M}\) (\(0.97\)) \(\to\) \(+\) (\(0.04\)): \(0.0388\), \(\quad \bar{M} \to -\) (\(0.96\)): \(0.9312\)

Check: \(0.0282 + 0.0018 + 0.0388 + 0.9312 = 1\) ✓

Table (path probability \(\cdot\) 20,000):

\(+\) \(-\) sum
\(M\) 564 36 600
\(\bar{M}\) 776 18,624 19,400
sum 1,340 18,660 20,000

Check: \(600 + 19{,}400 = 1{,}340 + 18{,}660 = 20{,}000\) ✓

Grading: 2 Pts. tree diagram with the complements, 4 Pts. inner fields and totals, 1 Pt. check.

b) Positive predictive value, the base is all 1,340 barrels with a positive result (Bayes’ theorem):

\[P(M \mid +) = \frac{P(M) \cdot P(+ \mid M)}{P(+)} = \frac{0.0282}{0.0282 + 0.0388} = \frac{564}{1340} \approx 0.4209 = 42.09\,\%\]

Explanation: the sensitivity \(P(+ \mid M)\) has the base “barrels mixed with syrup”, the predictive value the base “barrels with a positive result”. Syrup is rare (3 %), so the 4 % false positives among the many barrels of pure honey (776) outnumber the barrels with syrup that are found (564). More than half of the barrels that go back to the suppliers are pure honey (base rate effect).

Grading: 1 Pt. base 1,340 (Bayes’ theorem), 1 Pt. result, 2 Pts. explanation with the base rate and the false positives.

c) Joint probability, the base is all 20,000 barrels (one path of the tree):

\[P(M \cap -) = \frac{36}{20000} = 0.0018 = 0.18\,\%\]

Negative predictive value, the base is the 18,660 barrels with a negative result that are delivered:

\[P(\bar{M} \mid -) = \frac{18624}{18660} \approx 0.9981 = 99.81\,\%\]

So about 99.81 % of the delivered barrels are pure honey: 36 of the 18,660 delivered barrels contain syrup.

Grading: 1 Pt. joint probability with the base 20,000, 1 Pt. value, 1 Pt. negative predictive value with the base 18,660, 1 Pt. value.

d) \(X\) = number of barrels mixed with syrup in the sample. Very many barrels arrive, so each barrel of the sample is mixed with syrup with the constant probability \(0.03\), independently of the others: \(X \sim B(140;\ 0.03)\).

\[P(X = k) = \binom{140}{k} \cdot 0.03^k \cdot 0.97^{140 - k}\]

(1) Exactly 3:

\[P(X = 3) = 447{,}580 \cdot 0.03^3 \cdot 0.97^{137} \approx 447{,}580 \cdot 0.000027 \cdot 0.015407 \approx 0.1862\]

Events (2) and (3) have seven terms each, so use the cumulative distribution \(P(X \leq x)\) of the calculator.

Calculator: MENU → 7 → ▼ → 1 → 2 (Variable), then x, N = 140 and p = 0,03, each with =, and = again gives \(P(X \leq x)\). For \(P(X = x)\), use MENU → 7 → 4 (Binomial-Dichte) → 2 in the same way.

(2) Not more than 6 (0 to 6):

\[P(X \leq 6) \approx 0.8706\]

(3) At least 4 but not more than 10 (both endpoints included, so \(X \leq 3\) is taken away):

\[P(4 \leq X \leq 10) = P(X \leq 10) - P(X \leq 3) \approx 0.9965 - 0.3920 = 0.6045\]

Grading: 1 Pt. binomial model with \(n\) and \(p\), 1 Pt. exactly 3, 1 Pt. not more than 6 with the cumulative distribution, 2 Pts. between 4 and 10 as \(P(X \leq 10) - P(X \leq 3)\).

e) A tested barrel is mixed with syrup with \(p = 0.03\). Complement:

\[1 - 0.97^n \geq 0.97 \iff 0.97^n \leq 0.03\]

Logarithm: \(n \cdot \ln(0.97) \leq \ln(0.03)\). Since \(\ln(0.97) < 0\), dividing by it reverses the sign:

\[n \geq \frac{\ln(0.03)}{\ln(0.97)} \approx \frac{-3.5066}{-0.030459} \approx 115.12\]

Round up: 116 barrels. Check: \(1 - 0.97^{115} \approx 0.9699 < 0.97\), but \(1 - 0.97^{116} \approx 0.9708 \geq 0.97\) ✓.

Grading: 1 Pt. complement inequality, 1 Pt. logarithm, 1 Pt. sign reversal, 1 Pt. rounding up with the check.

f) Number of ways: 3 of 10 cooperatives, the order does not matter, no cooperative twice, so a combination:

\[\binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120\]

Exactly 2 that delivered syrup: 2 of the 4 cooperatives that delivered syrup and 1 of the 6 other cooperatives (counting principle):

\[P(\text{exactly } 2) = \frac{\binom{4}{2} \cdot \binom{6}{1}}{\binom{10}{3}} = \frac{6 \cdot 6}{120} = \frac{36}{120} = 0.3\]

At least one that delivered syrup: complement “all 3 from the 6 other cooperatives”:

\[P(\text{at least } 1) = 1 - \frac{\binom{6}{3}}{\binom{10}{3}} = 1 - \frac{20}{120} = \frac{5}{6} \approx 0.8333\]

The cooperatives are picked without replacement from only 10 cooperatives, so the binomial distribution does not apply here.

Grading: 1 Pt. combination with the reason, 2 Pts. exactly 2, 2 Pts. at least one with the complement.

g) Let \(g = P(- \mid M)\) be the false negative rate. The paths that end in \(-\) are \(P(M \cap -) = 0.03g\) and \(P(\bar{M} \cap -) = 0.97 \cdot 0.96 = 0.9312\).

Condition: at most 1 of 1,000 delivered barrels contains syrup, so the negative predictive value is at least 99.9 %:

\[P(\bar{M} \mid -) = \frac{0.9312}{0.03g + 0.9312} \geq 0.999\]

The denominator is positive, so multiplying by it keeps the sign:

\[0.9312 \geq 0.999 \cdot (0.03g + 0.9312) = 0.02997g + 0.9302688\]

\[0.0009312 \geq 0.02997g \iff g \leq \frac{0.0009312}{0.02997} \approx 0.031071\]

A maximum is rounded down and a minimum up, so that the rounded values still meet the demand. The false negative rate may be at most 3.10 %, so the sensitivity must be at least \(1 - 0.0310 = 96.90\,\%\).

Check with \(g = 0.031\): \(\frac{0.9312}{0.00093 + 0.9312} \approx 0.999002 \geq 0.999\) ✓. Today \(g = 6\,\%\) gives a negative predictive value of about 99.81 % (part c), below the demand: about 0.19 % of the delivered barrels contain syrup, almost twice the allowed 0.1 %. The importer needs a test that finds clearly more of the barrels with syrup.

Grading: 1 Pt. unknown \(g\) in the tree, 1 Pt. inequality with the negative predictive value, 2 Pts. solving (with the positive denominator), 1 Pt. largest false negative rate, 1 Pt. smallest sensitivity.

Problem 3: Season Tickets and Interval Drinks at a Concert Hall [35 pts. total]

A concert hall sells season tickets (one ticket for all concerts of a season) and single tickets. 2,000 visitors attend one concert. Let \(T\) be the event “the visitor has a season ticket” and \(B\) the event “the visitor buys a drink at the bar during the interval”. 30 % of the visitors buy a drink. Of the visitors who buy a drink, 60 % have a season ticket. 42 % of all visitors have no season ticket and buy no drink.

  1. Display the fractions in a contingency table. Fill in the blank fields. Then give the table of counts for the 2,000 visitors. [7 pts.]
season ticket (\(T\)) single ticket (\(\bar{T}\)) sum
buys a drink (\(B\))
buys no drink (\(\bar{B}\))
sum 100 %

(for verification: 46 % of the visitors have a season ticket)

  1. Compute with Bayes’ theorem the probability that a visitor with a season ticket buys a drink. Compute the same probability for a visitor with a single ticket. [5 pts.]

  2. The bar manager thinks that having a season ticket and buying a drink have nothing to do with each other. Check this with the expected count of the field \(T \cap B\). Explain what the result means for planning the bar. [4 pts.]

  3. Argue whether the following statements are true or false.

(1) If the bar manager were right, 324 visitors with a single ticket would buy a drink.

(2) 58 % of the visitors have a season ticket or buy a drink.

(3) 60 % of the visitors with a season ticket buy a drink.

(4) The events “has a season ticket” and “buys no drink” are mutually exclusive.

[4 pts.]

  1. For a survey about new drinks, the bar manager wants to interview visitors who have a single ticket and buy a drink. At the next concert, the manager speaks to randomly chosen visitors one after another. Set up an inequality for the number \(n\) of visitors the manager has to speak to so that, with a likelihood of at least 95 %, at least one of them has a single ticket and buys a drink, and solve it. Explain why rounding to the nearest whole number would give a wrong answer here. [4 pts.]

At the next concert, 20 visitors are chosen at random. \(X\) is the number of them who buy a drink during the interval. The shares from a) stay the same.

  1. State the four requirements of a binomial distribution and explain for each of them whether it holds for \(X\). Compute the expected value and the standard deviation of \(X\), and interpret the expected value. [5 pts.]

  2. Compute the probability for the following events:

(1) Exactly 5 of the 20 visitors buy a drink.

(2) Not more than 8 of the 20 visitors buy a drink.

(3) At least 5 but not more than 11 of the 20 visitors buy a drink.

[6 pts.]

a) Given: \(P(B) = 0.30\) (a total), \(P(T \mid B) = 0.60\) (“of the visitors who buy a drink”, a conditional probability) and \(P(\bar{T} \cap \bar{B}) = 0.42\) (a field, “of all visitors”).

Fields and totals, one after the other:

  1. \(P(T \cap B) = P(B) \cdot P(T \mid B) = 0.30 \cdot 0.60 = 0.18\) (multiplication rule)
  2. \(P(\bar{T} \cap B) = 0.30 - 0.18 = 0.12\)
  3. \(P(\bar{T}) = 0.12 + 0.42 = 0.54\), so \(P(T) = 1 - 0.54 = 0.46\)
  4. \(P(T \cap \bar{B}) = 0.46 - 0.18 = 0.28\) and \(P(\bar{B}) = 1 - 0.30 = 0.70\)
\(T\) \(\bar{T}\) sum
\(B\) 18 % 12 % 30 %
\(\bar{B}\) 28 % 42 % 70 %
sum 46 % 54 % 100 %

Check: \(0.28 + 0.42 = 0.70\) ✓. Counts (each field \(\cdot\) 2,000):

\(T\) \(\bar{T}\) sum
\(B\) 360 240 600
\(\bar{B}\) 560 840 1,400
sum 920 1,080 2,000

Grading: 1 Pt. \(P(T \cap B)\) with the multiplication rule, 1 Pt. \(P(\bar{T} \cap B)\), 1 Pt. \(P(\bar{T})\) from the given field, 1 Pt. remaining fields, 1 Pt. totals with the check, 2 Pts. table of counts.

b) Bayes’ theorem (the condition is now \(T\), the base is the 920 visitors with a season ticket):

\[P(B \mid T) = \frac{P(B) \cdot P(T \mid B)}{P(T)} = \frac{0.30 \cdot 0.60}{0.46} = \frac{0.18}{0.46} = \frac{9}{23} \approx 0.3913 = 39.13\,\%\]

Single ticket (base: the 1,080 visitors with a single ticket):

\[P(B \mid \bar{T}) = \frac{P(\bar{T} \cap B)}{P(\bar{T})} = \frac{0.12}{0.54} = \frac{240}{1080} = \frac{2}{9} \approx 0.2222 = 22.22\,\%\]

Grading: 1 Pt. Bayes’ theorem with the correct base, 1 Pt. \(P(T) = 0.46\) from a), 1 Pt. \(P(B \mid T)\), 1 Pt. base \(\bar{T}\), 1 Pt. \(P(B \mid \bar{T})\).

c) Expected count if \(T\) and \(B\) were independent:

\[E_{T,B} = \frac{\text{row total} \cdot \text{column total}}{\text{grand total}} = \frac{600 \cdot 920}{2000} = 276\]

Observed: 360 visitors with a season ticket buy a drink. \(360 \neq 276\), so \(T\) and \(B\) are not independent: the bar manager is wrong. (Equivalently, \(P(B \mid T) \approx 0.3913 \neq 0.30 = P(B)\).)

Meaning: visitors with a season ticket buy a drink much more often (39.13 %) than visitors with a single ticket (22.22 %), 84 more than expected under independence. For a concert with many season-ticket holders, the bar needs more staff and more drinks. It could also let season-ticket holders order their interval drink in advance.

Grading: 1 Pt. expected count 276 with the formula, 1 Pt. comparison and decision, 2 Pts. meaning with the two conditional shares.

d)

  1. True. If the manager were right, \(T\) and \(B\) would be independent (part c). The expected count of the field \(\bar{T} \cap B\) is then \(E_{\bar{T},B} = \frac{600 \cdot 1080}{2000} = 324\), i.e. 30 % of the 1,080 visitors with a single ticket. In fact only 240 of them buy a drink.
  2. True. “Season ticket or drink” is \(T \cup B\), and its complement “neither” is the given field \(\bar{T} \cap \bar{B}\): \(P(T \cup B) = 1 - P(\overline{T \cup B}) = 1 - 0.42 = 0.58\).
  3. False. 60 % is \(P(T \mid B)\), the share of season-ticket holders among the drink buyers. The share of drink buyers among the season-ticket holders is \(P(B \mid T) \approx 39.13\,\%\): the bases differ.
  4. False. Mutually exclusive would mean \(T \cap \bar{B} = \emptyset\). But \(P(T \cap \bar{B}) = 0.28 \neq 0\): 560 visitors have a season ticket and buy no drink.

Grading: 1 Pt. statement (1), 1 Pt. statement (2), 1 Pt. statement (3), 1 Pt. statement (4), each with the decision and the reason.

e) A randomly chosen visitor has a single ticket and buys a drink with (from the table)

\[p = P(\bar{T} \cap B) = \frac{240}{2000} = 0.12\]

Complement: \(1 - 0.88^n \geq 0.95 \iff 0.88^n \leq 0.05\)

Logarithm: \(n \cdot \ln(0.88) \leq \ln(0.05)\). Since \(\ln(0.88) < 0\), dividing by it reverses the sign:

\[n \geq \frac{\ln(0.05)}{\ln(0.88)} \approx \frac{-2.9957}{-0.12783} \approx 23.43\]

The manager must speak to at least 24 visitors.

Why not the nearest whole number: \(n\) must be a whole number with \(n \geq 23.43\). The nearest whole number 23 lies below this bound: \(1 - 0.88^{23} \approx 0.9471 < 0.95\), so 23 visitors are not enough. The probability \(1 - 0.88^n\) grows with \(n\), so the smallest whole number above the bound is the answer: \(1 - 0.88^{24} \approx 0.9535 \geq 0.95\) ✓.

Grading: 1 Pt. \(p = 0.12\) from the table, 1 Pt. complement inequality and logarithm with the sign reversal, 1 Pt. \(n = 24\), 1 Pt. explanation with the check of 23.

f) Requirements of the binomial distribution:

  1. Fixed number of trials: \(n = 20\) visitors ✓
  2. Two outcomes per trial: buys a drink or not ✓
  3. The same probability in every trial: each randomly chosen visitor buys a drink with \(p = P(B) = 0.3\) ✓
  4. Independent trials: the 20 visitors are chosen at random from a large audience, so one visitor’s decision does not change the probability for another. This is an assumption: friends who come together often buy together.

So \(X \sim B(20;\ 0.3)\):

\[\mu = n \cdot p = 20 \cdot 0.3 = 6, \qquad \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{20 \cdot 0.3 \cdot 0.7} = \sqrt{4.2} \approx 2.05\]

Meaning of \(\mu\): in many groups of 20 randomly chosen visitors, on average 6 per group buy a drink during the interval. The bar can plan with about 30 % of the audience.

Grading: 1 Pt. requirements 1 to 3, 1 Pt. independence as an assumption with a reason, 1 Pt. \(\mu\), 1 Pt. meaning of \(\mu\), 1 Pt. \(\sigma\).

g)

\[P(X = k) = \binom{20}{k} \cdot 0.3^k \cdot 0.7^{20 - k}\]

(1) Exactly 5:

\[P(X = 5) = 15{,}504 \cdot 0.3^5 \cdot 0.7^{15} \approx 15{,}504 \cdot 0.00243 \cdot 0.0047476 \approx 0.1789\]

(2) Not more than 8 (0 to 8, nine terms), with the cumulative distribution (N = 20, p = 0,3):

\[P(X \leq 8) \approx 0.8867\]

(3) At least 5 but not more than 11 (both endpoints included, seven terms):

\[P(5 \leq X \leq 11) = P(X \leq 11) - P(X \leq 4) \approx 0.9949 - 0.2375 = 0.7574\]

Grading: 1 Pt. binomial formula with \(n\) and \(p\), 1 Pt. exactly 5, 2 Pts. not more than 8 as \(P(X \leq 8)\) with the value, 2 Pts. between 5 and 11 as \(P(X \leq 11) - P(X \leq 4)\).

Exam Preparation Checklist

Before the exam, check that you can:

Show all work, name the base of every probability, and interpret your results in the business context.