N2StatisticsStretch

The Normal Distribution

Q: Why is the second parameter the variance rather than the standard deviation?

Convention, and it makes some theory tidier — variances add when independent variables are combined. Always take the square root before standardising.

Q: Why is P(X = a) = 0?

Probability is area under the curve, and a single point has zero width and therefore zero area. Only intervals carry probability for a continuous distribution.

Q: What does a negative z mean?

The value lies below the mean. Its size still gives the number of standard deviations, so z = -2 is as far from the mean as z = 2, on the other side.

Q: Why does the normal distribution appear so often?

Because quantities formed by adding many small independent effects tend towards a normal shape regardless of the individual effects — the reason it models heights, measurement errors and much else.

The normal distribution is the bell curve: symmetric about its mean, continuous, and the model for an enormous range of naturally varying quantities — heights, measurement errors, exam marks. It is defined entirely by two numbers, its mean $\mu$ and its standard deviation $\sigma$ .

40 min Statistical Distributions

The big picture

What makes the normal manageable is that every normal distribution is the same shape, just relocated and rescaled. $\textbf{Standardising}$ converts any value into the number of standard deviations it sits from its mean, which turns every problem into a question about one fixed curve. That is why a single table or calculator function handles all of them, and it is the same manoeuvre used later to test a sample mean. The reason the distribution appears so widely is deeper still: quantities produced by adding many small independent effects tend towards normality regardless of what those effects individually look like.

What you'll be able to do

State the properties of the normal distribution and use its notation
Standardise values using $z = \frac{x-\mu}{\sigma}$ and interpret $z$
Find probabilities and use the inverse normal to find values
Solve for an unknown $\mu$ or $\sigma$ from given probabilities

Shape and notation

A normal distribution is written $X \sim N\left(\mu, \sigma^{2}\right)$ — note the second parameter is the $\textbf{variance}$ , not the standard deviation, which is a routine source of error.

The curve is symmetric about $\mu$ , where the mean, median and mode coincide. The total area beneath it is 1, and probability is area: $P(X < a)$ is the area to the left of $a$ .

The distribution is continuous, so $P(X = a) = 0$ for any single value — a point has no width and therefore no area. That is why $<$ and $\leq$ are interchangeable here, unlike in the binomial.

Roughly 68% of values lie within one standard deviation of the mean, 95% within two and 99.7% within three. Those figures are worth knowing as a sanity check on any answer.

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

Second parameter is the variance.

P(X = a) = 0

Continuous: single values have no probability.

Why this works

Because $P(X=a)=0$ , "at least 70" and "more than 70" are the same probability for a normal variable. Carrying the binomial habit of adjusting by one across to the normal is a common and unnecessary error.

Standardising

The $z$ -score measures how many standard deviations a value lies from the mean. Subtracting $\mu$ recentres the distribution on zero; dividing by $\sigma$ rescales it so one unit is one standard deviation.

The result follows the $\textbf{standard normal}$ distribution $Z \sim N(0,1)$ , which is the single curve every normal problem is converted to. A positive $z$ sits above the mean, a negative $z$ below.

Standardising also makes values from different distributions comparable. A mark of 70 in one exam and 65 in another can be ranked by which has the larger $z$ .

z = \dfrac{x - \mu}{\sigma}

Standard deviations from the mean.

Z \sim N(0, 1)

The standard normal.

Example 1Heights are

N(170, 8^{2})

cm. Find the

z

-score of a height of 182 cm and interpret it.

z = \dfrac{182 - 170}{8} = \dfrac{12}{8}

2That is

1.5

Answer

z = 1.5

— the height is 1.5 standard deviations above the mean.

Example 2Student A scores 72 where

\mu = 60, \sigma = 10

; student B scores 55 where

\mu = 45, \sigma = 5

. Who performed better relative to their group?

1A:

z = \tfrac{72-60}{10} = 1.2

2B:

z = \tfrac{55-45}{5} = 2.0

AnswerB, despite the lower raw mark — 2 standard deviations above the mean beats 1.2.

Tip — Remember $\sigma$ is the standard deviation, so take the square root of the variance in the notation. $N(50, 16)$ has $\sigma = 4$ , not 16.

Finding probabilities and the inverse normal

Calculators give $P(X < a)$ directly from $\mu$ and $\sigma$ . For "greater than", use the complement; for a range, subtract the two cumulative values.

Symmetry is often quicker than a second calculation. Because the curve is symmetric about $\mu$ , $P(Z < -a) = P(Z > a)$ , and $P(X > \mu) = 0.5$ exactly.

The $\textbf{inverse normal}$ runs the process backwards: given a probability, it returns the value $a$ with that much area below it. Use it for questions phrased as "the top 10% score above what mark?" — noting that the top 10% means an area of 0.9 to the left.

P(X > a) = 1 - P(X < a)

Complement.

P(a<X<b) = P(X<b) - P(X<a)

Range — no

-1

adjustment, unlike the binomial.

Example 1For

X \sim N(500, 100^{2})

, find the value exceeded by the top 5%.

1Top 5% means 95% lies below, so use the inverse normal with area

0.95

2The standard normal value is

z = 1.6449

3Convert back:

x = \mu + z\sigma = 500 + 1.6449(100)

Answer

x \approx 664

Why this works

The inverse normal always works with the area to the $\textit{left}$ . Converting "top 5%" to "0.95 below" before touching the calculator prevents the most common inverse-normal error.

Finding an unknown mean or standard deviation

When $\mu$ or $\sigma$ is unknown, work backwards from the standardising formula. Convert the given probability into a $z$ -value with the inverse normal, then substitute into $z = \dfrac{x-\mu}{\sigma}$ and solve.

One unknown needs one probability statement; two unknowns need two, giving simultaneous equations. Watch the signs — a value below the mean has a negative $z$ .

x = \mu + z\sigma

Rearranged for solving.

Example 1For

X \sim N(\mu, 5^{2})

, it is given that

P(X < 32) = 0.10

. Find

\mu

1An area of

0.10

to the left gives

z = -1.2816

(negative, since 32 is below the mean).

2Substitute:

-1.2816 = \dfrac{32 - \mu}{5}

-6.408 = 32 - \mu

, so

\mu = 32 + 6.408

Answer

\mu = 38.4

(3 s.f.)

Tip — Sketch the curve and shade the region before solving. It tells you immediately whether $z$ should be positive or negative, which is where sign errors creep in.

Think like an examiner

Write the distribution in full: " $X \sim N(170, 8^{2})$ " — and remember the second parameter is the variance.
Sketch and shade the region. It is quick, often credited, and prevents sign and tail errors.
Convert "top $p\%$ " into an area to the left before using the inverse normal.
Do not apply continuity-style adjustments to a normal variable; $<$ and $\leq$ give the same probability.

Common misconceptions

Reading

N(60, 25)

as having standard deviation 25.

The second parameter is the variance, so

\sigma = 5

Using

z = \dfrac{\mu - x}{\sigma}

It is

\dfrac{x-\mu}{\sigma}

. Reversing it flips the sign of every

z

Entering 0.05 into the inverse normal for "the top 5%".

The inverse normal takes the area to the left, so use 0.95.

Treating

P(X \geq 70)

and

P(X > 70)

as different.

For a continuous distribution they are equal, since

P(X=70)=0

Normal distribution

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

Variance in the notation.

z = \tfrac{x-\mu}{\sigma}

Standardising.

x = \mu + z\sigma

Reversed, for unknowns.

Z \sim N(0,1)

Standard normal.

68\%,\ 95\%,\ 99.7\%

Within 1, 2, 3 standard deviations.

Stretch yourself

The masses of apples are normally distributed. It is known that 10% weigh less than 120 g and 5% weigh more than 190 g. Find $\mu$ and $\sigma$ .

Hint — Each statement gives one equation via standardising. Two unknowns need two equations.

Reveal solution

SolutionFrom

P(X < 120) = 0.10

, the inverse normal gives

z = -1.2816

, so

\dfrac{120-\mu}{\sigma} = -1.2816

, i.e.

120 - \mu = -1.2816\sigma

. From

P(X > 190) = 0.05

, the area to the left is

0.95

z = 1.6449

, giving

190 - \mu = 1.6449\sigma

. Subtracting the first equation from the second:

70 = \left(1.6449 + 1.2816\right)\sigma = 2.9265\sigma

, so

\sigma = 23.9

g (3 s.f.). Substituting back:

\mu = 190 - 1.6449(23.92) = 190 - 39.3 = 151

g (3 s.f.). Check:

120

\tfrac{120-150.7}{23.92} = -1.28

standard deviations below, matching the first condition ✓. Two probability statements are exactly what two unknowns require.

\blacksquare

Questions students ask

Why is the second parameter the variance rather than the standard deviation?

Convention, and it makes some theory tidier — variances add when independent variables are combined. Always take the square root before standardising.

Why is

P(X = a) = 0

Probability is area under the curve, and a single point has zero width and therefore zero area. Only intervals carry probability for a continuous distribution.

What does a negative

z

mean?

The value lies below the mean. Its size still gives the number of standard deviations, so

z = -2

is as far from the mean as

z = 2

, on the other side.

Why does the normal distribution appear so often?

Because quantities formed by adding many small independent effects tend towards a normal shape regardless of the individual effects — the reason it models heights, measurement errors and much else.

Key takeaways

$X \sim N(\mu,\sigma^{2})$ is symmetric about $\mu$ , with the variance in the notation.
It is continuous, so $P(X=a)=0$ and strict inequalities make no difference.
Standardising with $z = \tfrac{x-\mu}{\sigma}$ converts any normal to $N(0,1)$ .
The inverse normal works from the area to the left — convert "top $p\%$ " first.
Unknown parameters are found by turning probabilities into $z$ -values and solving.