S4 Statistical distributions — printable coverage pack

S4.1 · Understand and use simple, discrete probability distributions (mean and variance of discrete random variables excluded), including the binomial distribution as a model; calculate probabilities using the binomial distribution.

A discrete probability distribution lists possible values with probabilities between $0$ and $1$ whose total is $1$ ; a discrete uniform distribution assigns equal probability to each value.
Use $X\sim\operatorname{B}(n,p)$ for a fixed number $n$ of independent trials, each with two outcomes and constant success probability $p$ .
For a binomial variable, $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ ; cumulative probabilities are often most efficiently found with a calculator.
Translate inequalities carefully: for integer-valued $X$ , $P(X<r)=P(X\leq r-1)$ and $P(X\geq r)=1-P(X\leq r-1)$ .

1. A discrete random variable $X$ takes values $0$ , $1$ and $2$ with probabilities $k$ , $3k$ and $4k$ respectively. Find $k$ and $P(X\geq1)$ .[3 marks]
Answer
- $k=\frac18$ .
- $P(X\geq1)=\frac78$ .
Method: Probabilities sum to $1$ , so $k+3k+4k=8k=1$ and $k=1/8$ . Therefore $P(X\geq1)=3k+4k=7k=7/8$ .

1. Let $X\sim\operatorname{B}(14,0.4)$ . Find $P(X\leq4)$ to $4$ decimal places.[3 marks]
Answer
- $0.2793$
Method: Use the cumulative binomial probability $P(X\leq4)=\sum_{r=0}^{4}\binom{14}{r}(0.4)^r(0.6)^{14-r}=0.279256\ldots$ . Therefore $P(X\leq4)=0.2793$ .

1. A binomial random variable $X$ has $10$ trials and satisfies $P(X=0)=0.8^{10}$ . Find its success probability $p$ , then calculate $P(X\geq3)$ to $4$ decimal places.[5 marks]
Answer
- $p=0.2$ .
- $P(X\geq3)=0.3222$ .
Method: For $X\sim\operatorname{B}(10,p)$ , $P(X=0)=(1-p)^{10}$ . Hence $(1-p)^{10}=0.8^{10}$ , so $p=0.2$ . Then $P(X\geq3)=1-P(X\leq2)=1-[0.8^{10}+10(0.2)(0.8)^9+\binom{10}{2}(0.2)^2(0.8)^8]=0.322200\ldots$ , giving $0.3222$ .

Write $X\sim\operatorname{N}(\mu,\sigma^2)$ , where $\mu$ is the mean and $\sigma$ is the standard deviation; standardise with $Z=\frac{X-\mu}{\sigma}$ .
The Normal curve is continuous, symmetric about $\mu$ , has total area $1$ , and has points of inflection at $\mu-\sigma$ and $\mu+\sigma$ .
A Normal model is plausible for a roughly symmetric, unimodal histogram with no strong outliers, but context and the variable's possible values also matter.
A binomial distribution may be approximated by a Normal distribution when $n$ is large and $p$ is close to $0.5$ ; use a continuity correction.

1. A random variable has distribution $X\sim\operatorname{N}(50,6^2)$ . Find $P(X<56)$ to $4$ decimal places.[2 marks]
Answer
- $0.8413$
Method: Standardise: $z=(56-50)/6=1$ . Therefore $P(X<56)=P(Z<1)=0.841344\ldots$ , which is $0.8413$ .

1. The lifetime $L$ of a component, in hours, is modelled by $L\sim\operatorname{N}(72,8^2)$ . Find the lifetime exceeded by exactly $10\%$ of components, to $1$ decimal place.[4 marks]
Answer
- $82.3$ hours
Method: If $P(L>l)=0.10$ , then $P(L\leq l)=0.90$ . The $0.90$ standard Normal quantile is $z=1.28155\ldots$ . Hence $l=72+8(1.28155\ldots)=82.252\ldots$ , so the required lifetime is $82.3$ hours.

1. Let $X\sim\operatorname{B}(200,0.35)$ . Use a Normal approximation with a continuity correction to estimate $P(X\geq82)$ , giving your answer to $4$ decimal places.[5 marks]
Answer
- $0.0441$
Method: Use $Y\sim\operatorname{N}(np,np(1-p))=\operatorname{N}(70,45.5)$ . The continuity correction gives $P(X\geq82)\approx P(Y>81.5)$ . Thus $z=(81.5-70)/\sqrt{45.5}=1.7049\ldots$ , so the upper-tail probability is $1-\Phi(1.7049\ldots)=0.0441$ .

Select a distribution by matching its assumptions to the variable and data-generating process, not merely because its parameters can be estimated.
A binomial model needs a fixed number of trials, two outcomes per trial, independence and a constant success probability.
A Normal model is continuous and symmetric with unbounded tails, so it can be unsuitable for strongly skewed, bounded or discrete data.
Support a choice with contextual evidence such as histogram shape, stability over time and dependence; state how a failed assumption could affect predictions.

1. A manufacturer inspects $12$ independently chosen switches. Each switch has probability $0.04$ of being faulty. State a suitable distribution for the number $F$ of faulty switches and give its parameters.[2 marks]
Answer
- $F\sim\operatorname{B}(12,0.04)$
Method: There are $12$ fixed, independent trials, each switch is faulty or not faulty, and the fault probability is constant at $0.04$ . Therefore a binomial model with $n=12$ and $p=0.04$ is suitable.

1. The masses of loaves from a stable production line form a roughly symmetric, single-peaked histogram with no clear outliers. Explain why a Normal model may be suitable and why a binomial model is not.[3 marks]
Answer
- Mass is continuous and the observed distribution is approximately symmetric and unimodal, supporting a Normal model.
- A binomial model counts successes in a fixed number of two-outcome trials, so it does not model individual loaf masses.
Method: Match the continuous measurement and bell-shaped empirical pattern to the features of a Normal distribution. Reject the binomial model because the response is a measured mass, not a discrete success count.