Unit 04 — How to Solve Anything
The heaviest-scoring unit in the portion. Six topics, each opening with how to recognise that a question belongs to it — which matters more here than anywhere else, because the whole unit is four distributions that look alike until you know the tells — then the master method, then questions worked in full, then every exercise Levin sets, solved, with the page number against each one.
Of the 31 in-scope marks, 9 sit in this unit — more than any other. Both papers put a Poisson question in Q1 (b) and a normal distribution question in Q2 (b), and both times the shape was identical: "the average is λ per interval, find P(X = 0) or P(X ≥ 1)" and "find an interval probability, then a one-tail probability".
Both Poisson versions solved → · Both normal versions solved →
Every question in Unit 4 is answered by picking the right distribution and then substituting. Picking wrong costs the entire question, so the choice is worth more than the arithmetic:
Poisson: Rate \(\lambda\) per interval (no fixed \(n\)) \(\to P(x) = \frac{\lambda^x e^{-\lambda}}{x!}\)
Normal: Continuous measurement with \(\mu, \sigma \to z = \frac{x - \mu}{\sigma}\)
Normal approx: \(np \ge 5, nq \ge 5 \implies \mu = np, \sigma = \sqrt{npq}\) (requires \(\pm 0.5\) continuity correction)
Topic 6 is nothing but this diagram, drilled. If you can run it in ten seconds you have already earned the method marks on Q1 (b) and Q2 (b).
Random Variables, Probability Distributions and Expected Value
The foundation the other five topics stand on. A probability distribution is a frequency distribution whose frequencies have been divided by the total, and the expected value is a grouped mean where the weights are probabilities instead of counts — so if you can do Unit 1's \(\Sigma fx/\Sigma f\), you can already do \(\Sigma x\,P(x)\).
| If the question gives you… | then it is… |
|---|---|
| a fixed number of trials \(n\) and a success probability \(p\) | Topic 2 — binomial |
| an average rate "per hour / per page / per 1,000" | Topic 3 — Poisson |
| a mean and a standard deviation of a measured quantity | Topic 4 — normal |
| an explicit list of outcomes and their probabilities | Here |
(discrete)
probability
random variable
from raw counts
The last one is not a separate formula — it is what \(\Sigma x P(x)\) collapses to when every period is equally likely. Two of Levin's exercises (5-9, 5-10) are exactly that.
- List the outcomes in a column. If you were given frequencies, divide by the total to get probabilities and check they sum to 1.00.
- Build the \(x\cdot P(x)\) column. Row by row.
- Total it. That total is \(E(X)\). Write the unit: rupees, accidents per month, cars.
- Answer the business question. Compare \(E(X)\) against the threshold in the question — the price, the policy limit, the rival option — and state the decision.
- Say what "expected" means if there is a spare line: it is a long-run average, not a prediction of any single outcome. An expected 7.75 accidents a month never actually happens in a month.
- Averaging the outcomes instead of weighting them. \(E(X)\) is not the mean of the \(x\) column; each \(x\) is weighted by its own probability.
- Forgetting the decision. Levin's expected-value exercises almost all end in "should he?" — a bare number scores half.
- Probabilities that do not sum to 1. Always check. If they do not, you have misread a row.
- Treating \(E(X)\) as achievable. "Expected 5.28" when the outcomes are 2, 4, 6, 8… is fine — an expected value need not be a possible outcome.
Q. Construct a table and a graph of the probability distribution of the total number of dots showing when two dice are tossed.
| Sum x | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | 1 | 2 | 3 | 4 | 5 | 6 | 5 | 4 | 3 | 2 | 1 |
| P(x) | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
Q. Which of these statements about probability distributions are correct? (a) It provides information about the long-run or expected frequency of each outcome. (b) Its graph has the possible outcomes marked on the horizontal axis. (c) It lists the probabilities that each outcome is random. (d) It is always constructed from a set of observed frequencies like a frequency distribution. (e) It may be based on subjective estimates of the likelihood of certain outcomes.
(b) Correct — outcomes on the horizontal axis, probability on the vertical.
(c) Incorrect — meaningless wording. It lists the probability that each outcome occurs, not that an outcome "is random".
(d) Incorrect — the word always kills it. A distribution can be built a priori from theory (the two dice in 5-2 needed no observation at all).
(e) Correct — subjective probabilities are one of the three legitimate ways of assigning probability (Unit 3, Concept 1).
Q. Southport Autos offers four luxury option combinations. 1. V-8 · sun roof · halogen headlights 2. Leather · power locks · cassette deck 3. Halogen · V-8 · leather 4. Cassette deck · V-8 · power locks. Combinations 2, 3 and 4 are equally likely; combination 1 is twice as likely as any of these. (a) What is the probability a customer orders a car with a V-8 engine? (b) Two customers order — construct the probability distribution of the number of V-8 engines ordered.
Q. Ritwik believes the Auto King has a 70% chance of completely replacing the Secretariat's pool of 500 cars; one chance in five that only half the 500 are replaced; and one chance in 10 that all are replaced and the pool is expanded by 20%. Construct a table and graph of the probability distribution of sales.
Q. Construct a probability distribution from the frequency distribution below, then (a) draw its graph and (b) compute the expected value of the outcome.
| Outcome | 2 | 4 | 6 | 8 | 10 | 12 | 15 |
|---|---|---|---|---|---|---|---|
| Frequency | 24 | 22 | 16 | 12 | 7 | 3 | 1 |
Q. From the graph of a probability distribution with spikes at 8,000 → 0.05; 9,000 → 0.15; 10,000 → 0.25; 11,000 → 0.30; 12,000 → 0.20; 13,000 → 0.05: (a) construct the table, (b) find the expected value.
Q. Bill Johnson bought a $300 VCR and can buy a 5-year extended warranty for $100. He believes maintenance expenses over the next five years will be:
| Expense ($) | 0 | 50 | 100 | 150 | 200 | 250 | 300 |
|---|---|---|---|---|---|---|---|
| Probability | 0.35 | 0.25 | 0.15 | 0.10 | 0.08 | 0.05 | 0.02 |
Find the expected value of the anticipated maintenance costs. Should Bill pay $100 for the warranty?
Q. Raj Gandhi must decide whether to install a traffic light. Accidents at the intersection:
| Year | J | F | M | A | M | J | J | A | S | O | N | D |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1995 | 10 | 8 | 10 | 6 | 9 | 12 | 2 | 10 | 10 | 0 | 7 | 10 |
| 1996 | 12 | 9 | 7 | 8 | 4 | 3 | 7 | 14 | 8 | 8 | 8 | 4 |
SHA policy is to install a light where the monthly expected number of accidents exceeds 7. Should Raj recommend installation?
1996: \(12{+}9{+}7{+}8{+}4{+}3{+}7{+}14{+}8{+}8{+}8{+}4 = 92\)
Q. Ankit Desai will investigate whichever division loses more first-class couriers per month on average — trucking or air, but not both.
| Lost from | J | F | M | A | M | J | J | A | S | O | N | D |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Truck | 4 | 5 | 2 | 3 | 2 | 1 | 3 | 5 | 4 | 7 | 0 | 1 |
| Airplane | 5 | 6 | 0 | 2 | 1 | 3 | 4 | 2 | 4 | 7 | 4 | 0 |
Airplane: \(5{+}6{+}0{+}2{+}1{+}3{+}4{+}2{+}4{+}7{+}4{+}0 = 38\)
Q. Each hockey ticket costs ₹5 to print and sells for ₹25; unsold tickets are discarded. Estimated sales: 25,000 (p = 0.10), 40,000 (0.30), 55,000 (0.45), 70,000 (0.15). How many tickets should management print?
\(Q=40{,}000:\) ₹7,62,500
\(Q=55{,}000:\) ₹9,12,500
\(Q=70{,}000:\) ₹8,93,750
Q. Airport Rent-a-Car rents an economy car for ₹35/day, buys cars at ₹6,300, operates 312 days a year, has variable cost ₹2.50 per car per day, and recovers 50% of cost on resale after a year. Daily demand: 13 (0.08), 14 (0.15), 15 (0.22), 16 (0.25), 17 (0.21), 18 (0.09). Use the expected-loss method to find the optimal number of cars to buy.
Q. Outcomes 102, 105, 108, 111, 114, 117 occurred with frequencies 10, 20, 45, 15, 20, 15. Draw the probability distribution and find the expected outcome.
Q. Dharmendra buys a stock whenever the expected rate of return exceeds 10%. A ₹150 stock has returns 0 (p = 0.20), 10 (0.25), 15 (0.30), 25 (0.15), 50 (0.10). Will he buy?
Q. Mario sells 1–4 "everything but" pizzas a night. Each costs $7 to prepare and sells for $12; unsold pizzas are thrown out, and each pizza ordered but undeliverable costs $5 in lost future business. Demand: 1 (0.40), 2 (0.30), 3 (0.20), 4 (0.10). How many should he stock to minimise expected loss?
Stock 3: 14, 7, 0, 5 · Stock 4: 21, 14, 7, 0
Stock 2: \(2.8+0+1.0+1.0 = \mathbf{4.8}\)
Stock 3: \(5.6+2.1+0+0.5 = 8.2\)
Stock 4: \(8.4+4.2+1.4+0 = 14.0\)
Q. From a graph of a probability distribution over the outcomes 1–10, construct the corresponding table.
The Binomial Distribution
The distribution of "how many successes in a fixed number of tries". Everything about it follows from four conditions, and almost every mistake made on it is a failure to translate the English phrase — more than two, at least one, no more than four — into the right set of \(r\) values before any arithmetic starts.
| If the question… | then it is… |
|---|---|
| gives a rate per interval with no \(n\) — "8 coughs on average" | Topic 3 — Poisson |
| gives \(n\) and \(p\) but \(n\ge20,\ p\le0.05\) | Topic 3 — Poisson approximation to the binomial |
| gives \(n\) and \(p\) with \(np\ge5\) and \(nq\ge5\), \(n\) large | Topic 5 — normal approximation |
| gives μ and σ of something measured | Topic 4 — normal |
(the shortcut)
(the workhorse)
Define "success" as whatever the question counts, even when it is a bad thing. If the question asks about defectives, then a defective is a success and \(p = 0.02\). Getting this backwards is the most expensive error in the topic.
- Check the four conditions and say so in one line: "fixed \(n=10\), two outcomes, constant \(p=0.02\), independent trials — binomial."
- Write \(n\), \(p\), \(q\) explicitly. Decide what counts as a success and state it.
- Translate the English into a set of \(r\) values — this is where the marks are lost, so do it on its own line. See the table below.
- Decide direct or complement. If the set has more than about three values, compute the complement instead. "At least one" is always \(1-q^n\).
- Substitute and evaluate term by term, keeping four decimals.
- Answer in words with the context: "there is a 99.5% chance that at least one browsing customer buys something in a given hour."
| The question says | Which \(r\) | Compute as |
|---|---|---|
| exactly 5 | r = 5 | one term |
| at least 1 | r = 1,2,…,n | \(1 - P(0) = 1 - q^n\) |
| at least 4 | r = 4,…,n | \(1 - [P(0)+P(1)+P(2)+P(3)]\) |
| more than 2 | r = 3,…,n | \(1 - [P(0)+P(1)+P(2)]\) — not including 2 |
| no more than 4 / at most 4 | r = 0,1,2,3,4 | direct sum |
| fewer than 8, with n = 7 | r = 0,…,7 (all) | \(= 1\) — a free mark if you spot it |
| one or three | r = 1 and r = 3 | \(P(1)+P(3)\) — mutually exclusive, so add |
| none | r = 0 | \(q^n\) |
- "More than 2" ≠ "2 or more". More than 2 starts at 3. Two or more starts at 2. One word, one mark.
- Using \(p\) for the wrong event. If 96% of disputes are marital and the question asks about non-marital, then \(p = 0.04\).
- \(\sigma = npq\) instead of \(\sqrt{npq}\). The variance is \(npq\); the standard deviation is its square root.
- Grinding out eight terms when the complement needs three. Always count which side is shorter first.
- Assuming independence when sampling without replacement from a small population. Levin's exercises say "assume flaws occur independently" precisely because that assumption is doing work.
Q. For a binomial distribution with \(n = 7\) and \(p = 0.2\), find (a) \(P(r=5)\), (b) \(P(r>2)\), (c) \(P(r<8)\), (d) \(P(r\ge4)\).
\(4{:}\,0.0287\) \(5{:}\,0.0043\) \(6{:}\,0.0004\) \(7{:}\,0.0000\)
Q. For \(n = 15\), \(p = 0.2\), find (a) \(P(r=6)\), (b) \(P(r\ge11)\), (c) \(P(r\le4)\).
Q. Find the mean and standard deviation of the binomial distributions: (a) n=15, p=0.20 (b) n=8, p=0.42 (c) n=72, p=0.06 (d) n=29, p=0.49 (e) n=642, p=0.21.
| n | p | q | μ = np | σ = √(npq) | |
|---|---|---|---|---|---|
| (a) | 15 | 0.20 | 0.80 | 3.00 | 1.5492 |
| (b) | 8 | 0.42 | 0.58 | 3.36 | 1.3960 |
| (c) | 72 | 0.06 | 0.94 | 4.32 | 2.0151 |
| (d) | 29 | 0.49 | 0.51 | 14.21 | 2.6920 |
| (e) | 642 | 0.21 | 0.79 | 134.82 | 10.3203 |
Q. For \(n = 8\) trials, compute \(P(r\ge1)\) for (a) p = 0.1, (b) p = 0.3, (c) p = 0.6, (d) p = 0.4.
(b) \(1-(0.7)^8 = 1-0.0576 = \mathbf{0.9424}\)
(c) \(1-(0.4)^8 = 1-0.0007 = \mathbf{0.9993}\)
(d) \(1-(0.6)^8 = 1-0.0168 = \mathbf{0.9832}\)
Q. Harley Davidson checks 10 transmissions a month for defects. Historically 2% have flaws, and flaws occur independently. (a) What is the probability the sample contains more than two flawed transmissions? (b) That none is flawed?
Q. Mayor Bruns's assistants estimate 40% of unemployment beneficiaries are secretly employed. An aide investigates 10 at random. (a) Probability that more than eight have jobs? (b) That one or three of them have jobs?
Q. Among IT professionals, P(2 weeks' vacation) = 0.45, P(1 week) = 0.10, P(3 or more weeks) = 0.20. Twenty workers are chosen at random. Find: (a) P(8 have 2 weeks), (b) P(only one has 1 week), (c) P(at most 2 have 3+ weeks), (d) P(at least 2 have 1 week).
Q. A browsing customer buys with probability 0.3, and 15 customers browse each hour. Find the probability that (a) at least one buys, (b) at least four buy, (c) none buys, (d) no more than four buy.
Q. Binomial with \(n=12\), \(p=0.45\): find (a) \(P(r=8)\), (b) \(P(r>4)\), (c) \(P(r\le10)\).
Q. Find \(\mu\) and \(\sigma\) for (a) n=16, p=0.40 (b) n=10, p=0.75 (c) n=22, p=0.15 (d) n=350, p=0.90 (e) n=78, p=0.05.
| n | p | μ = np | σ = √(npq) | |
|---|---|---|---|---|
| (a) | 16 | 0.40 | 6.40 | 1.960 |
| (b) | 10 | 0.75 | 7.50 | 1.369 |
| (c) | 22 | 0.15 | 3.30 | 1.675 |
| (d) | 350 | 0.90 | 315.00 | 5.612 |
| (e) | 78 | 0.05 | 3.90 | 1.925 |
Q. A poll gives P(alliance ABC) = 0.55, P(PQR) = 0.30, P(XYZ) = 0.15. For 10 randomly chosen people find: (a) P(four are PQR), (b) P(none is ABC), (c) P(two are XYZ), (d) P(at least eight are PQR).
Q. A call centre finds that 25% of calls are resolved on first contact. In a random sample of 12 calls, state the four conditions that make this binomial, then find the probability that (a) exactly 3 are resolved first time, (b) at least 1 is, (c) at most 2 are.
Solution
Q. A supplier claims 95% of components meet specification. A buyer inspects 20. Find the probability that (a) none fails, (b) more than 2 fail. What would you conclude if 4 failed?
Solution
Q. A firm mails 400 offers; historically 15% respond. Find the mean and standard deviation of the number of responses, and the range within which the response count will fall about 95% of the time.
Solution
Q. For each, say whether you would use the binomial directly, the Poisson approximation, or the normal approximation, and give the parameter(s): (a) n = 12, p = 0.30; (b) n = 500, p = 0.004; (c) n = 200, p = 0.40; (d) n = 40, p = 0.02.
Solution
The Poisson Distribution
Both 2025-26 papers put a Poisson question in Q1 (b), and both had the same shape: an average rate per interval is given, and you are asked for P(X = 0) or P(X ≥ 1). One formula, one rescaling rule, and the complement trick — that is the whole 4 marks.
| If the question… | then it is… |
|---|---|
| fixes \(n\) and \(p\) with \(p\) moderate (0.1–0.9) | Topic 2 — binomial, done directly |
| fixes \(n \ge 20\) with \(p \le 0.05\) | Here — Poisson approximation, \(\lambda = np\) |
| gives \(\mu\) and \(\sigma\) separately for a measured quantity | Topic 4 — normal |
| gives a mean that equals the variance | Here — that identity is unique to Poisson |
(both equal λ)
to the binomial
\(P(x=0)=e^{-\lambda}\) because \(\lambda^0 = 1\) and \(0! = 1\). This one line is the most-used result in the entire unit — both papers' Q1 (b) reduce to it.
- Find \(\lambda\) and say what interval it belongs to. Write "λ = 5 accidents per week" — the unit matters more than the number.
- Rescale if the question changes the interval. Do it before anything else, and show the multiplication.
- If you were given \(n\) and \(p\) instead, check \(n\ge20,\ p\le0.05\), say the check passes, and set \(\lambda = np\).
- Translate the phrase into a set of \(x\) values — the same table as the binomial. "More than three" starts at 4.
- Compute \(e^{-\lambda}\) once and reuse it in every term. Every term shares it.
- Use the complement whenever the set is long. Poisson has infinitely many values, so "at least \(k\)" is always done as \(1 - P(x \le k-1)\) — you cannot sum upwards to infinity.
- Answer in the story's language, with the decision if one was asked for.
| Paper | The question |
|---|---|
| Final | Q1 (b), 4 marks — an average number of events per interval, then a probability of none / at least one QP p. 1 · worked in full |
| Re-Exam | Q1 (b), 4 marks — structurally identical, different story and rate QP p. 1 · worked in full |
| Class notes | The births-at-a-hospital problem, λ = 1.8, worked to 0.0723 notes pp. 1–9 |
- Not rescaling λ. The most expensive error in the topic. Rate per 10 minutes, question about an hour ⇒ multiply by 6.
- Summing upwards for "at least". There is no upper limit in a Poisson — you must go through the complement.
- Using the Poisson approximation when \(p\) is too large. The rule is \(p \le 0.05\); at \(p = 0.3\) it is simply wrong.
- Forgetting \(0! = 1\). \(P(0)=e^{-\lambda}\), not \(0\).
- Reporting \(\sigma = \lambda\). The variance equals λ; the standard deviation is \(\sqrt{\lambda}\).
Q. Given a binomial with \(n = 28\) and \(p = 0.025\), use the Poisson approximation to find (a) \(P(r\ge3)\), (b) \(P(r<5)\), (c) \(P(r=9)\).
Q. New car prices increase an average of four times every 3 years. Find the probability of (a) no price hikes in a randomly selected 3-year period, (b) two hikes, (c) four hikes, (d) five or more.
Q. Binomial with \(n=25\), \(p=0.032\). Use the Poisson approximation to find (a) \(P(r=3)\), (b) \(P(r=5)\), (c) \(P(r\le2)\).
Q. Given \(\lambda = 6.1\), find (a) \(P(x\le3)\), (b) \(P(x\ge2)\), (c) \(P(x=6)\), (d) \(P(1\le x\le4)\).
\(P(3)=0.0848\) \(P(4)=0.1294\)
Q. On average eight coughs occur just before Donna Prima begins to play. She has sworn that if she hears more than five coughs tonight she will refuse to play. What is the probability that she will play?
\(P(3)=0.0286\) \(P(4)=0.0573\) \(P(5)=0.0916\)
Q. On average five people a week have accidents on the Lucknow station lifts. Finance will release funds only if the chance of more than three accidents in any week exceeds 75%. Will the funds be allocated?
Q. 4% of new calculators develop mathematical inconsistencies. The VP selects 50 for a demonstration. Using the Poisson as an approximation to the binomial, find (a) the chance of at least three malfunctioning, (b) that none malfunctions.
Q. 96% of the disputes handled by the Family Court are marital. Out of 80 disputes, what is the probability that (a) exactly seven are non-marital, (b) none is non-marital?
Q. Only 0.5% of printed bills are too flawed for circulation. Out of a batch of 1,000 bills, what is the probability that (a) none is too flawed, (b) ten are, (c) fifteen are?
Q. Given \(\lambda = 4.2\), find (a) \(P(x\le2)\), (b) \(P(x\ge5)\), (c) \(P(x=8)\).
Q. Binomial with \(n=30\), \(p=0.04\). Use the Poisson approximation to find \(P(r=25)\), \(P(r=3)\) and \(P(r=5)\).
Q. A helpdesk receives an average of 3 calls per 10-minute interval. Find (a) P(no calls in a 10-minute interval), (b) P(at least one call in 10 minutes), (c) P(exactly 4 calls in 10 minutes), (d) P(no calls in a full hour).
Solution
Q. A bank branch experiences an average of 2.5 loan defaults per month. Find (a) P(no defaults in a month), (b) P(at least two in a month), (c) P(at least one in a quarter). The branch manager is rated "clean" only if a whole quarter passes with no default — comment on that target.
Solution
Q. A factory ships 300 items; each has a 1% chance of being damaged in transit. Find P(exactly 2 damaged) and P(at most 1 damaged) using the Poisson approximation, and say how good the approximation is.
Solution
Q. A proof-reader finds on average 1.5 typing errors per page. Find (a) P(a page is error-free), (b) P(a page has exactly 2 errors), (c) P(a 4-page document has at least 2 errors). What is the standard deviation of errors per page?
Solution
The Normal Distribution
The largest single slot in the portion. Every normal question reduces to the same three moves: standardise with \(z=(x-\mu)/\sigma\), draw the curve and shade, then read the table and combine. The drawing is not decoration — it is what tells you whether to add the two areas or subtract them.
| If the question… | then it is… |
|---|---|
| counts successes out of a fixed \(n\) | Topic 2 — binomial |
| gives a rate per interval | Topic 3 — Poisson |
| gives \(n\) and \(p\) but asks for a range of counts, \(np\ge5\) | Topic 5 — normal approximation, with ±0.5 |
| gives μ and σ of a sample mean \(\bar{x}\) | Units 5–6 — sampling distribution, σ becomes \(\sigma/\sqrt{n}\) |
one probability
out of N
Three formulas, one idea. \(z\) is how many standard deviations \(x\) is from the mean — nothing more. Everything else is table reading.
Rule 4 versus rule 5 is where marks are lost. The sketch settles it in one second: if the shaded band crosses the centre line, add; if it sits entirely on one side, subtract.
- Write \(\mu\) and \(\sigma\) and confirm the variable is continuous and normal.
- Draw the curve. Mark \(\mu\) on the centre line, mark the \(x\) value(s), shade the region the question asks for. Thirty seconds, and it decides every sign that follows.
- Standardise each \(x\), showing the substitution. Round \(z\) to two decimals because that is all the table has — and say so when it is not exact.
- Read A(z) for each z.
- Combine by the shading: band across the mean ⇒ add; band on one side ⇒ subtract; tail ⇒ 0.5 minus.
- Convert to the answer asked for — a percentage, or a count via \(N\times P\), or an \(x\) value via \(\mu + z\sigma\).
- Say it in words in the story's own language.
| Paper | The question |
|---|---|
| Final | Q2 (b), 5 marks — an interval probability, then a one-tail "highly successful" probability QP p. 2 · worked in full |
| Re-Exam | Q2 (b), 5 marks — identical structure; produces \(z=-0.625\) and \(z=1.875\), both exactly half-way between table rows QP p. 2 · worked in full |
On non-terminating z values: when \(z\) lands between table rows, state your rounding — "taking \(z\approx1.88\)" — and the answer is accepted. Never invent a third decimal the table does not have.
- Adding when you should subtract. Both \(z\) values negative (or both positive) ⇒ subtract. Draw the picture.
- Forgetting the 0.5. A(z) is only half the curve. A tail is \(0.5-A(z)\), not \(1-A(z)\).
- Losing the sign of z. \(A(-1.25)=A(1.25)\) for the area, but which side it is on decides whether you add or subtract.
- Mixing conventions mid-paper. If you start with A(z), never quote a Φ(z) value from memory.
- Answering with a probability when a count was asked for. "How many of the 19 books" needs \(19\times P\).
- Using \(\sigma\) when the question is about a sample mean. That is Unit 5 — the spread becomes \(\sigma/\sqrt{n}\).
Q. \(X\) is normal with mean 6.4 and standard deviation 2.7. Find (a) \(P(4.0
Q. In a normal distribution with \(\sigma = 5.0\), the probability that an observation exceeds 21 is 0.14. (a) Find the mean. (b) Find the value below which 4% of the distribution lies.
Q. Weekly demand for mailing tubes is normal with mean 100, and 90% of the time demand is below 115. (a) What is the standard deviation? (b) What stock level keeps the probability of running out at 0.05 or less?
Q. A drive shaft must have diameter \(5.1 \pm 0.05\) inches. Stock rods have mean diameter 5.07″ and standard deviation 0.07″. What is the probability that a rod from inventory fits?
Q. A shop orders 110 grease cartridges a week and runs out 1 week in 4. Average use is 95 per week, demand is normal. (a) Find the standard deviation. (b) How many should he order so the chance of running out is at most 0.2?
Q. Hat size is normal with mean 7″. Stock covers 6¼″ to 7¾″. Find the probability that a recruit is fitted using (a) \(\sigma = 0.75\), (b) \(\sigma = 0.875\).
Q. Training completion time is normal with mean 44 days and standard deviation 12. Find the probability an employee finishes (a) in 33 to 42 days, (b) in fewer than 30 days, (c) in fewer than 25 or more than 60 days.
Q. Publication takes on average 11 months with standard deviation 2.4 months. Out of 19 books this year, approximately how many will be finished in less than a year?
Q. Adult height is normal with mean 73 inches and standard deviation 6. Nobb wants doors tall enough for 95% of adults to pass without stooping. How tall should the doors be?
Q. With \(\mu = 850{,}000\) and \(\sigma = 200{,}000\), find \(P(x \ge 1{,}000{,}000)\).
Q. Monthly sales at a store are normally distributed with mean ₹85,000 and standard deviation ₹12,000. (a) What is the probability that sales in a month fall between ₹70,000 and ₹95,000? (b) A month is called "highly successful" if sales exceed ₹1,05,000 — what proportion of months are highly successful? (c) Comment for the manager.
Solution
Q. Delivery times are normal with mean 42 minutes and standard deviation 8. (a) The slowest 10% of deliveries trigger a review — above what time? (b) Within what symmetric interval about the mean do the middle 95% of deliveries fall?
Solution
Q. Bolt diameters are normal with mean 10.00 mm and standard deviation 0.06 mm. The specification is 9.90 to 10.10 mm. What proportion is accepted, and what is the reject rate per 10,000 bolts?
Solution
Q. Candidate A scored 78 on a test with mean 70 and standard deviation 5. Candidate B scored 82 on a different test with mean 75 and standard deviation 8. Which performed better relative to their group?
Solution
Q. Component life is normal with mean 500 hours and standard deviation 40. Give the intervals containing about 68%, 95% and 99.7% of components, and state what fraction fail before 420 hours.
Solution
\(\pm2\sigma\): 420–580 h (95.4%)
\(\pm3\sigma\): 380–620 h (99.7%)
The Normal Approximation to the Binomial
A binomial with a large \(n\) is unusable by hand — \(P(7 \le r \le 18)\) out of 200 is twelve terms, each with a factorial of 200 in it. The normal approximation replaces all of that with two \(z\) values. The price is one extra step, the continuity correction, and that single ±0.5 is where nearly every mark in this topic is won or lost.
n ≥ 20 and p ≤ 0.05 → Poisson approximation instead, \(\lambda=np\), no continuity correction.
Neither → you must do the binomial directly.
| If the question… | then… |
|---|---|
| gives μ and σ of a measured quantity | Topic 4 — no continuity correction, ever |
| gives \(n\), \(p\) with \(np < 5\) | normal approximation is invalid — use Poisson or exact binomial |
| gives \(n \ge 20\), \(p \le 0.05\) | Topic 3 — Poisson, \(\lambda = np\) |
| gives a small \(n\) (say 10 or 15) | Topic 2 — do the binomial directly |
(state it every time)
continuity correction
Write the validity check as a line of working, not in your head. It is a free mark and it is the only thing that justifies everything after it.
| The question asks | Integers wanted | Use the boundaries |
|---|---|---|
| exactly \(r\) | r | \(r-0.5\) to \(r+0.5\) |
| between \(a\) and \(b\) inclusive | a … b | \(a-0.5\) to \(b+0.5\) — stretch outward |
| at least \(a\) (\(r \ge a\)) | a, a+1, … | from \(a-0.5\) upward |
| more than \(a\) (\(r > a\)) | a+1, a+2, … | from \(a+0.5\) upward |
| at most \(b\) (\(r \le b\)) | …, b−1, b | up to \(b+0.5\) |
| fewer than \(b\) (\(r < b\)) | …, b−2, b−1 | up to \(b-0.5\) |
The reliable method: first rewrite the phrase as a list of whole numbers ("more than 10, out of 50" ⇒ 11, 12, …, 50). Then put the boundary half a unit outside the smallest and largest integers in that list. Doing it in that order makes the ±0.5 automatic instead of memorised.
- Confirm it is binomial and write \(n\), \(p\), \(q\).
- Check \(np \ge 5\) and \(nq \ge 5\) — both, in writing.
- Compute \(\mu=np\) and \(\sigma=\sqrt{npq}\).
- Convert the phrase to a list of integers, then to continuity-corrected boundaries.
- Standardise each boundary and read A(z).
- Combine by the sketch — same rules as Topic 4.
- Say the answer in words.
- Omitting the ±0.5 entirely. On a narrow range this can shift the answer by several percentage points.
- Correcting in the wrong direction. "At least 30" goes to 29.5, not 30.5 — 30 itself is wanted, so the boundary must fall below it.
- Applying a continuity correction to a genuinely continuous variable. Topic 4 questions never get one.
- Skipping the validity check and approximating when \(np<5\) — the answer can be badly wrong, and the check is a mark.
- Using \(\sigma = npq\). That is the variance.
Q. \(X\) is binomial with \(n=50\), \(p=0.25\). Use the normal approximation to find (a) \(P(x>10)\), (b) \(P(x<18)\), (c) \(P(x>21)\), (d) \(P(9 Q. Use the normal approximation for (a) n=35, p=0.15, between 7 and 10 inclusive; (b) n=29, p=0.25, at least 9; (c) n=84, p=0.42, at most 40; (d) n=63, p=0.11, 10 or more; (e) n=18, p=0.67, between 9 and 12 inclusive. Q. 5% of cars fail their annual inspection. Using the normal approximation, find the probability that between 7 and 18 of the next 200 cars fail. Q. Use the normal approximation for (a) n=30, p=0.35, between 10 and 15 inclusive; (b) n=42, p=0.62, 30 or more; (c) n=15, p=0.40, at most 7; (d) n=51, p=0.42, between 17 and 25 inclusive. Q. A fair coin is tossed 100 times. Find (a) P(at most 45 heads), (b) P(between 45 and 55 heads inclusive). Q. 30% of 400 customers who enter a store make a purchase. Find the probability that at least 110 buy something. Q. 20% of 150 items sampled are expected to be seconds. Find P(between 25 and 35 seconds, inclusive). Q. \(n = 60\), \(p = 0.40\). Find P(at most 20 successes). Q. An insurer expects 8% of 500 policies to produce a claim. Find P(more than 49 claims). Q. In a poll of 80 voters, 45% are expected to support a proposal. Find P(between 30 and 40 supporters, inclusive). Q. A batch of 1,000 units has a 2% defect rate. Find P(between 15 and 25 defectives inclusive). Both approximations are available here — which would you use, and does it matter? Q. \(n = 25\), \(p = 0.10\). A student uses the normal approximation to find P(at most 4 successes) and gets 0.9082. Is this acceptable? What should be done instead? The topic that decides the other five. In the exam nobody tells you "use Poisson" — the story does, and only if you can read it. This section is pure identification: no arithmetic, just the reflex that turns a paragraph into a named distribution and its parameters. Q. Which distribution is most likely appropriate — binomial, Poisson or normal? (a) The life span of a female born in 1977. (b) The number of autos passing through a tollbooth. (c) The number of defective radios in a lot of 100. (d) The water level in a reservoir. Q. What characteristics of a situation help to determine which distribution is appropriate? Q. Explain the difference between discrete and continuous random variables. What difference do such classifications make in determining the probabilities of future events? Q. In practice, the nature of a distribution is often not obvious. What alternatives are open to someone who wants to use probability distributions but is not sure which one fits? Q. A call centre receives an average of 12 calls per hour. What is the probability of more than 15 calls in an hour? Q. Of 15 machines inspected, each has a 20% chance of needing service. What is the probability that exactly 4 need service? Q. The weight of cereal in a box averages 500 g with a standard deviation of 8 g. What proportion of boxes fall below 490 g? Q. A batch of 400 components has a 1% defect rate. What is the probability of exactly 3 defectives? Q. In a survey of 500 voters, 45% are expected to favour a candidate. What is the probability that between 210 and 240 favour them? Q. A machine breaks down on average 2.5 times per month. Over the last 6 months, the variance in breakdowns was measured at 2.4. What does that tell you, and what is the probability of no breakdowns next month? Q. "The number of typing errors on a page" and "the number of pages containing an error, out of 50 pages". Are these the same distribution? Q. Name the distribution and parameters for each: (a) time to serve a customer, mean 4.2 min, s.d. 1.1; (b) number of accidents at a junction per week, average 3; (c) number of heads in 40 coin tosses; (d) number of faulty items in a shipment of 1,000 with a 0.3% fault rate; (e) height of adult males, mean 172 cm, s.d. 7 cm.
np, nq μ σ Boundaries z values Answer (a) 5.25, 29.75 ✓ 5.25 2.1125 6.5 – 10.5 0.59, 2.49 0.2712 (b) 7.25, 21.75 ✓ 7.25 2.3318 8.5 → 0.54 0.2946 (c) 35.28, 48.72 ✓ 35.28 4.5235 ← 40.5 1.15 0.8749 (d) 6.93, 56.07 ✓ 6.93 2.4835 9.5 → 1.03 0.1515 (e) 12.06, 5.94 ✓ 12.06 1.9949 8.5 – 12.5 −1.78, 0.22 0.5496
μ = np σ = √(npq) Boundaries z values A(z) Answer (a) 10.50 2.6125 9.5 – 15.5 −0.38, 1.91 0.1480, 0.4719 0.6199 (b) 26.04 3.1457 29.5 → 1.10 0.3643 0.1357 (c) 6.00 1.8974 ← 7.5 0.79 0.2852 0.7852 (d) 21.42 3.5247 16.5 – 25.5 −1.40, 1.16 0.4192, 0.3770 0.7962 Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Choosing the Correct Probability Distribution
Distribution Signature in the question Parameters Identity to check Binomial "out of 20…", "a sample of n", pass/fail, defective/good \(n,\ p\) \(\mu=np\), \(\sigma^2=npq\); \(\sigma^2 < \mu\) Poisson "on average λ per hour / page / km"; no n exists \(\lambda\) \(\mu = \sigma^2 = \lambda\) — unique to Poisson Normal "normally distributed, mean μ, s.d. σ"; a measured quantity \(\mu,\ \sigma\) continuous, symmetric, bell-shaped Poisson approx. binomial wording, but \(n\ge20\) and \(p\le0.05\) \(\lambda=np\) no continuity correction Normal approx. binomial wording, \(np\ge5\) and \(nq\ge5\) \(\mu=np,\ \sigma=\sqrt{npq}\) ±0.5 continuity correction required
2. Is the number of trials fixed and known in advance? Yes ⇒ binomial. No, only a rate per interval ⇒ Poisson.
3. Are there exactly two outcomes per trial? Success/failure, defective/good ⇒ binomial.
4. Is the probability of success constant across trials? If it drifts, the binomial does not apply.
5. Are the trials (or occurrences) independent? Required by both the binomial and the Poisson.
6. Does the mean equal the variance? If so, that is the signature of a Poisson.
7. Is the number of possible values finite or unlimited? Binomial stops at \(n\); Poisson has no upper limit.
8. Is the distribution symmetric and bell-shaped? Points to the normal — or to a binomial large enough to be approximated by one.
What follows from it — four consequences:
1. For a discrete variable, \(P(X = x)\) is a genuine, non-zero probability. For a continuous variable, \(P(X = x) = 0\) for any single value — only intervals have positive probability, because there are infinitely many possible values.
2. Discrete probabilities come from a formula evaluated at each value and summed; continuous probabilities come from the area under a curve between two limits.
3. Discrete distributions are drawn as spikes or bars; continuous ones as a smooth curve.
4. Moving between them needs the continuity correction: approximating a discrete count by a continuous curve requires the ±0.5 adjustment, exactly as in Topic 5.
2. Compare the mean and the variance. If they are approximately equal, the Poisson is indicated; if the variance is clearly smaller than the mean, the binomial.
3. Examine the process, not just the numbers. Ask whether there is a fixed number of independent trials with two outcomes — that is a structural test which does not need data at all.
4. Test the fit formally. A chi-square goodness-of-fit test compares observed frequencies with those a candidate distribution predicts (Unit 14).
5. Use an empirical distribution. If nothing standard fits, use the observed relative frequencies themselves as the probability distribution.
6. Consult published work or an expert in the field, where the underlying process is often already documented.
7. Test the sensitivity of the decision. Compute the answer under two candidate distributions — if the business decision is the same either way, the choice does not matter and you can proceed.Solution
Solution
Solution
Solution
Solution
Solution
Solution
"Errors on a page" — a count with no upper limit and a rate per page ⇒ Poisson, parameter λ = mean errors per page.
"Pages containing an error, out of 50" — a fixed \(n = 50\), each page either does or does not contain an error ⇒ binomial, with \(p = P(\text{a page has at least one error}) = 1 - e^{-\lambda}\).
The two are linked: the binomial's \(p\) is computed from the Poisson. Spotting that link is what the question is testing.Solution
(e) Normal, μ = 172 cm, σ = 7 cm.
Poisson: \(P(x) = \frac{\lambda^x e^{-\lambda}}{x!}, \qquad P(X=0) = e^{-\lambda}, \quad P(X \ge 1) = 1 - e^{-\lambda}\)
Normal: \(z = \frac{x - \mu}{\sigma}\)
Normal approx: \(np \ge 5, nq \ge 5 \implies \mu = np, \sigma = \sqrt{npq}\) (with \(\pm 0.5\) continuity correction)
Concepts behind all of this: Unit 04 notes. The two real exam questions from this unit — Poisson Q1 (b) and normal Q2 (b) — are worked in both their versions at Solved Papers. The z table your class uses, and how it converts to the cumulative one, is on the Formula & Table Sheet.
Next in exam weight: Units 5–6 carry 12 marks between them, and both lean on the normal table you have just drilled — the only change is that \(\sigma\) becomes \(\sigma/\sqrt{n}\).