Inverse Normal with Unknown Mean or SD

Usually you know the mean and standard deviation of a normal distribution and use them to find probabilities. Real problems often run the other way: a factory knows it wants only $1\%$ of bags to be underweight and needs to know what average fill to set the machine to. Here you’re given probabilities and have to find the mean, the standard deviation, or both. The tool that makes this possible is the z-value, which turns every normal distribution into the standard one.

Key ideas

Standardizing

If $X \sim N(\mu, \sigma^2)$, the standardized value of $x$ is

$$z = \frac{x - \mu}{\sigma}$$

It tells you how many standard deviations $x$ is from the mean, and $Z = \dfrac{X - \mu}{\sigma}$ follows the standard normal distribution $Z \sim N(0, 1)$. The key fact is that

$$P(X \lt x) = P\left(Z \lt \frac{x - \mu}{\sigma}\right)$$

(More on z-scores, with known $\mu$ and $\sigma$, in z-scores and the standard normal distribution.)

Why you need z when a parameter is unknown

Your GDC’s inverse normal function needs both $\mu$ and $\sigma$. If one of them is unknown, you can’t use $X$ directly. Instead:

1. Use the inverse normal on the standard normal ($\mu = 0$, $\sigma = 1$) to find the z-value that matches the given probability.
2. Set that z-value equal to $\dfrac{x - \mu}{\sigma}$ and solve for the unknown.

The IB expects these z-values from technology, not tables. Keep at least $4$ or $5$ significant figures in $z$ (or store it in your GDC) and only round the final answer to $3$ s.f.

Left areas and right areas

The inverse normal function works with the area to the left of a value (unless your GDC has a tail setting, which some do). Convert first:

Given	Area to the left of $x$
$P(X \lt x) = p$	$p$
$P(X \gt x) = p$	$1 - p$
$P(\mu - k \lt X \lt \mu + k) = p$	$\dfrac{1 + p}{2}$ for the upper value $\mu + k$

A quick sketch tells you the sign of $z$: a value below the mean has $z \lt 0$, a value above has $z \gt 0$.

The three types of question

• Unknown mean, known $\sigma$: $\ \mu = x - z\sigma$.
• Unknown $\sigma$, known mean: $\ \sigma = \dfrac{x - \mu}{z}$.
• Both unknown: two probabilities give two equations, $x_1 = \mu + z_1\sigma$ and $x_2 = \mu + z_2\sigma$. Solve them simultaneously (subtracting is quickest, or use your GDC’s equation solver).

Worked examples

Example 1: Unknown mean

A machine fills bottles with juice. The volume $V$ mL is normally distributed with standard deviation $4$ mL. Only $10\%$ of bottles contain less than $500$ mL. Find the mean volume.

Solution. $P(V \lt 500) = 0.1$, so $500$ is below the mean and its z-value is negative. From the GDC, inverse normal with area $0.1$, $\mu = 0$, $\sigma = 1$:

$$z = -1.28155\ldots$$

Then

$$\begin{aligned} \frac{500 - \mu}{4} &= -1.28155 \\ 500 - \mu &= -5.1262 \\ \mu &= 505.126\ldots \end{aligned}$$

The mean volume is $505$ mL (3 s.f.).

Check: on the GDC, normal cdf from $-\infty$ to $500$ with $\mu = 505.126$, $\sigma = 4$ gives $0.100$. ✓

Example 2: Unknown standard deviation

The times to run $5$ km in a fun run are normally distributed with mean $28$ minutes. $15\%$ of runners take longer than $32$ minutes. Find the standard deviation.

Solution. $P(T \gt 32) = 0.15$, so the area to the left of $32$ is $0.85$:

$$z = 1.03643\ldots \quad (\text{inverse normal, area } 0.85)$$ $$\frac{32 - 28}{\sigma} = 1.03643 \quad\Rightarrow\quad \sigma = \frac{4}{1.03643} = 3.859\ldots$$

The standard deviation is $3.86$ minutes (3 s.f.).

Example 3: Both unknown

The lengths of fish in a lake are normally distributed. $20\%$ of the fish are shorter than $30$ cm and $10\%$ are longer than $45$ cm. Find the mean and standard deviation.

$20\%$ of the area is below $30$ and $10\%$ is above $45$. The mean is between them, a little closer to $30$.

Solution. Find the two z-values:

• $P(L \lt 30) = 0.2$: $\ z_1 = -0.841621\ldots$
• $P(L \gt 45) = 0.1$, so the area to the left of $45$ is $0.9$: $\ z_2 = 1.281551\ldots$

Write $x = \mu + z\sigma$ for each:

$$\begin{aligned} \mu - 0.841621\sigma &= 30 \\ \mu + 1.281551\sigma &= 45 \end{aligned}$$

Subtract the first equation from the second:

$$2.123173\sigma = 15 \quad\Rightarrow\quad \sigma = 7.06489\ldots$$ $$\mu = 30 + 0.841621(7.06489) = 35.9459\ldots$$

The mean is $35.9$ cm and the standard deviation is $7.06$ cm (3 s.f.).

Check: with these values, the GDC gives $P(L \lt 30) = 0.200$ and $P(L \gt 45) = 0.100$. ✓

Example 4: Using the quartiles

The scores $X$ on a test are normally distributed. The upper quartile is $64$ and the interquartile range is $10$. Find the mean and standard deviation.

Solution. The normal curve is symmetric, so the mean (and median) is in the middle of the quartiles. $Q_1 = 64 - 10 = 54$, so

$$\mu = \frac{54 + 64}{2} = 59$$

The upper quartile has $75\%$ of the area to its left: $z = 0.674490\ldots$

$$\frac{64 - 59}{\sigma} = 0.674490 \quad\Rightarrow\quad \sigma = \frac{5}{0.674490} = 7.413\ldots$$

So $\mu = 59$ and $\sigma = 7.41$ (3 s.f.).

Common mistakes

Using the right-hand area in the inverse normal. If $P(X \gt 32) = 0.15$, the area to the left is $0.85$, not $0.15$. Using $0.15$ gives $z = -1.036$, and a negative standard deviation.

Getting the sign of z wrong. A sketch fixes this. A value below the mean must have a negative $z$. If you get $\sigma \lt 0$, a sign has gone wrong.

Rounding z too early. Using $z = -1.28$ instead of $-1.28155$ in Example 1 gives $\mu = 505.12$ instead of $505.126$. That’s harmless here, but in two-equation problems early rounding can change the third significant figure. Store the z-values in your GDC.

Plugging an unknown mean into the inverse normal function. You can’t put $\mu$ into the GDC if you don’t know it. Work with $Z \sim N(0, 1)$ to get $z$, then solve $\dfrac{x - \mu}{\sigma} = z$.

Mixing up variance and standard deviation. $X \sim N(\mu, \sigma^2)$ has variance $\sigma^2$. If you’re asked for the variance, square your $\sigma$. If a question gives $N(50, 16)$, then $\sigma = 4$, not $16$.

Forgetting to check. Put your answers back into the normal cdf. If you don’t get the given probabilities, something is wrong.

Practice

1. (Warm-up) $X \sim N(\mu, 3^2)$ and $P(X \lt 20) = 0.25$. Find $\mu$.

Solution

Inverse normal (area $0.25$): $z = -0.674490$.

$$\frac{20 - \mu}{3} = -0.674490 \quad\Rightarrow\quad \mu = 20 + 3(0.674490) = 22.0235\ldots$$

$\mu = 22.0$ (3 s.f.).

2. (Warm-up) $X \sim N(50, \sigma^2)$ and $P(X \gt 58) = 0.2$. Find $\sigma$.

Solution

The area to the left of $58$ is $0.8$: $z = 0.841621$.

$$\frac{58 - 50}{\sigma} = 0.841621 \quad\Rightarrow\quad \sigma = \frac{8}{0.841621} = 9.505\ldots$$

$\sigma = 9.51$ (3 s.f.).

3. (Core) Bags of rice are labelled $500$ g. The mass of rice in a bag is normally distributed with standard deviation $6$ g. The company wants only $1\%$ of bags to contain less than $500$ g. What mean mass should the machine be set to?

Solution

$P(M \lt 500) = 0.01$: $z = -2.326348$.

$$\frac{500 - \mu}{6} = -2.326348 \quad\Rightarrow\quad \mu = 500 + 6(2.326348) = 513.958\ldots$$

The machine should be set to a mean of $514$ g (3 s.f.).

4. (Core) The scores on an exam are normally distributed. $30\%$ of students scored less than $55$ and $15\%$ scored more than $80$. Find the mean and standard deviation of the scores.

Solution

$P(X \lt 55) = 0.3$: $z_1 = -0.524401$. $P(X \gt 80) = 0.15$, left area $0.85$: $z_2 = 1.036433$.

$$\begin{aligned} \mu - 0.524401\sigma &= 55 \\ \mu + 1.036433\sigma &= 80 \end{aligned}$$

Subtract: $1.560834\sigma = 25$, so $\sigma = 16.017\ldots$ Then $\mu = 55 + 0.524401(16.017) = 63.399\ldots$

Mean $63.4$, standard deviation $16.0$ (3 s.f.).

5. (Core) $X \sim N(40, \sigma^2)$ and $P(36 \lt X \lt 44) = 0.8$. Find $\sigma$.

Solution

The interval is symmetric about the mean, so $10\%$ of the area is in each tail. The area to the left of $44$ is $0.9$: $z = 1.281552$.

$$\frac{44 - 40}{\sigma} = 1.281552 \quad\Rightarrow\quad \sigma = 3.1212\ldots$$

$\sigma = 3.12$ (3 s.f.).

6. (Core) A bus journey time is normally distributed with standard deviation $5$ minutes. $90\%$ of journeys take less than $45$ minutes.

• (a) Find the mean journey time.
• (b) Find the probability that a journey takes less than $30$ minutes.

Solution

(a) Area $0.9$: $z = 1.281552$.

$$\frac{45 - \mu}{5} = 1.281552 \quad\Rightarrow\quad \mu = 45 - 6.40776 = 38.592\ldots$$

The mean is $38.6$ minutes (3 s.f.).

(b) Using the stored $\mu = 38.592$ and $\sigma = 5$, the GDC gives

$$P(T \lt 30) = 0.0429 \text{ (3 s.f.)}$$

(If you use the rounded mean $38.6$, you get $0.0427$; that’s why you keep full accuracy.)

7. (Core) For the juice bottles in Example 1, $10\%$ of bottles contain less than $500$ mL. A shop buys $12$ bottles, chosen at random. Find the probability that at most one of them contains less than $500$ mL.

Solution

The number $N$ of underfilled bottles follows a binomial distribution: $N \sim B(12, 0.1)$.

$$P(N \le 1) = (0.9)^{12} + 12(0.1)(0.9)^{11} = 0.6590\ldots$$

The probability is $0.659$ (3 s.f.). On a GDC, use binomial cdf with $n = 12$, $p = 0.1$, up to $1$.

8. (Challenge) The lengths of bolts are normally distributed. $5\%$ of bolts are longer than $52.0$ mm and $2\%$ are shorter than $47.5$ mm.

• (a) Find the mean and standard deviation of the lengths.
• (b) Bolts are rejected if they are shorter than $48$ mm or longer than $52$ mm. Find the proportion of bolts that are accepted.

Solution

(a) $P(L \gt 52) = 0.05$, left area $0.95$: $z_1 = 1.644854$. $P(L \lt 47.5) = 0.02$: $z_2 = -2.053749$.

$$\begin{aligned} \mu + 1.644854\sigma &= 52 \\ \mu - 2.053749\sigma &= 47.5 \end{aligned}$$

Subtract: $3.698603\sigma = 4.5$, so $\sigma = 1.21668\ldots$ and $\mu = 52 - 1.644854(1.21668) = 49.9987\ldots$

Mean $50.0$ mm, standard deviation $1.22$ mm (3 s.f.).

(b) With the stored values, the GDC gives $P(48 \lt L \lt 52) = 0.900$ (3 s.f.). About $90.0\%$ of bolts are accepted. (Rounded values $50.0$ and $1.22$ give $0.899$.)

9. (Challenge) The heights of a type of seedling are normally distributed with mean $\mu$ cm and standard deviation $0.2\mu$ cm. $10\%$ of seedlings are taller than $30$ cm. Find $\mu$.

Solution

The area to the left of $30$ is $0.9$: $z = 1.281552$.

$$\begin{aligned} \frac{30 - \mu}{0.2\mu} &= 1.281552 \\ 30 - \mu &= 0.256310\mu \\ 30 &= 1.256310\mu \\ \mu &= 23.879\ldots \end{aligned}$$

$\mu = 23.9$ cm (3 s.f.), so $\sigma = 0.2\mu \approx 4.78$ cm.

Check: $P(X \gt 30)$ with $\mu = 23.879$, $\sigma = 4.7759$ is $0.100$. ✓