● BS MIDTERM
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BUSINESS STATISTICSβ€ΊUNIT 04β€ΊPROBABILITY DISTRIBUTIONS

Unit 04 β€” Probability Distributions

Binomial distribution, Poisson distribution β€” use of Excel and SPSS functions. Normal distribution: properties, standardizing a random variable, applications.

SESSIONS: 2 TEXTBOOK: LEVIN CH. 5, pp. 207–265 EXAM WEIGHT: Q1 (b) + Q2 (b) β€” 9 MARKS, IN BOTH PAPERS Full PPT: Unit 4 deck 51 Class notes 11
PRACTISE THIS UNIT

Every question shape in this unit is drilled on Unit 04 β€” How to Solve Anything: six topics, each with a "how to recognise this topic" card, a master method, and every exercise Levin sets on that topic solved with its page number. This is the 9-mark unit β€” start here.

β—† What the papers ask from this unit

The most heavily examined unit in the mid-term portion β€” 9 marks in both papers, in two fixed slots:

  • Q1 (b) β€” 4 marks, Poisson. "On average Ξ» per interval… find P(no arrivals)" (Final) or "…find P(at least one breakdown)" (Re-Exam). Two lines of work.
  • Q2 (b) β€” 5 marks, Normal. Always in two parts: an interval probability, then a one-tail "highly successful" probability.

Final QP pp. 1–2, Re-Exam QP pp. 1–2. Both solved β†’

1. Random Variables and Expected Value Core syllabus concept

1Understand the Concept

A random variable takes on a defined set of values with different probabilities β€” its value is subject to random variation, so it is not deterministic slide 2. Roll a die and the outcome is random with six equally likely values.

TypeValuesExamples from the deck slide 12
DiscreteA countable set.Counts (number of cyclones), categories (much above average β†’ much below average), binary variables (rain / no rain).
ContinuousAny value in a range.Rainfall in mm in a season, temperature in a day.

2Expected value β€” the weighted average

"All probability distributions are characterized by an expected value (mean) and variance (standard deviation squared)" slide 3. The expected value is a weighted average β€” more frequent values of X are weighted more heavily.

Expected Value
\[ E(X) = \mu = \sum x \cdot P(x) \quad \text{(Discrete)} \qquad E(X) = \mu = \int x \cdot P(x)\,dx \quad \text{(Continuous)} \]
Variance
\[ \text{Var}(X) = \sigma^2 = E[(X - \mu)^2] = \sum (x - \mu)^2 P(x) \]

Written identically in your class notes Class notes p. 1 and p. 4.

3Worked example β€” the lottery, and why expected value is a decision tool

The deck's own example slides 6–9. A lottery picks 6 numbers from 1 to 49. It costs $1.00 to play; a win pays $2 million after taxes.

  1. Step 1 β€” Probability of winning one try Number of possible tickets = C(49,6) = 49! / (43! Β· 6!) = 13,983,816 P(win) = 1 / 13,983,816 = 7.2 Γ— 10⁻⁸
  2. Step 2 β€” Build the probability function
    x ($)p(x)
    βˆ’10.999999928
    +2,000,0007.2 Γ— 10⁻⁸

    Note that it sums to 1.0 β€” always check this.

  3. Step 3 β€” Expected value E(X) = (2,000,000)(7.2 Γ— 10⁻⁸) + (βˆ’1)(0.999999928) = 0.144 βˆ’ 1.000 = βˆ’$0.86
  4. Step 4 β€” Decide

    "Negative expected value is never good! You shouldn't play if you expect to lose money." Over ten years of weekly play: 520 Γ— (βˆ’0.86) = βˆ’$447.20 slide 9.

The general use: in any decision question, compute E(X) for each alternative and pick the largest. That is the entire content of "use of expected value in good decision making" slide 6.

4pmf, CDF, and pdf β€” three ways to write the same distribution

  • A probability mass function (pmf) lists p(x) for each discrete value. For a fair die, p(x) = 1/6 for x = 1…6, and Ξ£p(x) = 1 slide 15.
  • A cumulative distribution function (CDF) gives P(X ≀ x). For the die: P(X≀1)=1/6, P(X≀2)=2/6 … P(X≀6)=6/6 slide 17. It is a staircase for discrete variables.
  • A probability density function (pdf) is used for continuous variables, because you cannot assign probability to a single point. Probability is the area under the curve between two values, and the total area must equal 1 slide 18.
Sources
Primary Unit 4 deck β€” random variables, expected value, lottery, pmf/CDF/pdf slides 2–21
Class Expected value formulas, discrete and continuous p. 1, p. 4
Textbook Levin Β§5.1–5.3 Random variables, expected value pp. 207–220

2. The Binomial Distribution Core syllabus concept

1The four conditions β€” check them before you use the formula

The deck lists them and adds a warning worth memorising slide 23:

  1. The data arise from a sequence of n independent trials.
  2. At each trial there are only two possible outcomes, conventionally called success and failure.
  3. The probability of success, p, is the same in each trial.
  4. The random variable of interest is the number of successes, X, in the n trials.
⚠ The deck's own warning

"The assumptions of independence and constant p in points 1 and 3 are important. If they are invalid, so is the binomial distribution" slide 23. And more generally: "Do not use a particular distribution unless you are satisfied that the assumptions which underlie it are (at least approximately) satisfied" slide 22.

2The formula

Std. dev. Οƒ = √[ nΒ·p(1 βˆ’ p) ] EXCEL: =BINOMDIST(r, n, p, FALSE) for exactly r =BINOMDIST(r, n, p, TRUE) for r or fewer

Formula slide 24, characteristics slide 25, Excel slide 26. Your class notes write the same formula and the same mean/variance Class notes p. 1, p. 3.

Define "success" before you touch the formula. Your class notes make the point in yellow ink p. 1: in "getting 4 heads", a success is getting just 1 head on a single toss; r = 4 is how many of those you want. Writing "Let X = number of defective bulbs in the sample of 10" as your first line prevents nearly every mistake in this topic.

3Worked examples β€” the three your class did

  1. (a) Probability of exactly 4 heads when a fair coin is tossed 10 times X = number of heads, n = 10, p = 0.5, r = 4 P(X = 4) = ¹⁰Cβ‚„ (0.5)⁴ (0.5)⁢ = 210 Γ— 0.0625 Γ— 0.015625 = 0.205 β‰ˆ 20%

    Class notes p. 1

  2. (b) 10 bulbs, defect rate 10%; probability that exactly 3 are defective X = number of defective bulbs, n = 10, p = 0.1, r = 3 P(X = 3) = ¹⁰C₃ (0.1)Β³ (0.9)⁷ = 120 Γ— 0.001 Γ— 0.4783 = 0.0574 β‰ˆ 6%

    Class notes p. 2 records β‰ˆ 0.05.

  3. (c) 10 shoppers, 30% make a purchase; probability that exactly 4 buy X = number who make a purchase, n = 10, p = 0.30, r = 4 P(X = 4) = ¹⁰Cβ‚„ (0.30)⁴ (0.70)⁢ = 210 Γ— 0.0081 Γ— 0.117649 = 0.200 = 20%

    Class notes p. 2

ENSolved problems on binomial probability distribution Gourav Manjrekar ENDiscrete distributions: example problems jbstatistics

Practice E1–E4 β€” binomial β†’

Sources
Primary Unit 4 deck β€” binomial conditions, formula, characteristics, Excel slides 22–26
Class Three worked binomial problems pp. 1–3
Textbook Levin Β§5.4 The binomial distribution, p. 222 pp. 222–229

3. The Poisson Distribution Core syllabus concept Β· Q1 (b), both papers

1Understand the Concept

The Poisson distribution counts events that occur randomly in a given interval of time or space slide 28. Two assumptions, and they are exactly what the exam wording signals:

Deck examples: number of tropical cyclones in a season, occasions when river levels exceed a value, accidents at a junction in a month, births in a hospital in 10 hours slide 28; car accidents at a site, users in a wireless network, document requests on a web server, outbreaks of war, photons landing on a photodiode slide 27.

The trigger phrase. Both papers open Q1 (b) with the words "on average … per [interval]": "On average, 2 customers arrive at a ticket counter every 10 minutes" (Final); "machine breakdowns occur at an average rate of 1.5 per day" (Re-Exam). An average rate per interval, with no fixed number of trials, is always Poisson.

2The formula

Poisson Distribution Formula
\[ P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \qquad X \sim \text{Po}(\lambda) \]
Poisson Mean & Variance
\[ \text{Mean } \mu = \lambda, \qquad \text{Variance } \sigma^2 = \lambda, \qquad \text{Std Dev } \sigma = \sqrt{\lambda} \]

Formula slide 29, mean and variance slide 31, Excel slide 32. Your class notes write it identically Class notes p. 2.

⚠ Watch the mean/SD swap in your class notes

Page 4 of your class notes has the labels the wrong way round β€” it lists "For Poisson: Mean = np" and "For Binomial: Mean = Ξ», SD = √λ" Class notes p. 4. The correct pairing, confirmed by the deck slide 25 and slide 31 and by Levin, is:

Binomial Parameters
\[ \mu = n \cdot p, \qquad \sigma = \sqrt{n \cdot p (1-p)} \]
Poisson Parameters
\[ \mu = \lambda, \qquad \sigma = \sqrt{\lambda} \]

3Method β€” the three steps, every time

Step 1 β€” Identify & Scale \(\lambda\)
Scale \(\lambda\) proportionally to match the time/space interval in the question.
Step 2 β€” Identify target event count
Determine whether looking for exactly \(x\), at least \(x\), or at most \(x\).
Step 3 β€” "At Least One" Rule
\[ P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-\lambda} \]

4Worked examples

  1. (a) The deck's hospital example

    Births occur randomly at an average rate of 1.8 per hour. What is P(exactly 4 births in a given hour)?

    Ξ» = 1.8, x = 4 P(X = 4) = e^(βˆ’1.8) Β· 1.8⁴ / 4! = (0.1653)(10.4976) / 24 = 0.0723

    And P(2 or more births in an hour) = 0.537 slide 30.

  2. (b) Your class's mall example

    On average 8 customers arrive at a mall every 10 minutes. What is the probability that no customer arrives in the next 10 minutes?

    Ξ» = 8, x = 0 P(X = 0) = e^(βˆ’8) Β· 8⁰ / 0! = e^(βˆ’8) = 0.000335 β‰ˆ 0.0003

    Class notes p. 3

  3. (c) Your class's 4-per-week example

    Ξ» = 4 occurrences per week; find P(exactly 5).

    P(X = 5) = e^(βˆ’4) Β· 4⁡ / 5! = (0.018316)(1024) / 120 = 0.1563 = 15.6%

    Class notes pp. 2–3

5Poisson as an approximation to the binomial

"Poisson distributions also arise as an approximation to binomial when n is large and p is small. As a rule of thumb, when n β‰₯ 20 and p ≀ 0.05" slide 29. Set Ξ» = nΒ·p and use the Poisson formula. Note also that a Poisson variable can take any positive integer value, whereas the binomial always has a finite upper limit of n.

ENPoisson distribution β€” solved examples, easiest way Dr. Harish Garg ENPoisson vs binomial β€” concepts and solved examples Shrenik Jain HIPoisson distribution word problems (Hindi) Shrenik Jain

Practice E5–E8 β€” Poisson, including both exam shapes β†’

Sources
Primary Unit 4 deck β€” Poisson process, formula, examples, mean and variance slides 27–32
Class Poisson formula and two worked problems pp. 2–3
Textbook Levin Β§5.5 The Poisson distribution pp. 230–237
Exam Final Q1 (b) p. 1 Β· Re-Exam Q1 (b) p. 1

4. The Normal Distribution and Standardising Core syllabus concept Β· Q2 (b), both papers

1Properties β€” the list to open a theory answer with

2Why standardise

The deck states the problem and the fix precisely slide 37: "Conceptually, we need a table of areas for each normal curve. This is impossible because there are infinitely many different normal curves. The way out of this difficulty is standardizing, which transforms every normal distribution into one particular normal distribution, the standard normal distribution."

Standardizing Z Score
\[ z = \frac{x - \mu}{\sigma} \qquad (\text{Standard Normal } Z \sim N(0,1)) \]
⚠ Which z table is yours? This changes every answer

The table used in your class β€” Standard Normal table.jpg, headed "Areas of the Standard Normal Distribution" β€” gives the area between 0 and z, not the cumulative area from βˆ’βˆž. Its entries run 0.0000 to 0.4998, never above 0.5. Your class notes use it throughout: z = 1.58 β†’ .4429, z = 1.33 β†’ .4082, z = 2.05 β†’ .4798 Class notes p. 5, p. 5.

The official answer keys use the cumulative table instead (Ξ¦(1.20) = 0.8849, Ξ¦(βˆ’0.4) = 0.3446). Both reach the same answer; the arithmetic differs. Pick one and stay in it. The Formula & Table Sheet shows the conversion between them and reproduces your table in full.

3Method β€” the four moves that cover every normal question

Direct Reading
\[ P(0 < Z < a) = A(a) \]
Right / Left Tail
\[ P(Z > a) = 0.5 - A(a) \qquad P(Z < -a) = 0.5 - A(a) \]
Interval Spanning Mean
\[ P(-a < Z < b) = A(a) + A(b) \]
Interval One Side
\[ P(a < Z < b) = A(b) - A(a) \]

4Worked example β€” the class's Q8, both parts

X ~ N(ΞΌ = 44, Οƒ = 12). Class notes pp. 4–5

  1. Part (a) β€” P(33 ≀ X ≀ 42), an interval entirely on one side of the mean z₁ = (33 βˆ’ 44)/12 = βˆ’11/12 = βˆ’0.92    zβ‚‚ = (42 βˆ’ 44)/12 = βˆ’2/12 = βˆ’0.17 P(βˆ’0.92 ≀ Z ≀ βˆ’0.17) = A(0.92) βˆ’ A(0.17) = 0.3212 βˆ’ 0.0675 = 0.2537 = 25.37%

    Both z's are negative, so the region does not straddle the mean β€” subtract, don't add. Your class notes truncate βˆ’0.9167 to βˆ’0.91 instead of rounding, read 0.3186, and get 0.2511 Class notes p. 5. Either is accepted; just be consistent about whether you round or truncate the second decimal of z.

  2. Part (b) β€” P(X < 25 or X > 60), the two tails z₁ = (25 βˆ’ 44)/12 = βˆ’1.58    zβ‚‚ = (60 βˆ’ 44)/12 = +1.33 P(Z < βˆ’1.58) + P(Z > 1.33) = (0.5 βˆ’ 0.4429) + (0.5 βˆ’ 0.4082) = 0.0571 + 0.0918 = 0.1489 = 14.89%

    Class notes p. 5 β€” with the curve drawn twice, once in x units and once in z units. Copy that habit; it earns method marks and stops sign errors.

ENStandardizing normally distributed random variables jbstatistics ENStandard normal tables, z scores, probability & the empirical rule The Organic Chemistry Tutor ENEverything you need to know about z-tables JoeCMath

Practice F1–F6 β€” normal distribution, both exam shapes β†’

Sources
Primary Unit 4 deck β€” normal properties, standardising, the standard normal curve, worked applications slides 33–46
Class Two full normal problems with curves drawn pp. 4–5
Textbook Levin Β§5.6 The normal distribution pp. 238–254
Exam Final Q2 (b) p. 1 Β· Re-Exam Q2 (b) p. 1

5. Normal Approximation to the Binomial Taught in class

1Why you need it

Computing P(X = 390) for n = 500 by the binomial formula means evaluating 500C390, which no one wants to do by hand. When n is large, the binomial's shape is very close to a normal curve, so you can swap in the normal and read a table instead.

Step 1 β€” Rule of Thumb
\[ np \ge 5 \quad \text{and} \quad nq \ge 5 \]
Step 2 β€” Match Moments
\[ \mu = np, \qquad \sigma = \sqrt{npq} \]
Step 3 β€” Continuity Correction
\[ P(X = k) \to P(k - 0.5 \le X \le k + 0.5) \] \[ P(a \le X \le b) \to P(a - 0.5 \le X \le b + 0.5) \] \[ P(X \ge k) \to P(X \ge k - 0.5), \qquad P(X \le k) \to P(X \le k + 0.5) \]
Step 4 β€” Standardise & Read Z-Table
Calculate \(z\) for corrected boundary values and look up probabilities.

2Worked example β€” the class's coin check, showing both methods agree

A fair coin is tossed 10 times. Find P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8). Class notes pp. 6–7

  1. The exact binomial answer ¹⁰Cβ‚…(.5)¹⁰ + ¹⁰C₆(.5)¹⁰ + ¹⁰C₇(.5)¹⁰ + ¹⁰Cβ‚ˆ(.5)¹⁰ = (252 + 210 + 120 + 45)/1024 = 627/1024 = 0.6123
  2. Now by the normal approximation ΞΌ = np = 10 Γ— 0.5 = 5     Οƒ = √(npq) = √(10 Γ— 0.5 Γ— 0.5) = √2.5 = 1.58

    Apply the continuity correction to the range 5 to 8:

    P(X = 5 to 8) β†’ P(4.5 ≀ X ≀ 8.5) z₁ = (4.5 βˆ’ 5)/1.58 = βˆ’0.31    zβ‚‚ = (8.5 βˆ’ 5)/1.58 = +2.21 P(βˆ’0.31 ≀ Z ≀ 2.21) = A(0.31) + A(2.21) = 0.1217 + 0.4864 = 0.6081
  3. Compare

    0.6123 exact vs 0.61 approximate β€” agreement to two decimals, with n only 10. Your class notes record 0.61 p. 7. This is the check that proves the method to you.

3Worked example β€” the deck's life-expectancy problem

The probability is 0.80 that a person aged 20 will be alive at 65. 500 twenty-year-olds are selected at random. Find (a) P(exactly 390 alive) and (b) P(between 375 and 425 inclusive). slide 49, worked in your class notes pp. 7–9.

  1. Set up ΞΌ = np = 500 Γ— 0.8 = 400 Οƒ = √(npq) = √(500 Γ— 0.8 Γ— 0.2) = √80 = 8.94

    Check: np = 400 β‰₯ 5 and nq = 100 β‰₯ 5, so the approximation is safe.

  2. (a) Exactly 390 β€” a single point becomes a width-1 interval P(X = 390) β†’ P(389.5 ≀ X ≀ 390.5) z₁ = (389.5 βˆ’ 400)/8.94 = βˆ’1.17    zβ‚‚ = (390.5 βˆ’ 400)/8.94 = βˆ’1.06 = A(1.17) βˆ’ A(1.06) = 0.3790 βˆ’ 0.3554 = 0.0236

    The exact binomial value is 0.0234 Class notes p. 7 β€” the approximation is right to three decimals.

  3. (b) Between 375 and 425 inclusive P(375 ≀ X ≀ 425) β†’ P(374.5 ≀ X ≀ 425.5) z₁ = (374.5 βˆ’ 400)/8.94 = βˆ’2.85    zβ‚‚ = (425.5 βˆ’ 400)/8.94 = +2.85 = A(2.85) + A(2.85) = 0.4978 + 0.4978 = 0.9956

    Class notes p. 9

⚠ Common mistakes across this unit
  • Skipping the continuity correction. Without it, P(X = 390) becomes P(390 ≀ X ≀ 390) = 0.
  • Getting the correction direction backwards. Widen the interval to include the whole bar. "Between 375 and 425 inclusive" widens outward to 374.5 and 425.5; "more than 375" starts at 375.5.
  • Choosing binomial when the question gives a rate. No n β†’ not binomial. "8 per 10 minutes" is Poisson.
  • Choosing Poisson when the question gives a fixed number of trials. "10 bulbs, each 10% defective" has n = 10 and p = 0.1 β†’ binomial.
  • Forgetting to square-root the variance. npq is σ², not Οƒ.
The one-line distribution chooser
Fixed n trials, two outcomes, constant p β†’ Binomial.
An average rate per interval, no fixed n β†’ Poisson.
A continuous measurement with a mean and an SD β†’ Normal.
A binomial with a large n β†’ Normal, with the continuity correction.
ENSolved examples: normal approximation to the binomial, with continuity correction Joshua Emmanuel ENNormal approximation to binomial (continuity corrections) ExamSolutions ENBinomial, Poisson or hypergeometric? SawFin Mathematics

Practice F5–F6 β€” normal approximation with continuity correction β†’

Sources
Primary Unit 4 deck β€” normal as an approximation to the binomial, step-by-step method, life-expectancy problem slides 47–49
Class Coin check and the 500-person problem, both worked pp. 6–9
Textbook Levin Β§5.6 β€” the normal approximation to binomial pp. 249–253 Β· Β§5.7 Choosing the correct probability distribution p. 254