Normal probabilities

Sketch, shade, then the normal cdf

In 14 of 20 recent exams
1% of the marks
Paper 1 or 2
1 to 3 marks

How it is set in the exam

Paper 2 (calculator), 1–3 marks: usually the first part of a Section B normal question, and its value is reused later (as $p$ in a binomial, or as the denominator of a conditional). Very common (14 of 20 exam versions, 18 parts). Once on Paper 1 (no calculator), with no calculator: symmetry and a known area.

Spot it in the exam

Exam phrases to spot (5)

“find the probability that a randomly chosen … is less than …”
“what percentage of … lie between … and …”
“how many of a sample of $n$ are expected to …”
“is recorded to the nearest …”
“within $k$ standard deviations of the mean”

In disguise (3)

"what proportion of the population …": a probability in other words
an item "fails", "is rejected" or "is classed as large" beyond a cut-off: the probability of that tail
the mean and standard deviation are left as $\mu$ and $\sigma$ , and the bounds are written as $\mu \pm k\sigma$ : standardise, and $\mu$ and $\sigma$ cancel

Variants (4)

tail = lower, upper or between; the answer as a probability or a percentage
the expected number in a sample: $n \times$ the probability
a value recorded to a given precision: the interval half a unit either side
, no calculator: symmetry, tails + middle $= 1$ , and the 68–95–99.7 figures

Looks like it, but isn't (4)

a probability is given and a value is asked for: Find the value from a probability
"given that", with both events ranges of $X$ : Given that X is in a range
only "the two tails and the middle add to 1", with nothing else to find: the concept The normal distribution
the number of successes among $n$ items that each have this probability: Binomial probabilities (this type supplies its $p$ )

Worked example

A lower tail and a middle region

Paper 2 (calculator)

Question

The volume of lemonade in a bottle filled by a machine is normally distributed with mean $503$ ml and standard deviation $2.4$ ml.
(a) Find the probability that a randomly chosen bottle contains less than $500$ ml. [2]
(b) Find the percentage of bottles that contain between $498$ ml and $506$ ml. [2]

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See it move: Shade the area under the normal curve

$\mathrm{P}(V < 503) = 0.500$ , with $z = \frac{503 - 503}{2.4} = 0.00$ .
$V \sim \mathrm{N}(503, 2.4^{2})$ : the curve is centred on the mean, and every probability is an area under it. The whole area is 1.
Less than 500: 500 is below the mean, so it is the left tail, and the area must be less than 0.5: $0.106$ .
Between 498 and 506 contains the mean, so the area is large: $0.87573\ldots$ , or 87.6%.
506 is as far above 503 as 500 is below it, so the tail above 506 is also $0.106$ : a check by symmetry.

Steps

Write the model,
$X \sim \mathrm{N}(\mu, \sigma^2)$ , and have $\sigma$ (not $\sigma^2$ ) ready for the GDC.
Sketch and shade:
the curve with $\mu$ in the middle, each bound on the correct side of $\mu$ , the region shaded. Write the probability statement.
GDC:
normal probability. Write the statement with its value, then round to 3 s.f. (or give a percentage if asked).
4
Finish what was asked:
- an expected number: $n \times$ the probability (the unrounded one);
- a value recorded to the nearest unit $u$ : "recorded as $x$ " is $x - \frac{u}{2} \le X < x + \frac{u}{2}$ ;
- : symmetry about $\mu$ , tails + middle $= 1$ , and the 68–95–99.7 figures;
- $\mu$ and $\sigma$ unknown: standardise each bound, so that $\mu + k\sigma$ becomes $k$ .
Check
against the sketch: a region entirely on one side of $\mu$ has probability at most $0.5$ , and regions that don't overlap add up to at most $1$ .

Common mistakes

Entering the variance
for $\mathrm{N}(503, 5.76)$ the GDC needs $\sigma = 2.4$ , not $5.76$ .
No statement and no sketch
the probability statement, or a shaded sketch, earns the method mark. A wrong value on its own scores nothing.
The bounds the wrong way round
if the GDC gives $0.894$ for $\mathrm{P}(V < 500)$ when $500$ is below the mean, the region is the wrong side. The sketch catches it.

More mistakes (4)

Worrying about $<$ and $\le$
for a normal variable they give the same probability. For a binomial variable they don't.
Rounding a probability before reusing it
keep the full value (keep full-accuracy values) for an expected number, a binomial $p$ or a conditional later on.
A rounded measurement treated as exact
$\mathrm{P}(M = 3.4) = 0$ . "Recorded as $3.4$ " is $3.35 \le M < 3.45$ , and "recorded as at least $3.0$ " starts at $2.95$ .
Using values from a later part
in a "within $k$ standard deviations" part: it scores nothing. Standardise the bounds.

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