Unknown mean or standard deviation

Standardise with z

In 12 of 20 recent exams
2% of the marks
Paper 2, calculator
3 to 7 marks

How it is set in the exam

Paper 2 (calculator) only, 3–7 marks, often the longest normal part of a Section B question. Very common (12 of 20 exam versions, 16 parts).

Spot it in the exam

Exam phrases to spot (4)

“the mean is … and $p\,\%$ are heavier than …; find $\sigma$ ”
“the interquartile range is …; find $\sigma$ ”
“find the mean and standard deviation”
“ $p\,\%$ are below … and $q\,\%$ are above …”

In disguise (5)

$\mathrm{N}(\mu, \sigma^2)$ with the letters left in, then probabilities at numerical values given (such as $\mathrm{P}(X > 50) = 0.3$ ): the letters are the unknowns, even when a first part only asks for $k$ in $50 = \mu + k\sigma$
" $90\%$ of the values lie between $a$ and $b$ ", with $a$ and $b$ symmetric about a given mean: each tail is $5\%$
a second population "whose mean is $3$ less and whose standard deviation is twice as large": one pair of unknowns shared by two distributions
the second fact is a conditional probability, or the total of a tree: turn it into a plain tail first
no tail is given on its own: the middle percentage is, and one tail is a multiple of the other. Find both tails first ( $x + \text{middle} + kx = 1$ , the tail-ratio step of Find the value from a probability), then one $z$ -equation for each bound

Variants (2)

unknown = $\sigma$ (with $\mu$ known), $\mu$ (with $\sigma$ known), or both
given = one tail, a symmetric interval, the IQR, two tails, an interval and a cumulative probability, a tree total, a related population, or a conditional probability

Looks like it, but isn't (2)

the IQR or standard deviation of a data set: Averages and spread or Box plots
the probability is at a bound written as $\mu + k\sigma$ (such as $\mathrm{P}(X > \mu + k\sigma) = 0.1$ ), and $k$ is asked for, whether $\mu$ and $\sigma$ are numbers or letters: Find the value from a probability ( $k$ is a $z$ -value; $\mu$ and $\sigma$ are never found)

Worked example

σ from the mean and one tail

Paper 2 (calculator)

Question

The time taken to assemble a chair is normally distributed with mean $85$ minutes and standard deviation $\sigma$ minutes. $10\%$ of chairs take longer than $100$ minutes to assemble.
(a) Find the value of $\sigma$ . [3]
(b) Find the probability that a chair takes less than $75$ minutes to assemble. [2]

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

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See it move: Widen the curve until the tail fits

\sigma

6.0

$\sigma = 6.0$ : $z = \frac{100 - 85}{6.0} = 2.500$ , and $\mathrm{P}(T > 100) = 0.006$ . Too little in the tail: widen the curve.
A bigger $\sigma$ spreads the same area of 1 wider and lower, so more of it lies beyond 100.
$\sigma$ is unknown, so inverse normal on $T$ is impossible: standardise. $\mathrm{P}(T < 100) = 0.9$ gives $z = 1.2815\ldots$ on $Z$ .
$\frac{100 - 85}{\sigma} = 1.2815\ldots$ , so $\sigma = 11.704\ldots \approx 11.7$ minutes. Then $\mathrm{P}(T < 75) = 0.196$ , with $\sigma$ unrounded.

Which method?

Unknown	Method
$\sigma$ only (or $\mu$ only)	one $z$ -equation, $\frac{x - \mu}{\sigma} = z$ with $z$ from inverse normal on $Z$ ; or the GDC solver on a normal probability with the unknown left in (both score full marks)
$\mu$ and $\sigma$	two $z$ -equations, solved simultaneously (GDC). The equations must use $z$ -values, not probabilities

Steps

Sketch,
with each given value on the correct side of the mean.
2
Turn each fact into an area to the left of a known value:
- an upper tail: the complement;
- a symmetric interval with probability $p$ : $\frac{1 - p}{2}$ in each tail;
- the IQR: the quartiles are $\mu \pm \frac{\text{IQR}}{2}$ , with areas $0.25$ and $0.75$ to their left;
- an interval: add the tail below it (cumulate);
- a conditional or a tree total: solve for the plain tail first.
Inverse normal on $Z$
( $\mu = 0$ , $\sigma = 1$ ) for each area. Write each $z$ -value, then $\frac{x - \mu}{\sigma} = z$ , or $x = \mu + z\sigma$ . For a second population, use its own mean and standard deviation, written in $\mu$ and $\sigma$ : a population with mean $\mu - 4$ and standard deviation $2\sigma$ gives $\frac{x - (\mu - 4)}{2\sigma} = z$ .
Solve:
one equation by rearranging; two simultaneously.
State $\mu$ and $\sigma$ , labelled,
to 3 s.f. Check: $\sigma > 0$ , and the values give back the probabilities (normal probability).

Common mistakes

Probabilities in the equation instead of $z$ -values
$\mu + 0.85\sigma = 70$ scores . The equations need $z$ from inverse normal on $Z$ .
Inverse normal on $X$
when $\mu$ or $\sigma$ is unknown: the GDC can't do it. Use $Z$ , with $\mu = 0$ and $\sigma = 1$ .
An upper tail without the complement
"15% above 70" needs the area $0.85$ to the left of $70$ . Using $0.15$ gives $z = -1.036$ , the wrong sign: a value above the mean must have a positive $z$ .

More mistakes (5)

An interval used as a left tail
cumulate first.
The whole IQR as the distance from the mean
the quartiles are $\mu \pm \frac{\text{IQR}}{2}$ .
A second population standardised with $\mu$ and $\sigma$
its own mean and standard deviation go in, e.g. $\frac{x - (\mu - 4)}{2\sigma}$ , not $\frac{x - \mu}{\sigma}$ .
$z$ rounded too early
keep at least 4 significant figures (better, the stored value), or $\mu$ and $\sigma$ can come out wrong at 3 s.f.
$\mu$ and $\sigma$ not labelled
the final mark needs each value clearly identified.

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