Practice Problems

Q. A retailer records daily sales (in ₹ '000) for 60 days:

Find the mean daily sales.

1. Step 1 — Midpoints

   Midpoint = (lower + upper)/2 → 5, 15, 25, 35, 45.

2. Step 2 — Build the f·x column

   Class	x	f	f·x
   0–10	5	8	40
   10–20	15	14	210
   20–30	25	22	550
   30–40	35	10	350
   40–50	45	6	270
   Total	—	Σf = 60	Σfx = 1420

3. Step 3 — Apply the formula x̄ = Σfx / Σf = 1420 / 60 = 23.67 (₹ '000 per day)

Q. Number of complaints handled per shift by 100 agents:

Find the mean. Must the series be converted to exclusive form first?

1. Step 1 — Answer the conversion question first

   No. Your class's rule: an inclusive series need not be converted for the mean; conversion is required only for the median and mode Class notes p. 3. Converting 1–5 to 0.5–5.5 leaves the midpoint at 3 either way, which is exactly why the mean is unaffected.

2. Step 2 — Midpoints and f·x x = 3, 8, 13, 18    f·x = 36, 224, 455, 450    Σf = 100, Σfx = 1165
3. Step 3 — Mean x̄ = 1165 / 100 = 11.65 complaints per shift

Q. For each dataset, name the tabular summary and the graph you would use, and justify it in one line.

Q. 60 guests rate a hotel: Poor 6, Below Average 9, Average 15, Above Average 27, Excellent 3. Build the relative and percent frequency distributions, and give the pie-chart angle for each class.

1. Step 1 — Relative frequency = f / n, percent = ×100, angle = ×360°

   Rating	f	Relative	Percent	Pie angle
   Poor	6	0.10	10%	36°
   Below Average	9	0.15	15%	54°
   Average	15	0.25	25%	90°
   Above Average	27	0.45	45%	162°
   Excellent	3	0.05	5%	18°
   Total	60	1.00	100%	360°

   Always check the three totals: 1.00, 100%, 360°. If any is off, a class is wrong.

2. Step 2 — The insight sentence

   Half the guests rated "Above Average" or "Excellent", which should please management; but there were twice as many "Poor" ratings as "Excellent" ones, which should not. That two-clause structure is the model answer deck slide 35.

Q. Build a stem-and-leaf display for these 15 service times (minutes): 24, 31, 27, 45, 38, 22, 29, 41, 33, 36, 28, 47, 30, 25, 39.

1. Step 1 — Stems are the tens digit, leaves the units, in rank order
   Stem-and-Leaf Display

   2 | 2 4 5 7 8 9
   3 | 0 1 3 6 8 9
   4 | 1 5 7
   
   Leaf unit = 1

2. Step 2 — Read it

   The display shows the shape (a peak in the 20s–30s, thinning above 40) and keeps every original value, which a histogram would throw away. The 15 values sort themselves as a side effect, so the median — the 8th value — can be read straight off: 31 minutes.

Q. Eight machines produced these output counts in an hour: 42, 38, 45, 50, 41, 39, 47, 46. Compute the standard deviation (a) treating these as a sample, (b) treating them as the whole population.

1. Step 1 — Mean x̄ = (42+38+45+50+41+39+47+46)/8 = 348/8 = 43.5
2. Step 2 — Sum of squared deviations

   x	42	38	45	50	41	39	47	46
   x − x̄	−1.5	−5.5	1.5	6.5	−2.5	−4.5	3.5	2.5
   (x − x̄)²	2.25	30.25	2.25	42.25	6.25	20.25	12.25	6.25

   Σ(x − x̄)² = 122  ← check the deviations sum to 0 first
3. Step 3 — (a) As a sample: divide by n − 1 s² = 122 / 7 = 17.43    s = √17.43 = 4.17
4. Step 4 — (b) As a population: divide by N σ² = 122 / 8 = 15.25    σ = √15.25 = 3.91
5. Step 5 — Note the gap

   The sample SD is larger — 4.17 vs 3.91 — because dividing by the smaller n − 1 inflates it. That inflation is what makes s² an unbiased estimator of the population variance. With n = 8 the gap is 7%; with n = 100 it would be under 0.5%.

Q. Branch A has mean monthly sales ₹150 lakh with SD ₹12 lakh. Branch B has mean ₹450 lakh with SD ₹30 lakh. Which branch is more consistent?

1. Step 1 — Note why the raw SDs cannot be compared

   B's SD (30) is more than double A's (12), which would suggest B is far less consistent. But B's average is three times larger, so a bigger absolute swing is expected. The means differ, so use the CV.

2. Step 2 — Compute both CV_A = (12/150) × 100 = 8.00% CV_B = (30/450) × 100 = 6.67%
3. Step 3 — Conclude

   Branch B is more consistent — its sales vary by 6.67% of its own average, against 8.00% for A. The raw SDs pointed the opposite way, which is exactly the trap this question sets.

Q. A distribution has mean 48, median 50, mode 52 and standard deviation 8. Compute Pearson's first and second coefficients of skewness and describe the shape.

1. Step 1 — First coefficient Sk₁ = (Mean − Mode)/SD = (48 − 52)/8 = −4/8 = −0.50
2. Step 2 — Second coefficient Sk₂ = 3(Mean − Median)/SD = 3(48 − 50)/8 = −6/8 = −0.75
3. Step 3 — Describe

   Both are negative, so the distribution is negatively skewed with a long left tail. This matches the ordering Mean 48 < Median 50 < Mode 52. The two coefficients differ in magnitude — they are different measures, not two routes to one number; quote whichever the question's data supports (use Sk₂ when the mode is unclear or bimodal).

Q. Two delivery channels are rated 1–5:

Interpret, and recommend one action.

1. Step 1 — Central tendency

   App: Mean 4.2 < Median 4 < Mode 5 — hold on. Mean 4.2 is above the median of 4, yet the mode is 5 and skewness is −1.42. The mode sitting at the top of the scale with a strong negative skew means ratings pile up at 5, with a thin tail of low scores. Report what the skewness says and note that with a 1–5 integer scale the median is coarse — it can only take five values, so the mean/median comparison is unreliable and the skewness figure is the more trustworthy indicator.

   Phone: Mean 3.5 < Median 4 = Mode 4 — mild negative skew (−0.28); most people say 4, but enough low ratings drag the mean down to 3.5.

2. Step 2 — Dispersion and consistency

   App: SD 0.61 with kurtosis 4.6 → leptokurtic, tall and narrow. Consistent, predictable experience.
   Phone: SD 1.20 with kurtosis 2.2 → platykurtic, flat and wide. Nearly double the variability: the phone channel is a lottery.

3. Step 3 — Recommendation, tied to a statistic

   The phone channel's SD of 1.20 is the problem, not its mean. Standardise the phone-order script and train for first-call resolution — that attacks variability directly. Chasing the mean (3.5 → 4.2) without cutting the SD would leave the same unpredictable experience with a slightly better average.

Q. 400 call durations have mean 64 seconds and standard deviation 6 seconds. (a) Using Chebyshev's theorem, at least how many calls lie between 52 and 76 seconds? (b) If the durations are normally distributed, roughly how many lie in that range? (c) At least how many lie within μ ± 3σ?

1. Step 1 — Recognise the interval as μ ± kσ 52 = 64 − 12 = μ − 2σ    76 = 64 + 12 = μ + 2σ   ⟹   k = 2
2. Step 2 — (a) Chebyshev At least 1 − 1/k² = 1 − 1/4 = 75% of values 400 × 0.75 = at least 300 calls

   No assumption about shape is needed — this holds for any distribution.

3. Step 3 — (b) Empirical rule, assuming normality Around 95% within μ ± 2σ   ⟹   400 × 0.95 = about 380 calls
4. Step 4 — (c) Chebyshev at k = 3 μ ± 3σ = 64 ± 18 = (46, 82)    at least 1 − 1/9 = 88.9% ≈ 89% 400 × 0.89 = at least 356 calls
5. Step 5 — The point of the comparison

   Chebyshev guarantees 300; normality predicts 380. Chebyshev is always safe but loose; the normal assumption is precise but must be justified. That trade-off is the whole reason the rest of the syllabus exists.

Q. For two events, P(A) = 0.45, P(B) = 0.30 and P(A ∩ B) = 0.15. Find (a) P(A ∪ B), (b) P(A | B), (c) are A and B independent? (d) are they mutually exclusive?

1. (a) Addition law P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.45 + 0.30 − 0.15 = 0.60
2. (b) Conditional probability P(A | B) = P(A ∩ B)/P(B) = 0.15 / 0.30 = 0.50
3. (c) Independence test — compute both sides and compare P(A)·P(B) = 0.45 × 0.30 = 0.135    but    P(A ∩ B) = 0.15

   0.135 ≠ 0.15, so A and B are NOT independent. Cross-check: P(A|B) = 0.50 but P(A) = 0.45 — knowing B occurred changed A's probability, which is the definition of dependence.

4. (d) Mutually exclusive?

   No. P(A ∩ B) = 0.15 ≠ 0, so they can occur together. Note that (c) and (d) are different questions: mutually exclusive events are always dependent, so nothing can be both.

Q. A component fails with probability 0.20. Five components are installed and operate independently. What is the probability that at least one fails?

1. Step 1 — Spot the trigger and take the complement

   "At least one" ⟹ 1 − P(none). Adding P(exactly 1) + P(exactly 2) + … would take five terms; this takes one.

   P(at least one fails) = 1 − P(none fails)
2. Step 2 — P(none) with equal probabilities P(none) = (1 − 0.20)⁵ = (0.80)⁵ = 0.3277
3. Step 3 — Subtract P(at least one) = 1 − 0.3277 = 0.6723 or 67.23%

Q. Three suppliers deliver independently. Their probabilities of being late are 0.30, 0.25 and 0.40. Find (a) the probability that at least one is late, (b) the probability that all three are on time, (c) the expected number of late deliveries.

1. (b) first — it is the building block P(all on time) = (1−0.30)(1−0.25)(1−0.40) = 0.70 × 0.75 × 0.60 = 0.315
2. (a) At least one late P(at least one late) = 1 − 0.315 = 0.685 or 68.5%
3. (c) Expected number of late deliveries

   Because the probabilities differ, sum them rather than using n·p:

   E = 0.30 + 0.25 + 0.40 = 0.95 late deliveries

   Note that 0.95 and 0.685 both describe the same situation and are both correct: one is an expected count, the other a probability. Do not confuse them.

Q. A committee is formed from 10 managers. (a) In how many ways can 3 be chosen for a project team? (b) In how many ways can 3 be chosen and assigned to the distinct roles of chair, secretary and treasurer?

1. Step 1 — Decide whether order matters

   (a) A project team is just a set of three people — order does not matter → combination.
   (b) The three roles are distinct, so who gets which matters — order matters → permutation.

2. Step 2 — (a) Combination C(10,3) = 10! / (3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120 ways
3. Step 3 — (b) Permutation P(10,3) = 10! / 7! = 10 × 9 × 8 = 720 ways
4. Step 4 — Sanity check

   720 = 120 × 3! — every one of the 120 teams can be arranged into the three roles in 3! = 6 ways. The permutation is always the larger number.

Q. Over 40 days, a rental firm recorded polishers rented per day: 0 on 4 days, 1 on 6 days, 2 on 18 days, 3 on 10 days, 4 on 2 days. Assign probabilities and find (a) P(at least 2 rented) and (b) the expected number rented per day.

1. Step 1 — Relative frequency method: divide each count by the total

   Rented x	0	1	2	3	4	Total
   Days	4	6	18	10	2	40
   p(x)	.10	.15	.45	.25	.05	1.00

   The probabilities sum to 1.00 — always check this before going further deck slide 14.

2. Step 2 — (a) P(at least 2) P(X ≥ 2) = .45 + .25 + .05 = 0.75  or 1 − (.10 + .15) = 0.75
3. Step 3 — (b) Expected value E(X) = Σ x·p(x) = 0(.10) + 1(.15) + 2(.45) + 3(.25) + 4(.05) = 0 + 0.15 + 0.90 + 0.75 + 0.20 = 2.00 polishers per day

Q. Machine A makes 60% of a factory's output and Machine B makes 40%. 2% of A's output is defective and 5% of B's. A randomly chosen item is found defective. What is the probability it came from Machine B?

1. Step 1 — Define A = made by Machine A    B = made by Machine B    D = the item is defective
2. Step 2 — Given P(A) = 0.60   P(B) = 0.40   P(D|A) = 0.02   P(D|B) = 0.05
3. Step 3 — Total probability of a defect P(D) = P(A)P(D|A) + P(B)P(D|B) = (0.60)(0.02) + (0.40)(0.05) = 0.012 + 0.020 = 0.032
4. Step 4 — Bayes P(B | D) = 0.020 / 0.032 = 0.625    (and P(A|D) = 0.012/0.032 = 0.375)
5. Step 5 — Insight

   Machine B makes only 40% of output but is responsible for 62.5% of defects. The prior of 0.40 has been revised sharply upward. Management should prioritise servicing Machine B — it is the dominant source of defects despite the smaller share of production.

Q. 2% of a population has a condition. A screening test correctly detects it in 95% of cases but returns a positive for 8% of healthy people. A person tests positive. What is the probability they have the condition? What is the managerial implication?

1. Step 1 — Define and list C = has the condition   H = healthy   T = tests positive P(C) = 0.02   P(H) = 0.98   P(T|C) = 0.95   P(T|H) = 0.08

   Note the wording: "returns a positive for 8% of healthy people" is already the false-positive rate. No subtraction.

2. Step 2 — Total probability P(T) = (0.02)(0.95) + (0.98)(0.08) = 0.019 + 0.0784 = 0.0974
3. Step 3 — Bayes P(C | T) = 0.019 / 0.0974 = 0.195 ≈ 19.5%
4. Step 4 — Insight

   Fewer than 1 in 5 positives is genuine. The reason: the healthy group is 49 times larger, so its 8% false-positive rate produces 78 false alarms per 1,000 people against only 19 true detections. A positive on this screen should trigger a confirmatory test, never a diagnosis or a treatment decision on its own.

Q. 0.5% of card transactions are fraudulent. A detection system flags 98% of frauds and 3% of legitimate transactions. A transaction is flagged. What is the probability it is fraudulent?

1. Step 1 — Given P(F) = 0.005   P(L) = 0.995   P(flag|F) = 0.98   P(flag|L) = 0.03
2. Step 2 — Total probability P(flag) = (0.005)(0.98) + (0.995)(0.03) = 0.0049 + 0.02985 = 0.03475
3. Step 3 — Bayes P(F | flag) = 0.0049 / 0.03475 = 0.141 ≈ 14.1%
4. Step 4 — Insight

   A system that catches 98% of fraud produces flags that are wrong 86% of the time. With a base rate of 0.5%, the 3% false-positive rate applied to the huge legitimate population swamps the true positives. Blocking every flagged transaction would inconvenience roughly six genuine customers for every fraudster caught — flags should be routed to a review queue or a step-up authentication, not to an automatic block.

Q. 40% of incoming emails are spam. A filter marks 99% of spam and 5% of legitimate mail. An email is marked. What is the probability it is spam? Compare with D3 and explain the difference.

1. Step 1 — Total probability P(marked) = (0.40)(0.99) + (0.60)(0.05) = 0.396 + 0.030 = 0.426
2. Step 2 — Bayes P(spam | marked) = 0.396 / 0.426 = 0.930 ≈ 93.0%
3. Step 3 — Why so different from D3?

   The filter's accuracy (99%) is barely better than the fraud system's (98%), yet the posterior is 93% instead of 14%. The base rate did all the work. Spam is 40% of email; fraud is 0.5% of transactions. When the thing you are looking for is common, a positive is usually real; when it is rare, a positive is usually a false alarm.

   That single comparison is the most useful thing you can carry into the exam's Q5 (a) — it lets you say why the answer came out where it did, not just what it is.

Q. 25% of customers entering a store make a purchase. Eight customers enter. What is the probability that exactly 3 make a purchase?

1. Step 1 — Confirm it is binomial, and define success

   Fixed n = 8, two outcomes (buys / does not), constant p = 0.25, independent. Binomial.

   Let X = number of customers out of 8 who make a purchase    n = 8, p = 0.25, r = 3
2. Step 2 — Apply P(X = 3) = ⁸C₃ (0.25)³ (0.75)⁵ ⁸C₃ = 8!/(3!5!) = 56 = 56 × 0.015625 × 0.2373 = 0.2076 or 20.76%

Q. 40% of a bank's loan applications are approved. Twelve applications are received. Find (a) the mean and standard deviation of the number approved, (b) P(at least 2 are approved).

1. (a) Mean and SD μ = n·p = 12 × 0.40 = 4.8 approvals σ² = n·p·q = 12 × 0.40 × 0.60 = 2.88 σ = √2.88 = 1.70

   Remember to take the square root — npq is the variance.

2. (b) At least 2 — take the complement P(X ≥ 2) = 1 − P(X = 0) − P(X = 1) P(X = 0) = ¹²C₀ (0.4)⁰ (0.6)¹² = 0.6¹² = 0.00218 P(X = 1) = ¹²C₁ (0.4)¹ (0.6)¹¹ = 12 × 0.4 × 0.003628 = 0.01741 P(X ≥ 2) = 1 − 0.00218 − 0.01741 = 0.9804

   Two terms to subtract, against eleven to add. Always take the complement when "at least" involves a small number.

Q. A production line has a 5% defect rate. In a batch of 20 items, find P(exactly 2 defective) (a) exactly, by the binomial, and (b) by the Poisson approximation. Is the approximation allowed?

1. Step 1 — Check the rule of thumb

   The deck's condition is n ≥ 20 and p ≤ 0.05 slide 29. Here n = 20 and p = 0.05, so both hold — just barely. The approximation is permitted.

2. Step 2 — (a) Exact binomial P(X = 2) = ²⁰C₂ (0.05)² (0.95)¹⁸ = 190 × 0.0025 × 0.3972 = 0.1887
3. Step 3 — (b) Poisson approximation, λ = n·p λ = 20 × 0.05 = 1 P(X = 2) = e^(−1) × 1² / 2! = 0.3679 / 2 = 0.1839
4. Step 4 — Compare

   0.1887 vs 0.1839 — within 0.005. The approximation is good, and it gets better as n grows and p shrinks. It exists because ²⁰C₂ is manageable but ⁵⁰⁰C₂ is not.

Q. A helpdesk receives an average of 3 escalations per day. Find (a) P(no escalations tomorrow), (b) P(at least one), (c) P(exactly two).

1. Step 1 — Identify the distribution and λ

   An average rate per interval with no fixed n ⟹ Poisson, λ = 3 per day.

2. (a) None P(X = 0) = e^(−3) = 0.0498 ≈ 5%
3. (b) At least one — the complement P(X ≥ 1) = 1 − e^(−3) = 1 − 0.0498 = 0.9502 ≈ 95%
4. (c) Exactly two P(X = 2) = e^(−3) × 3² / 2! = 0.0498 × 9 / 2 = 0.2240
5. Note

   (a) and (b) are complements and must sum to 1: 0.0498 + 0.9502 = 1.000 ✓. Use that as a free check.

Q. Customers arrive at a service desk at an average of 4 per hour. What is the probability that at most one customer arrives in the next 30 minutes?

1. Step 1 — Rescale λ to the interval asked about

   This is the step the question is testing. The rate is per hour; the question asks about half an hour.

   λ = 4 per hour × 0.5 hour = 2 per 30 minutes
2. Step 2 — "At most one" means X = 0 or X = 1 P(X ≤ 1) = P(X = 0) + P(X = 1) P(X = 0) = e^(−2) = 0.1353 P(X = 1) = e^(−2) × 2¹ / 1! = 0.1353 × 2 = 0.2707
3. Step 3 — Add P(X ≤ 1) = 0.1353 + 0.2707 = 0.4060 or 40.6%
4. Common mistake

   Using λ = 4 gives 0.0916 — less than a quarter of the right answer. Always rescale λ to match the interval in the question.

Q. Machine stoppages occur at an average of 5 per week. (a) Find P(exactly 7 stoppages next week). (b) State the mean, variance and standard deviation. (c) Find P(at most 2).

1. (a) Exactly 7 P(X = 7) = e^(−5) × 5⁷ / 7! = 0.006738 × 78125 / 5040 = 0.1044
2. (b) The Poisson's signature property μ = λ = 5    σ² = λ = 5    σ = √5 = 2.24

   Mean and variance are equal — that is the Poisson's distinguishing feature and a common one-mark question on its own.

3. (c) At most 2 P(X ≤ 2) = e^(−5)[1 + 5 + 5²/2!] = 0.006738 × (1 + 5 + 12.5) = 0.006738 × 18.5 = 0.1247

   Factoring e^(−λ) out of the sum saves three separate exponential evaluations.

Q. Training programme completion times are normally distributed with mean 500 hours and SD 100 hours. Find (a) P(a participant takes more than 650 hours), (b) P(between 400 and 600 hours).

1. Step 1 — Set up (no n, so divide by σ) X ~ N(μ = 500, σ = 100)
2. (a) Right tail z = (650 − 500)/100 = 1.50 P(Z > 1.50) = 0.5 − A(1.50) = 0.5 − 0.4332 = 0.0668 ≈ 6.7%
3. (b) Symmetric interval z₁ = (400 − 500)/100 = −1.00    z₂ = (600 − 500)/100 = +1.00 P(−1 < Z < 1) = 2 × A(1.00) = 2 × 0.3413 = 0.6827

   This is the empirical rule's 68% appearing on cue, because 400 and 600 are exactly μ ± 1σ.

Q. Employee productivity scores are N(μ = 72, σ = 9). Find P(60 < X < 80).

1. Step 1 — Standardise both ends z₁ = (60 − 72)/9 = −12/9 = −1.33 z₂ = (80 − 72)/9 = 8/9 = +0.89
2. Step 2 — Decide add or subtract by the SIGNS

   One negative, one positive ⟹ the interval spans the mean ⟹ ADD.

   P(−1.33 < Z < 0.89) = A(1.33) + A(0.89) = 0.4082 + 0.3133 = 0.7215
3. Step 3 — Answer

   About 72.15% of employees score between 60 and 80.

Q. With the same N(72, 9) scores, what score marks the top 10% of employees?

1. Step 1 — Translate into an area, then find z

   "Top 10%" means 10% of the area is to the right, so 40% lies between 0 and z:

   A(z) = 0.4000   →   search the table body for 0.4000 → z ≈ 1.28

   This is a reverse table lookup: find the area first, then read off z. The nearest entries are A(1.28) = 0.3997 and A(1.29) = 0.4015, so 1.28 is the standard answer.

2. Step 2 — Un-standardise x = μ + z·σ = 72 + 1.28 × 9 = 72 + 11.52 = 83.52
3. Step 3 — Answer

   A score of about 83.5 or above puts an employee in the top 10%.

Q. Monthly electricity consumption per household is N(μ = 250, σ = 40) units. What proportion of households consume less than 200 units?

1. Step 1 — Standardise z = (200 − 250)/40 = −50/40 = −1.25
2. Step 2 — Left tail, so use symmetry P(Z < −1.25) = 0.5 − A(1.25) = 0.5 − 0.3944 = 0.1056
3. Step 3 — Answer

   About 10.6% of households consume under 200 units. Note the table has no negative rows — the curve is symmetric, so A(−1.25) = A(1.25) and only the tail arithmetic changes.

Q. 60% of voters support a proposal. In a random sample of 400, what is the probability that at least 250 support it?

1. Step 1 — Recognise the binomial and check the approximation is safe X ~ Binomial(n = 400, p = 0.60) n·p = 240 ≥ 5    n·q = 160 ≥ 5   ⟹ normal approximation valid

   Computing ⁴⁰⁰C₂₅₀ + ⁴⁰⁰C₂₅₁ + … by hand is out of the question, which is why this method exists.

2. Step 2 — Match the moments μ = n·p = 400 × 0.60 = 240 σ = √(n·p·q) = √(400 × 0.6 × 0.4) = √96 = 9.80
3. Step 3 — Continuity correction

   "At least 250" includes 250 itself, so the continuous interval starts at the lower edge of the 250 bar:

   P(X ≥ 250) → P(X ≥ 249.5)
4. Step 4 — Standardise and read z = (249.5 − 240)/9.80 = 9.5/9.80 = 0.97 P(Z ≥ 0.97) = 0.5 − A(0.97) = 0.5 − 0.3340 = 0.1660
5. Step 5 — Answer

   About a 16.6% chance that at least 250 of the 400 support the proposal.

Q. A fair coin is tossed 100 times. Use the normal approximation to find P(exactly 55 heads), and compare it with the exact binomial value of 0.0485.

1. Step 1 — Moments μ = 100 × 0.5 = 50    σ = √(100 × 0.5 × 0.5) = √25 = 5
2. Step 2 — A single value becomes a width-1 interval

   Without this step P(X = 55) would come out as exactly zero, because a continuous distribution assigns no probability to a single point.

   P(X = 55) → P(54.5 ≤ X ≤ 55.5)
3. Step 3 — Standardise both edges z₁ = (54.5 − 50)/5 = 0.90    z₂ = (55.5 − 50)/5 = 1.10
4. Step 4 — Both positive, so SUBTRACT P(0.90 ≤ Z ≤ 1.10) = A(1.10) − A(0.90) = 0.3643 − 0.3159 = 0.0484
5. Step 5 — Compare

   0.0484 approximate vs 0.0485 exact — agreement to three decimals. The continuity correction is what makes that work; without it the answer would have been 0.

Q. Adult male heights have mean 68 inches and SD 3 inches. A random sample of 36 men is taken. What is the probability the sample mean lies between 67 and 69 inches?

1. Step 1 — Note the words "sample mean" μ = 68   σ = 3   n = 36   ⟹ μ_x̄ = 68
2. Step 2 — Standard error σ_x̄ = σ/√n = 3/√36 = 3/6 = 0.5
3. Step 3 — Standardise z₁ = (67 − 68)/0.5 = −2.00    z₂ = (69 − 68)/0.5 = +2.00
4. Step 4 — Symmetric, so double P(−2 < Z < 2) = 2 × A(2.00) = 2 × 0.4772 = 0.9544
5. Step 5 — Contrast with a single man

   For one individual, 67 to 69 inches is only μ ± ⅓σ and the probability would be about 26%. For a mean of 36, it is 95.4%. Averaging shrinks the spread by a factor of √n — that is the entire content of the standard error.

Q. Monthly household expenditure has mean ₹1,200 and SD ₹150. A sample of 25 households is drawn. Find (a) the standard error, (b) P(the sample mean exceeds ₹1,260).

1. (a) Standard error σ_x̄ = 150/√25 = 150/5 = ₹30
2. (b) Standardise and read the right tail z = (1260 − 1200)/30 = 60/30 = +2.00 P(Z > 2.00) = 0.5 − A(2.00) = 0.5 − 0.4772 = 0.0228
3. Interpretation

   Only a 2.28% chance. So if a surveyor does report a sample mean of ₹1,260 from 25 households, either they got a very unusual sample or the true mean is higher than ₹1,200 — which is precisely the reasoning hypothesis testing formalises in Unit 9.

Q. Assembly times have mean 45 minutes and SD 8 minutes. A sample of 64 assemblies is timed. What is the probability the sample mean is below 43.5 minutes?

1. Step 1 — Standard error σ_x̄ = 8/√64 = 8/8 = 1.0 minute
2. Step 2 — Standardise z = (43.5 − 45)/1.0 = −1.50
3. Step 3 — Left tail P(Z < −1.50) = 0.5 − A(1.50) = 0.5 − 0.4332 = 0.0668 or 6.68%

Q. A company has 1,000 employees whose training scores have SD 20. A sample of 100 is taken. (a) Should the finite population multiplier be used? (b) Compute the standard error with and without it.

1. (a) Test the sampling fraction n/N = 100/1000 = 0.10

   0.10 > 0.05, so yes, use the FPM deck slide 21.

2. (b) Without the FPM σ_x̄ = 20/√100 = 2.00
3. (c) With the FPM FPM = √[(N − n)/(N − 1)] = √[(1000 − 100)/999] = √(900/999) = √0.9009 = 0.9492 σ_x̄ = 2.00 × 0.9492 = 1.898
4. Step 4 — Read what changed

   The correction shaved about 5% off the standard error. Sampling without replacement from a finite pool leaves less uncertainty, because each employee measured is one fewer you are unsure about. As n → N the multiplier → 0, which is right: measure everyone and there is no sampling error at all.

Q. A survey with n = 100 gives a standard error of 4 units. (a) What sample size would halve the standard error? (b) What would n = 400 give? (c) Explain the practical implication in one sentence.

1. (a) To halve the SE

   SE ∝ 1/√n, so halving it requires √n to double, which requires n to quadruple.

   n = 100 × 4 = 400
2. (b) Verify

   From SE = σ/√n = 4 at n = 100, we get σ = 4 × 10 = 40. Then:

   SE at n = 400  =  40/√400 = 40/20 = 2.0 ✓ exactly half
3. (c) Practical implication

   Halving the error costs four times the survey. Going from n = 100 to n = 400 buys a drop from 4 to 2; going from 400 to 1,600 would buy only 2 to 1. Past some point the extra precision is not worth the money — that is the law of diminishing returns in sampling, and it is why national polls stop at a few thousand respondents.

Q. 30% of a bank's customers use mobile banking. In a random sample of 200 customers, what is the probability that more than 35% use it?

1. Step 1 — Check the normal approximation is safe n·p = 200 × 0.30 = 60    n·q = 200 × 0.70 = 140

   Both well above 5, so the approximation holds. Write this check down — the deck makes it an explicit Step 1 slide 18.

2. Step 2 — Standard error of the proportion σ_p̂ = √(pq/n) = √(0.30 × 0.70 / 200) = √0.00105 = 0.0324
3. Step 3 — Standardise z = (p̂ − p)/σ_p̂ = (0.35 − 0.30)/0.0324 = 0.05/0.0324 = 1.54
4. Step 4 — Right tail P(Z > 1.54) = 0.5 − A(1.54) = 0.5 − 0.4382 = 0.0618 or 6.18%

Q. 55% of voters favour a candidate. In a poll of 150 voters, what is the probability the sample proportion lies between 50% and 60%?

1. Step 1 — Check and compute the standard error n·p = 82.5, n·q = 67.5 — both ≥ 5 ✓ σ_p̂ = √(0.55 × 0.45 / 150) = √0.00165 = 0.0406
2. Step 2 — Standardise both ends z₁ = (0.50 − 0.55)/0.0406 = −1.23    z₂ = (0.60 − 0.55)/0.0406 = +1.23
3. Step 3 — Symmetric, so double P(−1.23 < Z < 1.23) = 2 × A(1.23) = 2 × 0.3907 = 0.7814
4. Step 4 — What a pollster reads from this

   Roughly a 78% chance the poll lands within ±5 percentage points of the truth. Put another way, about one poll in five will land outside that band by pure sampling variation — which is why single polls should not be over-interpreted.

Q. 12% of items from a supplier are returned. In a shipment of 400 items, what is the probability that fewer than 10% are returned?

1. Step 1 — Check n·p = 400 × 0.12 = 48    n·q = 400 × 0.88 = 352 — both ≥ 5 ✓

   Worth noticing: with n = 30 instead of 400, n·p would be only 3.6 and the normal approximation would not be valid. The check is not a formality.

2. Step 2 — Standard error σ_p̂ = √(0.12 × 0.88 / 400) = √0.000264 = 0.0162
3. Step 3 — Standardise z = (0.10 − 0.12)/0.0162 = −0.02/0.0162 = −1.23
4. Step 4 — Left tail P(Z < −1.23) = 0.5 − A(1.23) = 0.5 − 0.3907 = 0.1093 or about 11%

Q. A researcher wants to estimate mean weekly study hours to within 3 hours with 95% confidence. Past data gives σ = 15 hours. What sample size is needed?

1. Step 1 — Mean or proportion?

   "Mean weekly study hours" is a measured average, and a σ is given → mean formula.

2. Step 2 — Inputs σ = 15    E = 3    95% ⟹ z = 1.96
3. Step 3 — Substitute n = (z·σ/E)² = (1.96 × 15 / 3)² = (29.4/3)² = (9.8)² = 96.04
4. Step 4 — Round up n = 97 students

   96.04 rounds up to 97, not down to 96. A minimum is a minimum.

Q. A quality manager wants to estimate mean fill volume to within 0.5 ml with 99% confidence. The process SD is 2.4 ml. Find the required sample size.

1. Step 1 — The right z for 99% 99% confidence ⟹ 0.5% in each tail ⟹ z = 2.576

   Not 1.96. The Final's Q3 (b) turns on exactly this, so drill it.

2. Step 2 — Substitute n = (2.576 × 2.4 / 0.5)² = (6.1824/0.5)² = (12.3648)² = 152.89
3. Step 3 — Round up n = 153 bottles
4. Step 4 — Compare with 95%

   At 95% the same problem needs (1.96 × 2.4/0.5)² = 88.5 → 89 bottles. Raising confidence from 95% to 99% costs 64 extra bottles — a 72% increase in sample for four extra percentage points of confidence. That is the precision/confidence trade-off in numbers.

Q. A firm wants to estimate the proportion of customers who would switch to a subscription plan, to within ±3 percentage points with 95% confidence. No prior estimate exists. Find n.

1. Step 1 — The p = 0.5 convention, with its reason

   With no prior estimate, take p = 0.5, because p·q is maximised there (0.25). Any other p gives a smaller n, so 0.5 guarantees the sample is big enough whatever the true value turns out to be. Write that sentence in the exam.

2. Step 2 — Inputs z = 1.96    E = 0.03    p = q = 0.5

   ±3 percentage points means E = 0.03, not 3.

3. Step 3 — Substitute n = z²pq/E² = (1.96)²(0.5)(0.5)/(0.03)² = 3.8416 × 0.25 / 0.0009 = 0.9604 / 0.0009 = 1067.1
4. Step 4 — Round up n = 1,068 customers

Q. Same firm, but a pilot study suggests p ≈ 0.30. Estimate to within ±5 percentage points at 95% confidence. Find n, and say what the prior estimate was worth.

1. Step 1 — Use the known p z = 1.96    E = 0.05    p = 0.30, q = 0.70    pq = 0.21
2. Step 2 — Substitute n = (1.96)² × 0.21 / (0.05)² = 3.8416 × 0.21 / 0.0025 = 0.8067 / 0.0025 = 322.7   ⟹   n = 323 customers
3. Step 3 — What did the prior estimate buy?

   With no prior estimate, p = 0.5 gives n = 3.8416 × 0.25/0.0025 = 384.2 → 385. So knowing p ≈ 0.30 saved 62 surveys — about 16%. The further p is from 0.5, the more a pilot study is worth: at p = 0.05 the requirement collapses to 73 deck slide 41.

Q. Estimate mean delivery time to within 5 minutes at 90% confidence, given σ = 20 minutes. (a) Find n. (b) What happens to n if the tolerance tightens to 2.5 minutes?

1. (a) At E = 5 90% ⟹ z = 1.645 n = (1.645 × 20 / 5)² = (32.9/5)² = (6.58)² = 43.3 ⟹ 44 deliveries
2. (b) At E = 2.5 — halve the tolerance n = (1.645 × 20 / 2.5)² = (13.16)² = 173.2 ⟹ 174 deliveries
3. Step 3 — The pattern

   Halving E quadrupled n, from 44 to 174, because E appears squared in the denominator. This is the same 1/√n relationship seen from the other direction in G5. Precision is expensive, and the price rises as the square.