Inverse Normal And Parameter Questions

Scope Label

Core 9758. This note covers the value-from-area side of normal-distribution work: inverse normal, unknown parameters, and probability inequalities involving boundary values.

What This Note Assumes

You should already know:

• $X\sim N(\mu ,{\sigma }^2)$ means the second parameter is variance;
• normal probabilities are areas;
• standardisation uses $z=(x-\mu )/\sigma$;
• right-tail and left-tail probabilities are different regions.

Direct Probability Versus Inverse Normal

Direct normal probability starts with a boundary value and asks for an area:

value -> area

Inverse normal starts with an area and asks for the boundary value:

area -> value

For example,

$$P(X<a)=0.90$$

means the value $a$ cuts off left-tail area $0.90$.

The central idea is:

Inverse normal reasoning locates a boundary from a known area.

How to read this figure. The shaded left-tail area $p$ is given, so inverse normal first finds the corresponding standard-normal quantile and then converts it to the boundary $a=\mu +z_p\sigma$. Endpoints carry zero probability.

Common Inverse Normal Forms

A left-tail form is already in calculator-friendly form:

$$P(X<a)=p.$$

Inverse quantiles require $0<p<1$.

A right-tail form should often be converted to a left-tail form:

$$P(X>a)=p$$

means

$$P(X<a)=1-p.$$

A central symmetric form may look like

$$P(\mu -a<X<\mu +a)=p.$$

Here $a\ge 0$ is a half-width.

If the region is symmetric about $\mu$, the remaining area is split equally between the two tails.

Do not split tail areas equally unless the interval is actually symmetric about the mean.

Core Example: Right-Tail Inverse Normal

Suppose $X\sim N(60,8^2)$ and find $a$ such that $P(X>a)=0.10$.

Inverse-normal functions use a left cumulative area, so convert first:

$$P(X<a)=\mathrm{0.90.}$$

The corresponding standard-normal quantile is $z_{0.90}\approx 1.282$, hence

$$a=60+1.282(8)\approx \mathrm{70.3.}$$

The boundary is above the mean, consistent with a small right tail.

Core Example: Symmetric Central Area

Suppose $X\sim N(100,12^2)$. Find the symmetric interval about $100$ containing probability $0.90$.

The total tail area is $1-0.90=0.10$, so each tail has area $0.05$. The upper endpoint therefore has cumulative area

$$0.05+0.90=0.95=\frac{1+0.90}{2}.$$

Since $z_{0.95}\approx 1.645$, the half-width is $1.645(12)\approx 19.7$. The interval is approximately

$$80.3<X<\mathrm{119.7.}$$

Equal tail splitting is valid here because the required interval is explicitly centred at the mean.

Core Example: A Non-Symmetric Interval

Let $X\sim N(50,10^2)$ and find $b$ such that $P(45<X<b)=0.60$.

The lower boundary is fixed and the interval is not centred at $\mu$, so do not split the remaining area equally. Instead,

$$P(X<b)=P(X<45)+\mathrm{0.60.}$$

Now $P(X<45)=\Phi (-0.5)\approx 0.30854$, so the cumulative area to $b$ is approximately $0.90854$. Thus

$$b=50+10{\Phi }^{-1}(0.90854)\approx \mathrm{63.3.}$$

This CDF-addition method works because the known area to the left of $45$ and the required interval area are adjacent and disjoint.

Core Example: Finding a Boundary Value

Suppose

$$X\sim N(80,10^2).$$

Find $a$ such that

$$P(X<a)=\mathrm{0.90.}$$

This is inverse normal because the probability is known and the boundary is unknown.

First find the standard normal value $z$ such that

$$P(Z<z)=\mathrm{0.90.}$$

Using a table or calculator,

$$z\approx \mathrm{1.282.}$$

Convert back to the $X$-scale:

$$a=\mu +z\sigma .$$

Thus

$$a\approx 80+1.282(10)=\mathrm{92.82.}$$

So

$$a\approx \mathrm{92.8.}$$

This means about $90\%$ of values lie below $92.8$ under this model.

Unknown Parameters

Some questions give probability statements and ask for $\mu$, $\sigma$, or both.

The key idea is:

A normal probability statement can be converted into a $z$-score equation.

How to read this figure. One probability statement normally produces one $z$-equation and can determine one unknown parameter. If both $\mu$ and $\sigma$ are unknown, two probability statements that produce non-redundant equations are normally needed; always retain $\sigma >0$.

Suppose

$$X\sim N(\mu ,{\sigma }^2)$$

and

$$P(X<85)=\mathrm{0.95.}$$

The probability $0.95$ corresponds to

$$P(Z<1.645)\approx \mathrm{0.95.}$$

Therefore,

$$\frac{85-\mu }{\sigma }=\mathrm{1.645.}$$

This gives one equation involving $\mu$ and $\sigma$.

Unknown-Parameter Workflow

1. Define $X$ and write its distribution.

2. Translate each probability statement into a standard normal statement.

3. Find the matching $z$-value.

4. Form an equation using

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

5. Solve the equation or simultaneous equations.

6. Check the sign: a boundary below $\mu$ should have negative $z$, while one above $\mu$ should have positive $z$.

7. Check that $\sigma >0$.

8. Interpret the result in context.

One probability statement usually gives one equation. If both $\mu$ and $\sigma$ are unknown, two probability statements that produce non-redundant equations are usually needed.

Core Example: Finding $\mu$ and $\sigma$

Suppose

$$X\sim N(\mu ,{\sigma }^2),$$

and

$$P(X<40)=0.10,P(X<70)=\mathrm{0.90.}$$

From standard normal values,

$$P(Z<-1.282)\approx 0.10$$

and

$$P(Z<1.282)\approx \mathrm{0.90.}$$

So

$$\frac{40-\mu }{\sigma }=-1.282$$

and

$$\frac{70-\mu }{\sigma }=\mathrm{1.282.}$$

These give

$$40-\mu =-1.282\sigma$$

and

$$70-\mu =1.282\sigma .$$

Subtract:

$$30\approx 2.564\sigma .$$

Thus

$$\sigma \approx \mathrm{11.7.}$$

Substituting back gives

$$\mu =55.$$

So

$$X\sim N(55,11.7^2)$$

approximately.

Ranges of Possible Values

Some questions ask for a range of boundary values, such as

$$P(X<k)>0.6$$

or

$$P(X>k)<\mathrm{0.2.}$$

These are not asking for a probability. They are asking which values of $k$ make the probability statement true.

Use this method:

1. Replace the inequality with equality.
2. Solve the boundary case.
3. Decide which side of the boundary satisfies the original inequality.

The monotonic directions can be summarised as follows. Let $q_p$ be the boundary satisfying $P(X<q_p)=p$.

Required condition	Equivalent boundary range
$P(X<k)>p$	$k>q_p$
$P(X<k)<p$	$k<q_p$
$P(X>k)>p$	$k<q_{1-p}$
$P(X>k)<p$	$k>q_{1-p}$

Left-tail area increases with $k$; right-tail area decreases with $k$.

Core Example: Range of Boundary Values

Suppose

$$X\sim N(20,4^2).$$

Find the range of $k$ such that

$$P(X<k)>\mathrm{0.75.}$$

First solve the equality:

$$P(X<k)=\mathrm{0.75.}$$

For the standard normal distribution,

$$P(Z<0.674)\approx \mathrm{0.75.}$$

So

$$\frac{k-20}{4}=\mathrm{0.674.}$$

Hence the unrounded boundary is

$$k_{\ast }=20+4{\Phi }^{-1}(0.75)\approx \mathrm{22.69796.}$$

As $k$ increases, the left-tail area $P(X<k)$ increases. Therefore,

$$P(X<k)>0.75$$

when $k>k_{\ast }$. Thus the precise calculator-based conclusion is

$$k>22.69796\dots .$$

If the question requests values to 3 significant figures, report the cutoff as $22.7$ while keeping the unrounded value for deciding the inequality.

Core Example: A Right-Tail Boundary Inequality

For $X\sim N(20,4^2)$, find $k$ such that $P(X>k)<0.20$.

Solve the boundary equality $P(X>k)=0.20$, equivalently $P(X<k)=0.80$. The unrounded cutoff is

$$k_{\ast }=20+4{\Phi }^{-1}(0.80)=23.3664849\dots .$$

As $k$ moves right, the right-tail area $P(X>k)$ decreases. Therefore the required range is

$$k>k_{\ast }=23.3664849\dots .$$

Round the cutoff only when the question specifies a precision, and use the unrounded value when deciding whether a candidate $k$ satisfies the strict inequality.

This direction is the opposite monotonic picture from a left-tail probability: $P(X<k)$ increases with $k$, whereas $P(X>k)$ decreases.

Symmetry in Maximum-Probability Interval Questions

Sometimes a question fixes the width of an interval and asks for the greatest possible probability inside it.

For a normal distribution, a fixed-width interval captures the greatest area when it is centred at the mean. For example, if

$$P(a<X<b)$$

is to be made as large as possible while $b-a$ is fixed, then the best position is usually the symmetric one:

$$\mu =\frac{a+b}{2}.$$

This works because the normal curve is highest at the mean and decreases symmetrically away from it. It is a symmetry argument, not a calculator shortcut.

Common Pitfalls

Pitfall	Better thinking
Treating inverse normal as another probability calculation	Inverse normal finds a value from an area.
Forgetting to convert a right tail	$P(X>a)=p$ means left-tail area $1-p$.
Splitting tails equally without symmetry	Equal tail split only works for symmetric central regions.
Using variance instead of standard deviation in $z=(x-\mu )/\sigma$	Use $\sigma$, not ${\sigma }^2$.
Accepting $\sigma <0$ from algebra	Standard deviation must be positive.
Solving inequality questions without checking direction	After finding the equality boundary, test which side satisfies the condition.

Revision Checklist

• Can you distinguish direct probability from inverse normal?
• Can you convert a right-tail inverse question into a left-tail area?
• Can you find a boundary value from a probability?
• Can you form a $z$-score equation from a probability statement?
• Can you solve for $\mu$ or $\sigma$ from normal probability information?
• Can you handle probability inequalities involving boundary values?