6. Probability & Statistics 2

Match λ to the interval length, check that events are countable and rare enough for the model, and use complements for “at least one” questions.

If a call centre averages 3 calls per minute, P(2 calls in one minute)=e^{−3}3²/2.

Use the equality as a model check, not as a statement that every sample has equal mean and variance. Estimate λ from appropriate exposure.

If the observed average is 4 events per hour, a two-hour interval has Po(8), not Po(4).

If flaws occur randomly at an average of 0.8 per metre, the number in 5 metres can be modelled by $Po(4)$, provided the rate is stable and flaws do not cluster.

Question the model when the rate varies systematically, one event triggers another, observations are capped by a fixed number of trials, or the exposure itself is unclear. State the modelling assumption when context makes it relevant.

If $X\sim B(200,0.01)$, then $n>50$ and $np=2<5$, so use $Y\sim Po(2)$. Thus $$P(X\ge2)\approx P(Y\ge2)=1-P(Y=0)-P(Y=1)=1-e^{-2}(1+2).$$

The approximation replaces many rare independent success opportunities by a random-event count with the same expected value $np$.

If $X\sim Po(25)$, the guide condition holds and $Y\sim N(25,25)$. Therefore $$P(X\ge30)\approx P(Y>29.5)=P\left(Z>\frac{29.5-25}{5}\right)=P(Z>0.9).$$

State the large-mean check, write the approximating normal distribution, correct the integer boundary, then standardise with $\sqrt{\lambda}$.

If independent $X\sim N(10,4)$ and $Y\sim N(6,9)$, then $$X-Y\sim N(10-6,\;4+9)=N(4,13).$$ The mean subtracts, but the variances add because the coefficient of $Y$ is squared: $(-1)^2=1$.

If independent counts $A\sim Po(2)$ and $B\sim Po(3.5)$ refer to compatible exposure, then $A+B\sim Po(5.5)$. A difference of Poisson variables is not Poisson by this result.

Define the required total, difference or cost as a linear combination; calculate its mean; calculate its variance with squared coefficients and an independence check; identify the resulting distribution; only then standardise or evaluate its probability.

For $f(x)=3/x^4$ on $x\ge1$, $$\int_1^\infty 3x^{-4}\,dx=1,$$ so it is a valid density. Also $P(X>2)=\int_2^\infty3x^{-4}\,dx=1/8$.

A density can exceed 1 over a narrow interval; only its total area must be 1. Density carries inverse units so that integrated probability is unitless.

Let $f(x)=kx$ for $0\le x\le2$. Normalisation gives $$1=\int_0^2kx\,dx=2k,$$ so $k=1/2$. Therefore $P(X>1)=\int_1^2x/2\,dx=3/4$.

For this density, $$E(X)=\int_0^2x(x/2)\,dx=4/3,$$ $$E(X^2)=\int_0^2x^2(x/2)\,dx=2,$$ so $$Var(X)=2-(4/3)^2=2/9.$$

The median $m$ divides the total area in half: $$\int_0^m x/2\,dx=1/2.$$ Hence $m^2/4=1/2$ and, since $m$ lies in the support, $m=\sqrt2$. Other percentiles use the same direct-area equation.

Define the target population before sampling and consider coverage, non-response and selection bias. Larger samples reduce random error but do not automatically remove systematic bias.

A survey of 500 randomly chosen voters is a sample; the proportion supporting a policy in all eligible voters is a population parameter.

Choosing random numbers from a school email list gives random selection from that list, but it cannot represent pupils omitted from the list. Explain the missing group and the likely direction of distortion when context supports it.

The syllabus asks for simple criticism and elementary random-number use. Knowledge of named schemes such as quota or stratified sampling is not required.

Distinguish variability of individual observations from variability of means, and account for finite-population or dependence conditions when relevant.

If σ=12 and n=36, the standard error of X̄ is 2, even though individual values vary with standard deviation 12.

If individual values are $N(100,15^2)$ and $n=25$, then $\bar X\sim N(100,9)$. Thus $$P(\bar X>106)=P\left(Z>\frac{106-100}{3}\right)=P(Z>2).$$

A more skewed or heavy-tailed population generally needs a larger sample before the approximation is convincing. Randomness and independence still matter; CLT does not remove sampling bias or dependence.

A non-normal population has mean 10 and standard deviation 4. For an independent sample of size 64, use $\bar X\approx N(10,0.25)$, so $$P(\bar X>11)\approx P\left(Z>\frac{11-10}{0.5}\right)=P(Z>2).$$

For data $2,4,6$, $\bar x=4$ and $\sum(x-\bar x)^2=4+0+4=8$, so the unbiased variance estimate is $s^2=8/(3-1)=4$.

Unbiased means the estimation procedure gives the true population value on average over repeated random samples. One particular estimate can still be above or below the parameter.

A normal population has known $\sigma=12$. From $n=36$ observations, $\bar x=50$. A 95% interval is $$50\pm1.96\left(\frac{12}{6}\right)=50\pm3.92,$$ giving $(46.08,53.92)$.

Interpret in context: we are 95% confident that the population mean lies between 46.08 and 53.92. Repeated use of this method captures the fixed population mean in about 95% of intervals.

If 40 of 100 sampled people support a proposal, $\hat p=0.40$ and $SE\approx\sqrt{0.4(0.6)/100}=0.0490$. A 95% interval is $$0.40\pm1.96(0.0490)=0.40\pm0.096,$$ giving approximately $(0.304,0.496)$.

We are 95% confident that the population proportion supporting the proposal lies between about 30.4% and 49.6%, assuming the sample was random and the large-sample approximation is appropriate.

Calculate a null tail probability (p-value) or compare the statistic with a pre-set rejection region. Reject $H_0$ when the p-value is at most $\alpha$ or the statistic lies in the rejection region; otherwise do not reject $H_0$.

A two-sided p-value of 0.04 gives sufficient evidence against $H_0$ at 5% but not at 1%. The final statement must name the contextual claim in $H_1$, such as evidence that a mean has changed.

To test whether a coin has probability greater than $0.5$ of heads, use $H_0:p=0.5$, $H_1:p>0.5$. If 15 heads occur in 20 tosses, the direct p-value is $$P(X\ge15\mid X\sim B(20,0.5))=\sum_{r=15}^{20}\binom{20}{r}(0.5)^{20}.$$ Compare this value with the stated significance level.

For a suitable binomial null model use $N(np_0,np_0(1-p_0))$; for a suitable Poisson null model use $N(\lambda_0,\lambda_0)$. Apply continuity correction to the observed/critical integer boundary before standardising.

For a two-tailed discrete test, construct both extreme regions under $H_0$ so their combined null probability does not exceed the significance level. Discreteness may make the actual significance smaller than the nominal level.

A normal population has known $\sigma=10$. Test $H_0:\mu=50$ against $H_1:\mu>50$ using $n=100$ and $\bar x=51.8$. Then $SE=1$ and $z=1.8$, giving upper-tail p-value $P(Z\ge1.8)\approx0.0359$. Reject at 5% and conclude there is evidence that the population mean exceeds 50.

The same $z=1.8$ would not reject a two-sided test at 5%, because the two-sided 5% critical values are approximately $\pm1.96$. Always formulate $H_1$ before using the data.

Type I is a false positive: evidence is declared when the null state is actually true. Type II is a false negative: the test fails to detect a specified departure from the null.

For a fixed test rule, the Type I probability is evaluated under the null parameter. The Type II probability depends on which alternative parameter value is actually true; it is not one universal number.

Suppose $X_1,\ldots,X_{25}$ are normal with known $\sigma=10$, and a 5% upper-tail test of $H_0:\mu=50$ rejects when $\bar X>53.29$ (since $50+1.645(10/5)=53.29$). Under $H_0$, $$P(\text{Type I})=P(\bar X>53.29\mid\mu=50)\approx0.05.$$

If the true mean is instead 54, the same fixed rule makes a Type II error when $\bar X\le53.29$. Hence $$P(\text{Type II at }\mu=54)=P\left(Z\le\frac{53.29-54}{2}\right)=P(Z\le-0.355)\approx0.361.$$

For a binomial rule ‘reject $H_0:p=p_0$ when $X\ge c$’, Type I is $P_{p_0}(X\ge c)$ and Type II at $p=p_1$ is $P_{p_1}(X<c)$. For a Poisson test, use the same regions with $Po(\lambda_0)$ and the stated alternative $Po(\lambda_1)$. Evaluate these discrete probabilities directly when required.