Statistical Distributions

Board Coverage

Board	Paper	Notes
AQA	Paper 1, 2	Binomial and normal in P1; Poisson in P2
Edexcel	P1, P2	Similar
OCR (A)	Paper 1, 2	Binomial in P1; normal and Poisson in P2
CIE (9709)	P1, P6	Binomial in P1; normal and Poisson in P6

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The formula booklet gives the probability mass function for the Binomial and Poisson distributions, and the normal distribution function. You must know when to use each distribution and how to find probabilities.

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1. Discrete Random Variables

1.1 Definition

Definition. A discrete random variable $X$ takes values from a countable set with probabilities $P(X = x_i) = p_i$ satisfying:

• $p_i \geq 0$ for all $i$
• $\sum_i p_i = 1$

1.2 Expectation and variance

$E(X) = \mu = \sum x_i\,p_i$ $\mathrm{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2 = \sum x_i^2\,p_i - \mu^2$

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2. The Binomial Distribution

2.1 Derivation from Bernoulli trials

A Bernoulli trial is an experiment with exactly two outcomes: success (probability $p$) and failure (probability $1-p$).

If we perform $n$ independent Bernoulli trials, the number of successes $X$ follows a Binomial distribution: $X \sim B(n, p)$.

Derivation of the PMF. Each sequence of $k$ successes and $n-k$ failures has probability $p^k(1-p)^{n-k}$. The number of such sequences is $\binom{n}{k}$ (choosing which $k$ of the $n$ trials are successes). Therefore:

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

2.2 Proof that $E(X) = np$

Proof. Let $X_i$ be the indicator variable for the $i$-th trial: $X_i = 1$ if success, $0$ if failure.

$X = X_1 + X_2 + \cdots + X_n$.

$E(X_i) = 1 \cdot p + 0 \cdot (1-p) = p$.

By linearity of expectation: $E(X) = \sum E(X_i) = np$. $\blacksquare$

2.3 Proof that $\mathrm{Var}(X) = np(1-p)$

Proof. $E(X_i^2) = 1^2 \cdot p + 0^2 \cdot (1-p) = p$.

$\mathrm{Var}(X_i) = E(X_i^2) - [E(X_i)]^2 = p - p^2 = p(1-p)$.

Since the $X_i$ are independent: $\mathrm{Var}(X) = \sum \mathrm{Var}(X_i) = np(1-p)$. $\blacksquare$

2.4 Properties

• The distribution is symmetric when $p = 0.5$.
• It is skewed left when $p \gt{} 0.5$ and skewed right when $p \lt{} 0.5$.
• The mode is at $\lfloor(n+1)p\rfloor$.

2.5 Direct derivation of $E(X) = np$ from the PMF

The proofs in Sections 2.2 and 2.3 use indicator variables. Here we derive the same results directly from the probability mass function using algebraic identities.

Proof. Starting from the definition of expectation applied to the binomial PMF:

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

The $k=0$ term vanishes, so begin the sum at $k=1$. Apply the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$:

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

Substitute $j = k - 1$:

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

By the binomial theorem, $\sum_{j=0}^{n-1}\binom{n-1}{j}p^j(1-p)^{n-1-j} = [p + (1-p)]^{n-1} = 1$.

Therefore $E(X) = np$. $\blacksquare$

2.6 Direct derivation of $\mathrm{Var}(X) = np(1-p)$ from the PMF

Proof. First compute $E(X(X-1))$:

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

Terms with $k = 0, 1$ are zero. Apply the identity $k(k-1)\binom{n}{k} = n(n-1)\binom{n-2}{k-2}$:

$E(X(X-1)) = \sum_{k=2}^{n} n(n-1)\binom{n-2}{k-2}p^k(1-p)^{n-k} = n(n-1)p^2\sum_{k=2}^{n}\binom{n-2}{k-2}p^{k-2}(1-p)^{(n-2)-(k-2)}$

Substitute $j = k - 2$:

$E(X(X-1)) = n(n-1)p^2\sum_{j=0}^{n-2}\binom{n-2}{j}p^j(1-p)^{n-2-j} = n(n-1)p^2$

The final equality follows from the binomial theorem: $\sum_{j=0}^{n-2}\binom{n-2}{j}p^j(1-p)^{n-2-j} = 1$.

Now $E(X^2) = E(X(X-1)) + E(X) = n(n-1)p^2 + np$.

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = n(n-1)p^2 + np - n^2p^2 = np - np^2 = np(1-p) \quad \blacksquare$

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3. The Normal Distribution

3.1 Motivation from the Central Limit Theorem

The Central Limit Theorem (CLT) states that the sum (or mean) of a large number of independent, identically distributed random variables is approximately normally distributed, regardless of the original distribution.

This is why the normal distribution appears so widely in nature: any quantity that is the sum of many small independent effects (height, measurement error, etc.) will be approximately normal.

3.2 Definition

$X \sim N(\mu, \sigma^2)$ has PDF

$f(x) = \frac◆LB◆1◆RB◆◆LB◆\sigma\sqrt◆LB◆2\pi◆RB◆◆RB◆\,e^{-\frac◆LB◆(x-\mu)^2◆RB◆◆LB◆2\sigma^2◆RB◆}$

3.3 Properties

• Bell-shaped, symmetric about $\mu$.
• $E(X) = \mu$, $\mathrm{Var}(X) = \sigma^2$.
• Approximately 68% of data within $\mu \pm \sigma$, 95% within $\mu \pm 2\sigma$, 99.7% within $\mu \pm 3\sigma$.

3.4 Standard normal

If $X \sim N(\mu, \sigma^2)$, then $Z = \dfrac◆LB◆X - \mu◆RB◆◆LB◆\sigma◆RB◆ \sim N(0, 1)$.

Probabilities are found using the standard normal table or a calculator's inverse normal function.

3.5 Finding probabilities

$P(a < X < b) = P\!\left(\frac◆LB◆a-\mu◆RB◆◆LB◆\sigma◆RB◆ < Z < \frac◆LB◆b-\mu◆RB◆◆LB◆\sigma◆RB◆\right) = \Phi\!\left(\frac◆LB◆b-\mu◆RB◆◆LB◆\sigma◆RB◆\right) - \Phi\!\left(\frac◆LB◆a-\mu◆RB◆◆LB◆\sigma◆RB◆\right)$

3.6 Normal approximation to Binomial

For large $n$ with $np \gt{} 5$ and $n(1-p) \gt{} 5$:

$B(n, p) \approx N(np, np(1-p))$

with continuity correction: $P(X \leq k) \approx P\!\left(Z \lt{} \frac◆LB◆k + 0.5 - np◆RB◆◆LB◆\sqrt{np(1-p)}◆RB◆\right)$.

warning

warning (Binomial) with a continuous one (Normal). Add or subtract 0.5 depending on the inequality direction.

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4. The Poisson Distribution

4.1 Definition

$X \sim \mathrm{Po}(\lambda)$ models the number of events in a fixed interval when events occur independently at a constant average rate $\lambda$.

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

4.2 Derivation as a limit of the Binomial

Theorem. If $n \to \infty$ and $p \to 0$ such that $np = \lambda$ remains constant, then $B(n, p) \to \mathrm{Po}(\lambda)$.

Proof.

$$\begin{aligned} P(X = k) &= \binom{n}{k}p^k(1-p)^{n-k} \\ &= \frac◆LB◆n(n-1)\cdots(n-k+1)◆RB◆◆LB◆k!◆RB◆ \cdot \frac◆LB◆\lambda^k◆RB◆◆LB◆n^k◆RB◆ \cdot \left(1-\frac◆LB◆\lambda◆RB◆◆LB◆n◆RB◆\right)^{n-k} \end{aligned}$$

Consider each factor as $n \to \infty$:

• $\dfrac◆LB◆n(n-1)\cdots(n-k+1)◆RB◆◆LB◆n^k◆RB◆ \to 1$ (each term $n-i \approx n$)
• $\left(1 - \dfrac◆LB◆\lambda◆RB◆◆LB◆n◆RB◆\right)^{n-k} \to e^{-\lambda}$ (using $\lim_{n\to\infty}(1+a/n)^n = e^a$)

Therefore:

$P(X = k) \to \frac{1}{k!} \cdot \lambda^k \cdot e^{-\lambda} = \frac◆LB◆e^{-\lambda}\lambda^k◆RB◆◆LB◆k!◆RB◆ \quad \blacksquare$

4.3 Proof that $E(X) = \lambda$

Proof.

$$\begin{aligned} E(X) &= \sum_{k=0}^{\infty}k \cdot \frac◆LB◆e^{-\lambda}\lambda^k◆RB◆◆LB◆k!◆RB◆ = \sum_{k=1}^{\infty}\frac◆LB◆e^{-\lambda}\lambda^k◆RB◆◆LB◆(k-1)!◆RB◆ \\ &= \lambda e^{-\lambda}\sum_{k=1}^{\infty}\frac◆LB◆\lambda^{k-1}◆RB◆◆LB◆(k-1)!◆RB◆ = \lambda e^{-\lambda}\sum_{j=0}^{\infty}\frac◆LB◆\lambda^j◆RB◆◆LB◆j!◆RB◆ \\ &= \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda \quad \blacksquare \end{aligned}$$

4.4 Proof that $\mathrm{Var}(X) = \lambda$

Proof. First compute $E(X(X-1))$:

$$\begin{aligned} E(X(X-1)) &= \sum_{k=2}^{\infty}k(k-1)\frac◆LB◆e^{-\lambda}\lambda^k◆RB◆◆LB◆k!◆RB◆ = \sum_{k=2}^{\infty}\frac◆LB◆e^{-\lambda}\lambda^k◆RB◆◆LB◆(k-2)!◆RB◆ \\ &= \lambda^2 e^{-\lambda}\sum_{j=0}^{\infty}\frac◆LB◆\lambda^j◆RB◆◆LB◆j!◆RB◆ = \lambda^2 e^{-\lambda} \cdot e^{\lambda} = \lambda^2 \end{aligned}$$

$E(X^2) = E(X(X-1)) + E(X) = \lambda^2 + \lambda$.

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$. $\blacksquare$

4.5 Additivity

If $X \sim \mathrm{Po}(\lambda)$ and $Y \sim \mathrm{Po}(\mu)$ are independent, then $X + Y \sim \mathrm{Po}(\lambda + \mu)$.

4.6 Conditions for the Poisson model

The Poisson distribution is appropriate when all of the following hold:

• Events occur independently of one another.
• Events occur at a constant average rate $\lambda$ in a fixed interval of time, space, or volume.
• The probability of more than one event occurring in a sufficiently small sub-interval is negligible.

These are sometimes called the Poisson postulates. When they are satisfied, the number of events in any interval of length $t$ follows $\mathrm{Po}(\lambda t)$.

Typical applications include: calls arriving at a call centre per hour, typing errors per page, radioactive decays per second, and cars passing a checkpoint per minute.

tip

tip constant over the interval and that events do not cluster. If events tend to occur in bursts, the Poisson model is not appropriate.

4.7 Poisson approximation to the Binomial

Practical rule. When $n \gt{} 50$ and $p \lt{} 0.1$, we may approximate $B(n, p)$ by $\mathrm{Po}(\lambda)$ where $\lambda = np$.

Justification. The theoretical result in Section 4.2 shows that as $n \to \infty$ and $p \to 0$ with $np = \lambda$ held constant, the binomial PMF converges pointwise to the Poisson PMF. The conditions $n \gt{} 50$ and $p \lt{} 0.1$ are practical thresholds that ensure:

1. $n$ is large enough that the discrete binomial is well-approximated by a limit distribution.
2. $p$ is small enough that the "rare event" assumption of the Poisson model is satisfied.
3. $\lambda = np$ is moderate (typically $0 \lt{} \lambda \lt{} 10$), so that neither distribution is heavily concentrated at a single point.

The approximation improves as $n$ increases and $p$ decreases while $\lambda = np$ remains fixed.

warning

warning and $n$ is large, use the normal approximation (Section 3.6) instead. The two approximations are complementary: Poisson handles the case of many trials with rare success, while normal handles the case of many trials with moderate success probability.

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5. Choosing the Right Distribution

Situation	Distribution
Fixed $n$ trials, success/failure	Binomial $B(n,p)$
Events in continuous interval, rare events	Poisson $\mathrm{Po}(\lambda)$
Continuous, bell-shaped	Normal $N(\mu,\sigma^2)$

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6. Coding of Random Variables

6.1 Definition

A coding (or linear transformation) of a discrete random variable $X$ is a new random variable $Y = aX + b$ where $a$ and $b$ are constants with $a \neq 0$.

Coding arises naturally when changing units (e.g. centimetres to metres, or Celsius to Fahrenheit) or when shifting and scaling a distribution.

6.2 Effect on expectation

Theorem. If $Y = aX + b$, then $E(Y) = aE(X) + b$.

Proof. Applying the definition of expectation to $Y$:

$E(Y) = \sum (ax_i + b)\,p_i = a\sum x_i\,p_i + b\sum p_i = aE(X) + b \cdot 1 = aE(X) + b \quad \blacksquare$

The key step is $\sum p_i = 1$, since the probabilities sum to 1.

6.3 Effect on variance

Theorem. If $Y = aX + b$, then $\mathrm{Var}(Y) = a^2\mathrm{Var}(X)$.

Proof.

$$\begin{aligned} \mathrm{Var}(Y) &= E(Y^2) - [E(Y)]^2 \\ &= E[(aX + b)^2] - [aE(X) + b]^2 \\ &= E[a^2X^2 + 2abX + b^2] - \\{a^2[E(X)]^2 + 2abE(X) + b^2\\} \\ &= a^2E(X^2) + 2abE(X) + b^2 - a^2[E(X)]^2 - 2abE(X) - b^2 \\ &= a^2\\{E(X^2) - [E(X)]^2\\} \\ &= a^2\mathrm{Var}(X) \quad \blacksquare \end{aligned}$$

Note how the terms $2abE(X)$ and $b^2$ cancel between $E(Y^2)$ and $[E(Y)]^2$.

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Adding a constant $b$ (a location shift) has no effect on variance. Only multiplying by $a$ (a scale change) affects variance, and it does so by a factor of $a^2$. This is why variance is measured in squared units of the original variable.

6.4 Effect on standard deviation

Since $\mathrm{Var}(Y) = a^2\mathrm{Var}(X)$, taking square roots gives:

$\mathrm{SD}(Y) = |a|\,\mathrm{SD}(X)$

The absolute value ensures the standard deviation remains non-negative regardless of the sign of $a$.

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Problem Set

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Problem 1

$X \sim B(10, 0.3)$. Find $P(X = 4)$, $P(X \leq 3)$, and $P(X \geq 7)$.

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Solution 1

$P(X=4) = \binom{10}{4}(0.3)^4(0.7)^6 = 210 \times 0.0081 \times 0.1176 \approx 0.2001$.

$P(X \leq 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3) \approx 0.0282 + 0.1211 + 0.2335 + 0.2668 \approx 0.6496$.

$P(X \geq 7) = P(X=7)+P(X=8)+P(X=9)+P(X=10) \approx 0.0090 + 0.0014 + 0.0001 + 0.0000 \approx 0.0106$.

If you get this wrong, revise: The Binomial Distribution — Section 2.

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Problem 2

Heights of men are normally distributed with mean 175 cm and standard deviation 8 cm. Find the probability that a randomly chosen man is taller than 185 cm.

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Solution 2

$X \sim N(175, 64)$. $P(X \gt{} 185) = P\!\left(Z \gt{} \dfrac{185-175}{8}\right) = P(Z \gt{} 1.25) = 1 - \Phi(1.25) \approx 1 - 0.8944 = 0.1056$.

If you get this wrong, revise: The Normal Distribution — Section 3.

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Problem 3

A call centre receives an average of 4.5 calls per minute. Find the probability of receiving exactly 6 calls in a given minute, and the probability of receiving more than 8 calls.

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Solution 3

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

$P(X=6) = \dfrac{e^{-4.5}(4.5)^6}{6!} = \dfrac◆LB◆0.01111 \times 8303.77◆RB◆◆LB◆720◆RB◆ \approx 0.1281$.

$P(X \gt{} 8) = 1 - P(X \leq 8) = 1 - \sum_{k=0}^{8}\dfrac{e^{-4.5}(4.5)^k}{k!} \approx 1 - 0.9804 = 0.0196$.

If you get this wrong, revise: The Poisson Distribution — Section 4.

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Problem 4

$X \sim B(100, 0.04)$. Use the Poisson approximation to find $P(X \leq 2)$.

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Solution 4

$\lambda = np = 4$. $X \approx \mathrm{Po}(4)$.

$P(X \leq 2) = e^{-4}\left(1 + 4 + \dfrac{16}{2}\right) = e^{-4}(1 + 4 + 8) = 13e^{-4} \approx 0.2381$.

If you get this wrong, revise: Derivation as a Limit — Section 4.2.

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Problem 5

Find $c$ such that $P(-c \lt{} Z \lt{} c) = 0.95$ where $Z \sim N(0,1)$.

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Solution 5

$P(-c \lt{} Z \lt{} c) = 2\Phi(c) - 1 = 0.95 \implies \Phi(c) = 0.975$.

From tables: $c \approx 1.96$.

If you get this wrong, revise: Standard Normal — Section 3.4.

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Problem 6

The number of emails received per hour follows $\mathrm{Po}(12)$. Find the probability of receiving between 10 and 15 emails (inclusive) in a given hour.

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Solution 6

$X \sim \mathrm{Po}(12)$.

$P(10 \leq X \leq 15) = P(X \leq 15) - P(X \leq 9)$.

$P(X \leq 15) \approx 0.7728$, $P(X \leq 9) \approx 0.2424$.

$P(10 \leq X \leq 15) \approx 0.7728 - 0.2424 = 0.5304$.

If you get this wrong, revise: The Poisson Distribution — Section 4.

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Problem 7

A machine produces bolts with lengths $X \sim N(50, 0.04)$ cm. Bolts with length less than 49.7 cm or greater than 50.3 cm are rejected. Find the proportion of bolts rejected.

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Solution 7

$\sigma = \sqrt{0.04} = 0.2$.

$P(X \lt{} 49.7) = P(Z \lt{} (49.7-50)/0.2) = P(Z \lt{} -1.5) = 0.0668$.

$P(X \gt{} 50.3) = P(Z \gt{} 1.5) = 0.0668$.

Proportion rejected $= 0.0668 + 0.0668 = 0.1336$ (13.36%).

If you get this wrong, revise: Finding Probabilities — Section 3.5.

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Problem 8

Prove that $E(aX + b) = aE(X) + b$ and $\mathrm{Var}(aX + b) = a^2\mathrm{Var}(X)$.

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Solution 8

$E(aX+b) = \sum(a x_i + b)p_i = a\sum x_i p_i + b\sum p_i = aE(X) + b$. ✓

$\mathrm{Var}(aX+b) = E[(aX+b)^2] - [E(aX+b)]^2 = E[a^2X^2 + 2abX + b^2] - [aE(X)+b]^2$ $= a^2E(X^2) + 2abE(X) + b^2 - a^2[E(X)]^2 - 2abE(X) - b^2$ $= a^2[E(X^2) - (E(X))^2] = a^2\mathrm{Var}(X)$. ✓

If you get this wrong, revise: Expectation and Variance — Section 1.2.

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Problem 9

$X \sim B(200, 0.15)$. Use the normal approximation with continuity correction to approximate $P(X \gt{} 35)$.

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Solution 9

$\mu = 200(0.15) = 30$, $\sigma^2 = 200(0.15)(0.85) = 25.5$, $\sigma \approx 5.05$.

$P(X \gt{} 35) \approx P\!\left(Z \gt{} \dfrac{35.5 - 30}{5.05}\right) = P(Z \gt{} 1.089) \approx 1 - 0.8621 = 0.1379$.

If you get this wrong, revise: Normal Approximation to Binomial — Section 3.6.

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Problem 10

If $X \sim \mathrm{Po}(3)$ and $Y \sim \mathrm{Po}(5)$ are independent, find $P(X + Y = 6)$.

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Solution 10

By additivity: $X + Y \sim \mathrm{Po}(3+5) = \mathrm{Po}(8)$.

$P(X + Y = 6) = \dfrac{e^{-8}(8)^6}{6!} = \dfrac◆LB◆e^{-8} \times 262144◆RB◆◆LB◆720◆RB◆ \approx \dfrac◆LB◆0.000335 \times 262144◆RB◆◆LB◆720◆RB◆ \approx 0.1221$.

If you get this wrong, revise: Additivity — Section 4.5.

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Problem 11

Starting from the definition $E(X) = \sum_{k=0}^{n} k\binom{n}{k}p^k(1-p)^{n-k}$, derive $E(X) = np$ using the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ and the binomial theorem.

Solution 11

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

$= np\sum_{k=1}^{n}\binom{n-1}{k-1}p^{k-1}(1-p)^{(n-1)-(k-1)} = np\sum_{j=0}^{n-1}\binom{n-1}{j}p^j(1-p)^{n-1-j}$

By the binomial theorem: $\sum_{j=0}^{n-1}\binom{n-1}{j}p^j(1-p)^{n-1-j} = [p+(1-p)]^{n-1} = 1$.

Therefore $E(X) = np$.

If you get this wrong, revise: Direct derivation of $E(X) = np$ from the PMF — Section 2.5.

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Problem 12

$X \sim \mathrm{Po}(7)$. Let $Y = 3X - 2$. Find $E(Y)$ and $\mathrm{Var}(Y)$.

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Solution 12

For $X \sim \mathrm{Po}(7)$: $E(X) = 7$ and $\mathrm{Var}(X) = 7$.

Using the coding formulae $E(aX+b) = aE(X)+b$ and $\mathrm{Var}(aX+b) = a^2\mathrm{Var}(X)$:

$E(Y) = 3(7) - 2 = 19$.

$\mathrm{Var}(Y) = 3^2 \times 7 = 63$.

Note that the additive constant $-2$ affects the mean but not the variance.

If you get this wrong, revise: Coding of Random Variables — Section 6.

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Problem 13

$X \sim B(80, 0.03)$. State whether the Poisson approximation is valid, giving reasons. If valid, use it to find $P(X \leq 1)$.

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Solution 13

Check conditions: $n = 80 \gt{} 50$ and $p = 0.03 \lt{} 0.1$. Both conditions are satisfied, so the Poisson approximation is valid with $\lambda = np = 80 \times 0.03 = 2.4$.

$X \approx \mathrm{Po}(2.4)$.

$P(X \leq 1) = P(X=0) + P(X=1) = e^{-2.4}\left(1 + 2.4\right) = 3.4\,e^{-2.4} \approx 3.4 \times 0.0907 \approx 0.3085$.

If you get this wrong, revise: Poisson approximation to the Binomial — Section 4.7.

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Problem 14

A discrete random variable $X$ has $E(X) = 5$ and $\mathrm{Var}(X) = 4$. Let $W = 2X + 3$. Find $E(W)$ and $\mathrm{Var}(W)$.

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Solution 14

$E(W) = 2E(X) + 3 = 2(5) + 3 = 13$.

$\mathrm{Var}(W) = 2^2 \times \mathrm{Var}(X) = 4 \times 4 = 16$.

$\mathrm{SD}(W) = \sqrt{16} = 4$.

If you get this wrong, revise: Coding of Random Variables — Section 6.

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Problem 15

Starting from $E(X(X-1)) = \sum_{k=0}^{n} k(k-1)\binom{n}{k}p^k(1-p)^{n-k}$, derive $\mathrm{Var}(X) = np(1-p)$ for $X \sim B(n,p)$.

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Solution 15

Using $k(k-1)\binom{n}{k} = n(n-1)\binom{n-2}{k-2}$:

$E(X(X-1)) = \sum_{k=2}^{n} n(n-1)\binom{n-2}{k-2}p^k(1-p)^{n-k} = n(n-1)p^2\sum_{j=0}^{n-2}\binom{n-2}{j}p^j(1-p)^{n-2-j} = n(n-1)p^2$

Then $E(X^2) = E(X(X-1)) + E(X) = n(n-1)p^2 + np$.

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = n(n-1)p^2 + np - n^2p^2 = np - np^2 = np(1-p)$.

If you get this wrong, revise: Direct derivation of $\mathrm{Var}(X) = np(1-p)$ from the PMF — Section 2.6.

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Problem 16

$X \sim B(120, 0.025)$. (a) Show that the Poisson approximation is appropriate. (b) Use it to find $P(X = 5)$. (c) State why the normal approximation would not be appropriate here.

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Solution 16

(a) $n = 120 \gt{} 50$ and $p = 0.025 \lt{} 0.1$, so the Poisson approximation is appropriate. $\lambda = np = 120 \times 0.025 = 3$.

(b) $X \approx \mathrm{Po}(3)$.

$P(X = 5) = \frac◆LB◆e^{-3} \times 3^5◆RB◆◆LB◆5!◆RB◆ = \frac◆LB◆e^{-3} \times 243◆RB◆◆LB◆120◆RB◆ = 2.025\,e^{-3} \approx 2.025 \times 0.0498 \approx 0.1008$

(c) For the normal approximation we need $np \gt{} 5$ and $n(1-p) \gt{} 5$. Here $np = 3 \lt{} 5$, so the normal approximation is not appropriate. The Poisson approximation is the correct choice since $p$ is small.

If you get this wrong, revise: Poisson approximation to the Binomial — Section 4.7.

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Problem 17

Temperatures in a city are modelled by $X \sim N(15, 9)$ in degrees Celsius. The temperature in Fahrenheit is $F = \frac{9}{5}X + 32$. Find $E(F)$, $\mathrm{Var}(F)$, and $P(F \gt{} 68)$.

Details

Solution 17

$E(F) = \frac{9}{5}E(X) + 32 = \frac{9}{5}(15) + 32 = 27 + 32 = 59^\circ\mathrm{F}$.

$\mathrm{Var}(F) = \left(\frac{9}{5}\right)^2 \times 9 = \frac{81}{25} \times 9 = \frac{729}{25} = 29.16$.

$\mathrm{SD}(F) = \sqrt{29.16} = 5.4$.

$P(F \gt{} 68) = P\!\left(Z \gt{} \dfrac{68 - 59}{5.4}\right) = P(Z \gt{} 1.667) \approx 1 - 0.9522 = 0.0478$.

If you get this wrong, revise: Coding of Random Variables — Section 6.

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