Statistical Distributions (Extended)

## Statistical Distributions (Extended Treatment)

This document provides rigorous coverage of the binomial, normal, and Poisson distributions, their Approximations, and hypothesis testing applications.

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A random variable $X$ has a binomial distribution with parameters $n$ and $p$ (written $X \sim B(n, p)$) if:

$P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}, \quad r = 0, 1, 2, \ldots, n$

Conditions for a binomial distribution:

1. A fixed number $n$ of trials.
2. Each trial has exactly two outcomes (success/failure).
3. The probability of success $p$ is constant for each trial.
4. Trials are independent.

$E(X) = np, \qquad \mathrm{Var}(X) = np(1-p)$

Proof of $E(X) = np$.

$E(X) = \sum_{r=0}^{n} r\binom{n}{r}p^r(1-p)^{n-r} = \sum_{r=1}^{n} r\binom{n}{r}p^r(1-p)^{n-r}$

Using $r\binom{n}{r} = n\binom{n-1}{r-1}$:

$= np\sum_{r=1}^{n}\binom{n-1}{r-1}p^{r-1}(1-p)^{n-r} = np\sum_{k=0}^{n-1}\binom{n-1}{k}p^k(1-p)^{n-1-k} = np \cdot 1 = np \quad \blacksquare$

$P(X \leq r) = \sum_{k=0}^{r}\binom{n}{k}p^k(1-p)^{n-k}$

$P(X \geq r) = 1 - P(X \leq r-1)$

Problem. A fair coin is tossed 12 times. Find the probability of getting: (a) exactly 7 heads; (b) at most 4 heads; (c) between 5 and 9 heads inclusive.

$X \sim B(12, 0.5)$.

(a) $P(X = 7) = \dbinom{12}{7}(0.5)^{12} = \dfrac{792}{4096} = \dfrac{99}{512} \approx 0.1934$

(b) $P(X \leq 4) = \displaystyle\sum_{k=0}^{4}\dbinom{12}{k}(0.5)^{12} = \dfrac{1 + 12 + 66 + 220 + 495}{4096} = \dfrac{794}{4096} \approx 0.1938$

(c) $P(5 \leq X \leq 9) = P(X \leq 9) - P(X \leq 4) = 1 - P(X \leq 4) - P(X \geq 10)$

$P(X \geq 10) = P(X \leq 2)$ (by symmetry of $p = 0.5$) $= \dfrac{1 + 12 + 66}{4096} = \dfrac{79}{4096}$

$P(5 \leq X \leq 9) = 1 - \dfrac{794}{4096} - \dfrac{79}{4096} = \dfrac{3223}{4096} \approx 0.7869$

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A random variable $X$ has a normal distribution with parameters $\mu$ and $\sigma^2$ (written $X \sim N(\mu, \sigma^2)$) if its probability density function is:

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x - \mu)^2}{2\sigma^2}\right), \quad x \in \mathbb{R}$

• The distribution is symmetric about $x = \mu$.
• The mean, median, and mode are all equal to $\mu$.
• $E(X) = \mu$, $\mathrm{Var}(X) = \sigma^2$.
• Approximately 68% of data lies within $\mu \pm \sigma$.
• Approximately 95% of data lies within $\mu \pm 2\sigma$.
• Approximately 99.7% of data lies within $\mu \pm 3\sigma$.

To find probabilities, we standardise to the standard normal $Z \sim N(0, 1)$:

$Z = \frac{X - \mu}{\sigma}$

$P(X \leq x) = P\!\left(Z \leq \frac{x - \mu}{\sigma}\right) = \Phi\!\left(\frac{x - \mu}{\sigma}\right)$

Where $\Phi(z)$ denotes the cumulative distribution function of the standard normal.

Problem. The masses of bags of sugar are normally distributed with mean $1.02\;\mathrm{kg}$ and Standard deviation $0.03\;\mathrm{kg}$. Find: (a) the probability a randomly selected bag has mass Less than $1.00\;\mathrm{kg}$; (b) the probability the mass is between $0.98$ and $1.05\;\mathrm{kg}$; (c) the value $m$ such that 90% of bags have mass less than $m$.

$X \sim N(1.02, 0.03^2)$.

(a) $P(X \lt 1.00) = P\!\left(Z \lt \dfrac{1.00 - 1.02}{0.03}\right) = P(Z \lt -0.667) = 1 - \Phi(0.667) \approx 1 - 0.7476 = 0.2524$

(b) $P(0.98 \lt X \lt 1.05) = P\!\left(\dfrac{0.98 - 1.02}{0.03} \lt Z \lt \dfrac{1.05 - 1.02}{0.03}\right) = P(-1.333 \lt Z \lt 1.000)$

$= \Phi(1.000) - \Phi(-1.333) = 0.8413 - (1 - 0.9088) = 0.8413 - 0.0912 = 0.7501$

(c) We need $\Phi\!\left(\dfrac{m - 1.02}{0.03}\right) = 0.90$So $\dfrac{m - 1.02}{0.03} = 1.282$.

$m = 1.02 + 0.03 \times 1.282 = 1.058\;\mathrm{kg}$

If $X \sim B(n, p)$ and $n$ is large, then $X$ is approximately normal with:

$X \approx N(np, np(1-p))$

Continuity correction. Since the binomial is discrete and the normal is continuous, apply a Continuity correction:

• $P(X \leq k) \approx P(Y \lt k + 0.5)$
• $P(X \geq k) \approx P(Y \gt k - 0.5)$
• $P(X = k) \approx P(k - 0.5 \lt Y \lt k + 0.5)$

The approximation is reasonable when $np \gt 5$ and $n(1-p) \gt 5$.

Problem. $X \sim B(80, 0.45)$. Use a normal approximation to find $P(X \gt 35)$.

\mu = 80 \times 0.45 = 36$$\sigma^2 = 80 \times 0.45 \times 0.55 = 19.8$$\sigma = 4.45.

$X \approx N(36, 19.8)$.

$P(X \gt 35) \approx P(Y \gt 34.5) = P\!\left(Z \gt \dfrac{34.5 - 36}{4.45}\right) = P(Z \gt -0.337)$

$= 1 - \Phi(-0.337) = \Phi(0.337) \approx 0.632$

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A random variable $X$ has a Poisson distribution with parameter $\lambda$ (written $X \sim \mathrm{Po}(\lambda)$) if:

$P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}, \quad r = 0, 1, 2, \ldots$

Conditions:

1. Events occur independently at a constant average rate.
2. The probability of more than one event in a sufficiently small interval is negligible.
3. Events occur singly in continuous time or space.

$E(X) = \lambda, \qquad \mathrm{Var}(X) = \lambda$

The equality of mean and variance is a distinguishing feature of the Poisson distribution.

Problem. A call centre receives an average of 4.5 calls per minute. Assuming a Poisson model, Find: (a) the probability of exactly 6 calls in a minute; (b) the probability of at most 2 calls In a minute; (c) the probability of more than 8 calls in a two-minute period.

$X \sim \mathrm{Po}(4.5)$.

(a) $P(X = 6) = \dfrac{e^{-4.5}(4.5)^6}{6!} = \dfrac{e^{-4.5} \times 8303.8}{720} \approx 0.1271$

(b) $P(X \leq 2) = e^{-4.5}\!\left(1 + 4.5 + \dfrac{4.5^2}{2}\right) = e^{-4.5}(1 + 4.5 + 10.125) = 15.625\,e^{-4.5} \approx 0.1736$

(c) For two minutes, $Y \sim \mathrm{Po}(9)$.

$P(Y \gt 8) = 1 - P(Y \leq 8) = 1 - e^{-9}\displaystyle\sum_{r=0}^{8}\dfrac{9^r}{r!} \approx 1 - 0.4557 = 0.5443$

If $X \sim B(n, p)$ where $n$ is large and $p$ is small (so that $np$ is moderate), then:

$X \approx \mathrm{Po}(np)$

This is valid when $n \geq 50$ and $p \leq 0.1$ (and $np \leq 10$ as a rough guideline).

Problem. A machine produces items with a defect rate of 0.02. In a batch of 200 items, find the Probability that exactly 3 are defective.

$X \sim B(200, 0.02)$. Since $n = 200$ is large and $p = 0.02$ is small, $X \approx \mathrm{Po}(4)$.

$P(X = 3) = \frac{e^{-4} \cdot 4^3}{3!} = \frac{64e^{-4}}{6} = \frac{32}{3}e^{-4} \approx 0.1954$

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Situation	Distribution
Fixed trials, two outcomes, const $p$	Binomial $B(n, p)$
Rare events, constant rate	Poisson $\mathrm{Po}(\lambda)$
Continuous, symmetric, bell-shaped	Normal $N(\mu, \sigma^2)$

Theorem. If $X \sim \mathrm{Po}(\lambda_1)$ and $Y \sim \mathrm{Po}(\lambda_2)$ are independent, Then $X + Y \sim \mathrm{Po}(\lambda_1 + \lambda_2)$.

Proof sketch. Using MGFs or direct convolution:

$P(X + Y = r) = \sum_{k=0}^{r}P(X = k)P(Y = r-k) = \sum_{k=0}^{r}\frac{e^{-\lambda_1}\lambda_1^k}{k!} \cdot \frac{e^{-\lambda_2}\lambda_2^{r-k}}{(r-k)!}$

$= \frac{e^{-(\lambda_1+\lambda_2)}}{r!}\sum_{k=0}^{r}\binom{r}{k}\lambda_1^k\lambda_2^{r-k} = \frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^r}{r!} \quad \blacksquare$

Problem. A shop receives orders at an average rate of 3 per hour from online and 2 per hour From walk-in customers. Find the probability of receiving more than 7 orders in a two-hour period.

Total rate per hour $= 3 + 2 = 5$. For two hours, $X \sim \mathrm{Po}(10)$.

$P(X \gt 7) = 1 - P(X \leq 7) = 1 - e^{-10}\displaystyle\sum_{r=0}^{7}\dfrac{10^r}{r!} \approx 1 - 0.2202 = 0.7798$

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$X \sim B(15, 0.35)$. Find: (a) $P(X = 5)$; (b) $P(3 \leq X \leq 7)$; (c) the most likely value Of $X$.

Solution

(a) $P(X = 5) = \dbinom{15}{5}(0.35)^5(0.65)^{10} \approx 0.2123$.

(b) $P(3 \leq X \leq 7) = P(X \leq 7) - P(X \leq 2) \approx 0.9506 - 0.0355 = 0.9151$.

(c) Mode $\approx (n+1)p = 16 \times 0.35 = 5.6$So check $r = 5$ and $r = 6$.

$P(X = 5) \approx 0.2123$, $P(X = 6) \approx 0.2186$. The mode is $X = 6$.

The heights of men are normally distributed with mean $175\;\mathrm{cm}$ and standard deviation $8\;\mathrm{cm}$. Find the probability that a randomly selected man is: (a) taller than $190\;\mathrm{cm}$; (b) between $168\;\mathrm{cm}$ and $182\;\mathrm{cm}$; (c) what height is exceeded by only 5% of men?

Solution

(a) $P(X \gt 190) = P\!\left(Z \gt \dfrac{15}{8}\right) = P(Z \gt 1.875) = 1 - 0.9696 = 0.0304$.

(b) $P(168 \lt X \lt 182) = P(-0.875 \lt Z \lt 0.875) = 2\Phi(0.875) - 1 = 2(0.8092) - 1 = 0.6184$.

(c) $P(Z \gt z) = 0.05 \implies z = 1.645$. Height $= 175 + 1.645 \times 8 = 188.2\;\mathrm{cm}$.

The number of emails received per hour follows a Poisson distribution with mean 6. Find the Probability that: (a) exactly 4 emails are received in an hour; (b) more than 10 emails in two Hours.

Solution

(a) $P(X = 4) = \dfrac{e^{-6} \cdot 6^4}{4!} = \dfrac{1296}{24}e^{-6} = 54e^{-6} \approx 0.1335$.

(b) For two hours, $Y \sim \mathrm{Po}(12)$.

$P(Y \gt 10) = 1 - P(Y \leq 10) \approx 1 - 0.6528 = 0.3472$.

A die is rolled 60 times. Use a suitable approximation to find the probability that the number of Sixes is between 8 and 14 inclusive.

Solution

$X \sim B(60, 1/6)$. $\mu = 10$, $\sigma^2 = 60 \times \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{25}{3} \approx 8.333$.

$\sigma \approx 2.887$.

$P(8 \leq X \leq 14) \approx P(7.5 \lt Y \lt 14.5)$ where $Y \sim N(10, 25/3)$.

$= P\!\left(\dfrac{7.5 - 10}{2.887} \lt Z \lt \dfrac{14.5 - 10}{2.887}\right) = P(-0.866 \lt Z \lt 1.558)$

$= \Phi(1.558) - \Phi(-0.866) = 0.9404 - 0.1931 = 0.7473$.

Statistical distributions are models that describe how probability mass is spread across possible outcomes. The binomial distribution counts successes in fixed trials like flipping a coin ten times and counting heads. The normal distribution emerges logically when many small random effects combine, like measuring heights where genetics and nutrition each contribute tiny variations. The Poisson distribution captures rare events in continuous time, like customers arriving at a shop. These three are deeply connected: the Poisson is a limiting case of the binomial when trials are numerous but success is rare, and the normal approximates both when conditions are right.

1. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

2. Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.

3. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

4. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

Problem. A fair coin is tossed 100 times. Use the normal approximation to estimate the probability of getting between 45 and 55 heads inclusive.

Solution. $X \sim B(100, 0.5)$. $\mu = np = 50$, $\sigma^2 = np(1-p) = 25$, $\sigma = 5$.

Continuity correction: $P(45 \leq X \leq 55) \approx P(44.5 < Y < 55.5)$ where $Y \sim N(50, 25)$.

$z_1 = \frac{44.5 - 50}{5} = -1.1, \quad z_2 = \frac{55.5 - 50}{5} = 1.1$

$P(-1.1 < Z < 1.1) = \Phi(1.1) - \Phi(-1.1) = 2\Phi(1.1) - 1 = 2(0.8643) - 1 = 0.7286$

$\blacksquare$

Problem. Calls arrive at a switchboard at an average rate of 4 per hour. Find the probability of exactly 6 calls in one hour and the probability of fewer than 3 calls.

Solution. $X \sim \mathrm{Po}(4)$.

$P(X = 6) = \frac{e^{-4} \times 4^6}{6!} = \frac{0.01832 \times 4096}{720} = \frac{75.08}{720} \approx 0.1043$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = e^{-4}(1 + 4 + 8) = 13e^{-4} \approx 0.2381$

$\blacksquare$

• Probability — The probability axioms and conditional probability theory underpin all distributions covered here.
• Hypothesis Testing — Binomial and normal distributions are the test statistics used in hypothesis testing.
• Data Representation — Frequency tables and grouped data are approximated by these theoretical distributions.