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The three standard continuous shapes, the stories that pick them out, memorylessness, and standardising a normal variable.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will choose between the uniform, exponential and normal families from the story a quantity comes with, compute probabilities and quantiles for a uniform variable, state what memorylessness says about an exponential waiting time, and standardise a normal value. You will also name what each choice of family has assumed away.
The previous lesson said what a density is and how to compute with one, but every density in it was invented for the question. This lesson stocks the shelf: three densities that between them model most continuous quantities, and the stories that decide which of the three a situation calls for.
Uniform on $[a, b]$: density $\dfrac{1}{b-a}$ on the interval and zero outside. Exponential with rate $\lambda$: density $\lambda e^{-\lambda x}$ for $x \ge 0$. Normal with mean $\mu$ and standard deviation $\sigma$: the bell-shaped density, written $N(\mu, \sigma^{2})$. Standardising a variable means subtracting its mean and dividing by its standard deviation, which turns any normal variable into $N(0, 1)$.
Uniform. No part of an interval is favoured over another of the same length. Its density is flat, its cumulative function is a straight line, and every question about it is answered with a ruler: $P(c \le X \le d) = \dfrac{d - c}{b - a}$.
Exponential. The gap between arrivals of a steady stream. Its survival function is $P(X > x) = e^{-\lambda x}$, and its defining property is memorylessness: $$P(X > s + t \mid X > s) = P(X > t).$$ The wait remaining after any wait already spent has the distribution it had at the start. It is the exact continuous partner of the Poisson model — if the counts over a stretch are Poisson with rate $\lambda$, the gaps between them are exponential with the same $\lambda$.
Normal. The shape a sum of many small independent contributions takes, whatever shape each contribution had. That claim is the central limit theorem and is proved at the end of this course; here it is the reason the family matters. Every normal variable becomes $N(0, 1)$ once standardised, so $z = \dfrac{x - \mu}{\sigma}$ is the only computation the family needs before a table is consulted.
Every limit theorem in this course is a statement about a long-run average, and about nothing else. The coin has no memory, no debt and no plan; the average settles down because later trials outnumber the early ones, not because anything corrects for them.
Another way: table
Three shapes, and what each one refuses to say.
| Family | Support | Shape | What it assumes away |
|---|---|---|---|
| Uniform | $[a, b]$ | flat | that any part of the range is special |
| Exponential | $[0, \infty)$ | falls away | that anything ages |
| Normal | the whole line | symmetric bell | that the tails are heavy or lopsided |
Another way: steps
A component whose lifetime is exponential with a mean of five years, at three ages.
| Age now | Chance of surviving five more years |
|---|---|
| New | $e^{-1} \approx 0.368$ |
| Five years old | $e^{-1} \approx 0.368$ |
| Twenty years old | $e^{-1} \approx 0.368$ |
The column is constant, and that is the model, not an accident of the numbers. It is a reasonable description of a fuse blowing from a random surge and a poor one of a bearing wearing out. Choosing the exponential is choosing to say that ageing does not happen, and that choice should be made out loud.
Expecting a long wait to make the next moment more likely. Under the exponential model it does not; the wait remaining is as long as ever. This is the gambler's fallacy in continuous dress.
Using the exponential for something that wears. Memorylessness is a strong assumption and it is false of most mechanical parts.
Choosing the normal because the histogram looks bell-shaped. The reason to reach for it is the story — many small independent contributions added together. A quantity with a floor at zero and a long right tail is not normal however much of it clusters in the middle.
Forgetting to subtract before dividing when standardising. $z = (x - \mu)/\sigma$, in that order.
A train leaves every $12$ minutes and you arrive without consulting the timetable, so your wait is uniform on $[0, 12]$.
Name the family from the story.
$P(\text{wait} \le 3) = 3/12 = 0.25$.
A length over a width.
The median wait is $6$ minutes, half way along.
A straight line inverts by division.
Heights are modelled as normal with mean $170$ and standard deviation $8$ centimetres.
Two parameters.
A height of $186$ standardises to $(186 - 170)/8 = 2$.
Subtract, then divide.
So it is two standard deviations above the mean, and the table for $N(0,1)$ answers every question about it.
One table serves every normal variable.
Subtract the mean: $42 - 50 = -8$.
The distance in the original units.
Divide by the standard deviation: $-8 / 4$.
Now in standard deviations.
So $z = -2$: two standard deviations below the mean.
Match each family to the situation it is the standard model for.
| The point at which a rod is snapped, with no part favoured | The time until the next call reaches a steady switchboard | The total weight of $304$ independently packed sacks | The number of faulty items in a batch of $304$ | |
|---|---|---|---|---|
| Uniform | ||||
| Exponential | ||||
| Normal | ||||
| Binomial |
A variable is uniform on the interval from $1$ to $17$. Give the value below which a quarter of the probability lies, and the median.
Lower quartile: p. Median: q.
A bus may arrive at any moment in the next $20$ minutes, with no moment favoured. Give the three probabilities.
| Probability | |
|---|---|
| Arrives within $4$ minutes | |
| Arrives between $4$ and $12$ minutes | |
| Arrives after $12$ minutes |
A component's lifetime is modelled as exponential. It has already run for $4$ years without failing. What does the model say about the time it has left?
A measurement is modelled as normal with mean $40$ and standard deviation $4$. A reading of $49$ is how many standard deviations from the mean?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Calls arrive at $8$ an hour and the gap until the next one is modelled as exponential. Give the set of values that waiting time can take, as an interval.
This task has no paper form; do it on a device.
You can pick the family a story calls for, work with a uniform variable directly, and standardise a normal value. Say in your own words what the exponential model claims about a component that has already run for years without failing.
9. Your turn: a variable is normal with mean $50$ and standard deviation $4$; standardise the value $42$, step 3