Practise: Averages, Spread and Surprises

The values of $x$ are 0, 1 and 2, and the sequence of two rolls fixes which one you get. Counting pairs gives $$P(x=0)=\frac{25}{36},P(x=1)=\frac{10}{36},P(x=2)=\frac{1}{36}.$$ The three entries add to 36/36, so the table is a distribution. The event $x\ge 1$ collects the last two, so $$P(x\ge 1)=\frac{10+1}{36}=\frac{11}{36}=0.30555\dots ,$$ which rounds to 0.306. Section 12.1 makes the point that the rule is yours to choose: keep the same 36 pairs and record their sum instead, and the values run from 2 to 12 with a different table. Both tables are correct; they answer different questions.

Only one of the six equally likely counts gives $x=0$, so $$P(x=1)=1-\frac{1}{6}=\frac{5}{6}=0.8333\dots ,$$ and $P(x=0)=1/6$. The expectation is $$E(x)=1\cdot \frac{5}{6}+0\cdot \frac{1}{6}=\frac{5}{6},$$ which rounds to 0.833. The count itself has expectation $(0+1+2+3+4+5)/6=2.5$. Both numbers describe the same six nights, and neither is more correct: the count answers "how many rings", the recode answers "did it ring". Section 12.1 warns that the coding fixes what the average can recover, and this pair of charts shows it in two lines.

The expectation is $$E(x)=0(0.2)+2(0.5)+4(0.3)=0+1+1.2=\mathrm{2.2.}$$ The squared deviations and their weighted contributions are

| value | probability | squared deviation | contribution | |---|---|---|---| | 0 | 0.2 | 4.84 | 0.968 | | 2 | 0.5 | 0.04 | 0.020 | | 4 | 0.3 | 3.24 | 0.972 |

The contributions add to $1.96$, so the variance is 1.96 square units and the standard deviation is $\sqrt{1.96}=1.4$ units. Section 12.2 reports the standard deviation rather than the variance because the square root returns to the units of the values; the variance is the quantity that adds cleanly when you combine independent pieces.

The total waiting time is $$5\times 2+68=10+68=78$$ days over six parcels, so the mean is $78/6=13$ days. The median is 2 days, because the sorted times are 2, 2, 2, 2, 2 and 68. Without the late parcel the mean would be 2 days. Section 12.2's pumpkin garden shows the same arithmetic with weights: one large value moves the mean and leaves the other five rows untouched. The late parcel also carries $68/78=87.2$ percent of the total waiting time, which is why the mean sits far from every parcel you can point at.

Section 12.3 gives $$\text{standard error}=\frac{\sigma }{\sqrt{n}}=\frac{9}{\sqrt{225}}=\frac{9}{15}=\mathrm{0.6.}$$ The average of 225 readings sits about 0.6 units from the population mean in repeated samples of this size. To halve that wobble you need four times the data, 900 readings, because the standard error shrinks with the square root rather than with the sample. The formula needs independent readings from one fixed population; 225 rows that influence one another do not earn it.

One die has mean $7/2=3.5$ and second moment $91/6$, so its variance is $$\frac{91}{6}-{\left(\frac{7}{2}\right)}^2=\frac{91}{6}-\frac{147}{12}=\frac{35}{12}=\mathrm{2.9167.}$$ The sum of 400 independent dice has variance $400\times 35/12=1166.67$ and standard deviation $$\sqrt{1166.67}=34.156\dots ,$$ which rounds to 34.16. The sum has mean $400\times 3.5=1400$. Dividing by 400, the average has mean 3.5 and spread $34.156/400=0.0854$. Section 12.4 states the two rules used here: the sum's spread grows like $\sigma \sqrt{n}$ and the average's spread shrinks like $\sigma /\sqrt{n}$. The shape claim is separate, and it says the sum is approximately normal whatever the flat shape of one die.

The count $S$ of heads has mean $400\times \frac{1}{2}=200$ and variance $400\times \frac{1}{2}\times \frac{1}{2}=100$. The gap from the mean to the target is $100$, so Chebyshev gives $$P(S\ge 300)\le \frac{100}{100^2}=\frac{1}{100}=\mathrm{0.01.}$$ The exact probability is about $1.3\times 10^{-24}$, and Hoeffding's bound is $e^{-50}=1.9\times 10^{-22}$. Both sharper numbers sit far below 0.01, which is section 12.5's lesson: Chebyshev uses only the variance, so it cannot tell a coin from a distribution with the same variance and a much longer tail. Its bound is honest and loose at the same time.

The variances are $4^2=16$ and $9^2=81$, and the covariance is $$0.5\times 4\times 9=18.$$ The matrix is $$\left(\begin{array}{cc}16& 18\\ 18& 81\end{array}\right),$$ whose determinant is $$16\times 81-18^2=1296-324=972.$$ The determinant is positive, so the matrix is positive definite and the cloud is an ellipse rather than a line. Its square root is $31.18$, and the normalising constant of the two-dimensional density carries $1/(2\pi \sqrt{972})=0.0051$. Section 12.6 reads the ellipse from the entries: the positive off-diagonal entry tilts it upward to the right, and setting it to zero would leave the ellipse upright.