Z-Scores — Mathslive.ie

Section 1 of 9

The Bell Curve and Standard Score

Lots of things in the real world — heights, exam results, IQ, weights of crisp packets — pile up around an average and thin out either side. Plot the count and you get a bell curve. The bell curve is also called the normal distribution.

The Empirical Rule from last topic told us how the data is spread out — $68\%$ within one $\sigma$, $95\%$ within two $\sigma$, $99.7\%$ within three $\sigma$. Here's the picture, this time with IQ scores ($\mu = 100$, $\sigma = 10$) on top and z-scores below — same curve, two scales:

The numbers inside each band are the percentage of values that fall in that strip. They're the same on the top and bottom curves — the bands have the same areas no matter what scale we use. The IQ axis and the z-axis are just two different rulers for the same picture. Notice the link: $\text{IQ} = 110$ on top sits over $z = 1$ on the bottom; $\text{IQ} = 80$ sits over $z = -2$. Each step of one $\sigma$ on the IQ scale ($+10$ in IQ points) is one step on the z-scale ($+1$).

(i) Why standardise?

Every real bell curve has a different $\mu$ and a different $\sigma$. IQ uses $\mu = 100$, $\sigma = 15$. Maths test marks might use $\mu = 61$, $\sigma = 3$. The shape is the same — but the numbers on the axis are different every time.

Instead of carrying $\mu$ and $\sigma$ around forever, we rescale any bell curve to one master version called the standard normal:

The standard normal — must learn

•Mean = $\mu = 0$. Standard deviation = $\sigma = 1$.

•The horizontal axis is in units of $\sigma$, called z-scores.

•The total area under the curve is exactly $1$ (= 100% of everything).

•It is symmetric about $z = 0$.

To convert a real-world value $x$ (from a distribution with mean $\mu$ and standard deviation $\sigma$) into a z-score, we use:

The standard score

z$\,=\, \dfrac{x - \mu}{\sigma}$ (how many standard deviations is $x$ above the mean?)

The z-score tells you where a value sits on the bell curve. $z = 0$ means dead average. $z = 1$ means one $\sigma$ above the mean (about $84\%$ of values are below you). $z = -2$ means two $\sigma$ below the mean. We'll use the formula properly in Section 7 — first we learn to read the curve.

You try

Without looking anything up, what percentage of values in a normal distribution lie within 2 standard deviations of the mean?

This is the Empirical Rule — middle band.

The Empirical Rule says $\mu \pm 2\sigma$ contains about $95\%$ of values.

$\approx 95\%$

You try

IQ scores have $\mu = 100$ and $\sigma = 15$. Between what two IQ values do roughly $68\%$ of people lie?

Within 1 standard deviation means $\mu \pm \sigma$.

$68\%$ lies within one $\sigma$ of the mean.

$\mu - \sigma \,=\, 100 - 15 \,=\, 85$.

$\mu + \sigma \,=\, 100 + 15 \,=\, 115$.

Between $85$ and $115$.

You try

A z-score of $z = 1.5$ means a value is how far above the mean — in units of $\sigma$?

z-scores are just standard deviations counted from the mean.

By definition, $z = \dfrac{x - \mu}{\sigma}$ counts standard deviations from $\mu$.

$1.5$ standard deviations above the mean.

$1.5\,\sigma$ above the mean.

Section 2 of 9

Reading P(z ≤ a) for Positive z

All questions on this topic are about areas under the bell curve. Probability = area. The standard z-table (in your Formulae & Tables, p. 36) gives you the area to the left of any positive z-value. We read it like a normal lookup table.

How to read P(z ≤ a)

1.Sketch the curve. Mark $a$ on the z-axis. Shade everything to the left of $a$.

2.Look up $a$ in the body of the z-table — row gives the units & tenths, column gives the hundredths.

3.The number you read is $P(z \leq a)$ — the shaded area, which is the probability.

(i) Two free ones to start

$P(z \leq 0)$ — area to the left of zero. The curve is symmetric, so half the area is on each side:

$P(z \leq 0) \,=\, 0.5$

$P(z \leq 1)$ — most of the curve. Read $z = 1.00$ in the table:

$P(z \leq 1) \,=\, 0.8413$

(ii) Worked example — P(z ≤ 0.6)

Question. Find $P(z \leq 0.6)$.

Sketch first — shaded everything left of $0.6$. Now look up $0.6$ in the table (row $0.6$, column $0.00$):

$P(z \leq 0.6) \,=\, 0.7257$

That's about $72.6\%$ of the area. Sense-check: $0.6$ is just to the right of the mean, so the shaded area should be a little more than half. $0.7257$ fits.

(iii) Worked example — P(z ≤ 0.31)

Question. Find $P(z \leq 0.31)$.

Row $0.3$, column $0.01$ in the z-table:

$P(z \leq 0.31) \,=\, 0.6217$

You try

Find $P(z \leq 0.45)$.

Sketch — vertical at $0.45$, shade to the left. Then row $0.4$, column $0.05$.

Shade everything left of $0.45$:

From the z-table: row $0.4$, column $0.05$.

$P(z \leq 0.45) = 0.6736$

You try

Find $P(z \leq 1.2)$.

Row $1.2$, column $0.00$.

$1.2$ is well above the mean, so expect something close to $0.9$:

$P(z \leq 1.2) = 0.8849$

You try

Find $P(z \leq 1.96)$. (This is a famous one — comes up in 95% confidence intervals later.)

Row $1.9$, column $0.06$.

Almost the whole curve is to the left of $1.96$:

$P(z \leq 1.96) = 0.9750$

Section 3 of 9

P(z ≥ a): The Complement Rule

The z-table only gives areas to the left. But often you want the area to the right — the values above a certain z-score. Two regions split the curve in two, and the total area is $1$:

a + b = 1

•Total area under the bell curve $\,=\, 1$.

•If you shade everything to the left of $z = a$ and call that area $a$, and the rest is $b$, then $a + b = 1$.

•So $P(z \geq a) \,=\, 1 - P(z \leq a)$. Look up the left tail, subtract from 1.

(i) Worked example — P(z ≥ 0.4)

Question. Find $P(z \geq 0.4)$.

Use the complement — get the left from the table, subtract from $1$:

$P(z \geq 0.4) \,=\, 1 - P(z \leq 0.4)$

$\hphantom{P(z \geq 0.4)} \,=\, 1 - 0.6554$

$\hphantom{P(z \geq 0.4)} \,=\, 0.3446$

(ii) Worked example — P(z ≥ 1.28)

Question. Find $P(z \geq 1.28)$.

$P(z \geq 1.28) \,=\, 1 - P(z \leq 1.28)$

$\hphantom{P(z \geq 1.28)} \,=\, 1 - 0.8997$

$\hphantom{P(z \geq 1.28)} \,=\, 0.1003$

That's the top $10\%$ of the curve — a number you'll see again when we do "top 10%" questions.

You try

Find $P(z \geq 0.3)$.

$1 - P(z \leq 0.3)$.

$P(z \geq 0.3) \,=\, 1 - P(z \leq 0.3)$

$\hphantom{P(z \geq 0.3)} \,=\, 1 - 0.6179$

$\hphantom{P(z \geq 0.3)} \,=\, 0.3821$

$0.3821$

You try

Find $P(z \geq 1.5)$.

Complement of $P(z \leq 1.5)$.

$P(z \geq 1.5) \,=\, 1 - 0.9332$

$\hphantom{P(z \geq 1.5)} \,=\, 0.0668$

$0.0668$

You try

Find $P(z \geq 2)$. (Should match your Empirical-Rule intuition for the top tail.)

$1 - P(z \leq 2)$. Empirical Rule says ~2.5% is above $\mu + 2\sigma$.

$P(z \geq 2) \,=\, 1 - 0.9772 \,=\, 0.0228$.

That is about $2.28\%$ — close to the Empirical Rule's $2.5\%$.

$P(z \geq 2) = 0.0228$

Section 4 of 9

Negative z: Using Symmetry

The standard z-table only gives positive z-values. So how do we handle $P(z \leq -0.4)$ or $P(z \geq -1)$? Easy — the bell curve is symmetric about $z = 0$. Flip the picture.

The two key tricks — together they handle every negative-z question:

Symmetry — must learn

1.$P(z \leq -a) \,=\, P(z \geq a) \,=\, 1 - P(z \leq a)$ (left tail of $-a$ has the same area as the right tail of $+a$)

2.$P(z \geq -a) \,=\, P(z \leq a)$ (everything above $-a$ is the same area as everything below $+a$ — by symmetry)

Picture rule 1 — shaded left tail of $-a$ mirrors the right tail of $+a$:

$P(z \leq -a)$

=

$1 - P(z \leq a)$

Picture rule 2 — shaded right tail of $-a$ mirrors the left tail of $+a$:

$P(z \geq -a)$

=

$P(z \leq a)$

(i) Worked example — P(z ≤ −0.2)

Question. Find $P(z \leq -0.2)$.

The shaded tail is on the negative side. By symmetry it has the same area as the right tail above $+0.2$:

$P(z \leq -0.2) \,=\, 1 - P(z \leq 0.2)$

$\hphantom{P(z \leq -0.2)} \,=\, 1 - 0.5793$

$\hphantom{P(z \leq -0.2)} \,=\, 0.4207$

(ii) Worked example — P(z ≥ −0.6)

Question. Find $P(z \geq -0.6)$.

Big shaded area to the right of $-0.6$ — by rule 2, the same as $P(z \leq 0.6)$:

$P(z \geq -0.6) \,=\, P(z \leq 0.6)$

$\hphantom{P(z \geq -0.6)} \,=\, 0.7257$

Notice: when both signs are negative, the two minuses cancel — $P(z \geq -0.6) = P(z \leq 0.6)$.

You try

Find $P(z \leq -0.5)$.

Same area as $1 - P(z \leq 0.5)$ by symmetry.

$P(z \leq -0.5) \,=\, 1 - P(z \leq 0.5) \,=\, 1 - 0.6915$.

$P(z \leq -0.5) = 0.3085$

$0.3085$

You try

Find $P(z \geq -1.2)$.

Two minuses cancel: $P(z \geq -a) = P(z \leq a)$.

$P(z \geq -1.2) \,=\, P(z \leq 1.2) \,=\, 0.8849$.

$P(z \geq -1.2) = 0.8849$

$0.8849$

You try

Find $P(z \leq -0.85)$.

Mirror to $1 - P(z \leq 0.85)$.

$P(z \leq -0.85) \,=\, 1 - 0.8023 \,=\, 0.1977$.

$P(z \leq -0.85) = 0.1977$

$0.1977$

Section 5 of 9

Intervals: P(a ≤ z ≤ b)

Now the question gives us a strip between two z-values. The trick: any "between" question = "big area left of the top" minus "smaller area left of the bottom".

The interval rule

$P(a \leq z \leq b) \,=\, P(z \leq b) \,-\, P(z \leq a)$ (top area, take the bottom area off)

•If either $a$ or $b$ is negative, convert that one to a positive lookup first using Section 4 symmetry.

(i) Worked example — P(−0.4 ≤ z ≤ 0.8)

Question. Find $P(-0.4 \leq z \leq 0.8)$.

Subtract the left-of-bottom area from the left-of-top area:

$P(z \leq 0.8)$

−

$P(z \leq -0.4)$

$P(-0.4 \leq z \leq 0.8) \,=\, P(z \leq 0.8) \,-\, P(z \leq -0.4)$

$\hphantom{P(-0.4 \leq z \leq 0.8)} \,=\, 0.7881 \,-\, [\,1 - P(z \leq 0.4)\,]$

$\hphantom{P(-0.4 \leq z \leq 0.8)} \,=\, 0.7881 \,-\, [\,1 - 0.6554\,]$

$\hphantom{P(-0.4 \leq z \leq 0.8)} \,=\, 0.7881 \,-\, 0.3446$

$\hphantom{P(-0.4 \leq z \leq 0.8)} \,=\, 0.4435$

(ii) Worked example — P(−0.33 ≤ z ≤ 0.67)

Question. Find $P(-0.33 \leq z \leq 0.67)$.

$P(-0.33 \leq z \leq 0.67) \,=\, P(z \leq 0.67) \,-\, P(z \leq -0.33)$

$\hphantom{=} \,=\, 0.7486 \,-\, [\,1 - 0.6293\,]$

$\hphantom{=} \,=\, 0.7486 \,-\, 0.3707$

$\hphantom{=} \,=\, 0.3779$

You try

Find $P(0.2 \leq z \leq 0.9)$. (Both ends positive — quick subtraction.)

$P(z \leq 0.9) - P(z \leq 0.2)$.

$P(0.2 \leq z \leq 0.9) \,=\, P(z \leq 0.9) \,-\, P(z \leq 0.2)$

$\hphantom{P(0.2 \leq z \leq 0.9)} \,=\, 0.8159 \,-\, 0.5793$

$\hphantom{P(0.2 \leq z \leq 0.9)} \,=\, 0.2366$

$0.2366$

You try

Find $P(-0.5 \leq z \leq 1.2)$.

Negative bottom — convert via symmetry. $P(z \leq -0.5) = 1 - P(z \leq 0.5)$.

$P(-0.5 \leq z \leq 1.2) \,=\, P(z \leq 1.2) \,-\, P(z \leq -0.5)$

$\hphantom{=} \,=\, 0.8849 \,-\, [\,1 - 0.6915\,]$

$\hphantom{=} \,=\, 0.8849 \,-\, 0.3085$

$\hphantom{=} \,=\, 0.5764$

$0.5764$

You try

Find $P(-1 \leq z \leq -0.3)$. (Both ends negative.)

Convert both: $P(z \leq -0.3) - P(z \leq -1) = (1-0.6179) - (1-0.8413)$.

$P(-1 \leq z \leq -0.3) \,=\, P(z \leq -0.3) \,-\, P(z \leq -1)$

$\hphantom{=} \,=\, [\,1 - 0.6179\,] \,-\, [\,1 - 0.8413\,]$

$\hphantom{=} \,=\, 0.3821 \,-\, 0.1587$

$\hphantom{=} \,=\, 0.2234$

$0.2234$

Section 6 of 9

Inverse Lookups: From Probability to z

So far the question gave us a z-value and asked for an area. Now we flip it — the question gives an area and asks which z-value matches it. This is the same z-table — used backwards.

Inverse lookup — must learn

1.If $P(z \leq z_1) = $ some number, find that number in the body of the z-table, then read the z-value from the row and column.

2.If the number is less than $0.5$, then $z_1$ is negative. Use symmetry: find $z_2$ such that $P(z \leq z_2) = 1 - $(given number), then $z_1 = -z_2$.

(i) Worked example — find z₁ given P = 0.6145

Question. $P(z \leq z_1) \,=\, 0.6145$. Find $z_1$.

Hunt for $0.6145$ in the body of the z-table. The closest entry is $0.6141$ at row $0.2$, column $0.09$.

$z_1 \,=\, 0.29$

Sense-check: the probability $0.6145$ is just a bit above $0.5$, so $z_1$ should be just above $0$. $0.29$ fits.

(ii) Worked example — find z₁ given P = 0.3515 (negative answer)

Question. $P(z \leq z_1) \,=\, 0.3515$. Find $z_1$.

$0.3515$ is less than $0.5$, so $z_1$ is to the left of the mean — negative. Mirror the shaded region to the right tail:

$1 - 0.3515 \,=\, 0.6485$ (area on the positive side)

$P(z \leq z_2) \,=\, 0.6485 \;\Rightarrow\; z_2 \,=\, 0.38$ (from the table)

$z_1 \,=\, -z_2 \,=\, -0.38$

You try

$P(z \leq z_1) \,=\, 0.8023$. Find $z_1$.

Big probability — $z_1$ is positive. Look up $0.8023$ in the table.

$0.8023$ sits at row $0.8$, column $0.05$ in the z-table.

$z_1 = 0.85$

You try

$P(z \leq z_1) \,=\, 0.2546$. Find $z_1$.

Less than $0.5$ — answer is negative. Use $1 - 0.2546 = 0.7454$.

$0.2546 < 0.5$, so $z_1$ is negative. Flip:

$1 - 0.2546 \,=\, 0.7454 \;\Rightarrow\; P(z \leq z_2) \,=\, 0.7454$.

From the table: $z_2 \,=\, 0.66$.

$z_1 \,=\, -0.66$

You try

$P(z \leq z_1) \,=\, 0.9750$. Find $z_1$.

Famous value — the $97.5\%$ point.

$0.9750$ corresponds exactly to row $1.9$, column $0.06$.

$z_1 = 1.96$

Section 7 of 9

The Standard Score Formula

Up to now everything has been on the standard curve — z is already a z-score. But real-world data isn't in z-scores; it's in €, $kg$, percentages, IQ points. We need to convert our raw value $x$ into a z-score first. Then we can use the table.

The standard score — the master formula

z$\,=\, \dfrac{x - \mu}{\sigma}$ where $\mu$ is the population mean and $\sigma$ is the population standard deviation.

•$x$ = the raw value you're interested in.

•$z$ = the standard score — how many $\sigma$ above the mean $x$ is.

•If $x > \mu$ then $z > 0$. If $x < \mu$ then $z < 0$.

(i) Worked example — Rent

Question. Cost of rent has a mean of €$600$ per month and standard deviation €$40$ per month. What percentage of rents are €$620$ or less?

$\mu = 600 \qquad \sigma = 40 \qquad x = 620$

$z \,=\, \dfrac{x - \mu}{\sigma} \,=\, \dfrac{620 - 600}{40} \,=\, \dfrac{20}{40} \,=\, 0.5$

$P(z \leq 0.5) \,=\, 0.6915$

$\approx 69.15\%$ of rents are €$620$ or less

(ii) Worked example — A bag of sweets

Question. In a bag of sweets the mean weight is $35\text{ g}$ and standard deviation is $2\text{ g}$. From $5000$ bags, how many will be $36\text{ g}$ or more?

$\mu = 35 \qquad \sigma = 2 \qquad x = 36$ (want $x \geq 36$)

$z \,=\, \dfrac{x - \mu}{\sigma} \,=\, \dfrac{36 - 35}{2} \,=\, 0.5$

$P(z \geq 0.5) \,=\, 1 - P(z \leq 0.5) \,=\, 1 - 0.6915 \,=\, 0.3085$

Number of bags $\,=\, 0.3085 \times 5000 \,=\, 1542.5$

$\approx 1543$ bags

Rounding tip. When the answer is a count of physical objects, round up to the next whole number — you can't have half a bag of sweets.

You try

A class has mean test score $\mu = 70$ and standard deviation $\sigma = 5$. Find the z-score of a student who scored $78$.

Plug into $z = (x - \mu)/\sigma$.

$z \,=\, \dfrac{x - \mu}{\sigma} \,=\, \dfrac{78 - 70}{5} \,=\, \dfrac{8}{5}$

$z = 1.6$

So the student is $1.6$ standard deviations above the mean.

$z = 1.6$

You try

Heights of adult men in a town have $\mu = 170\text{ cm}$ and $\sigma = 8\text{ cm}$. Find the z-score of a man who is $158\text{ cm}$ tall.

$x$ is below the mean, so expect a negative $z$.

$z \,=\, \dfrac{x - \mu}{\sigma} \,=\, \dfrac{158 - 170}{8} \,=\, \dfrac{-12}{8}$

$z = -1.5$

$1.5$ standard deviations below the mean.

$z = -1.5$

Section 8 of 9

Applications: Counts, Percentages and Ranks

Most exam questions on z-scores come in three flavours: (a) what fraction / how many, (b) what percentage, (c) where in the rankings. Three flavours, one recipe — convert to z, look up the area, scale up.

The four-step recipe

1.Write down $\mu$, $\sigma$ and the raw value(s) of $x$.

2.Compute $z = (x - \mu)/\sigma$. If there are two endpoints, compute two z-values.

3.Sketch the curve. Find the required area using Sections 2–5.

4.Multiply by the population size for a count; or read off as a percentage / rank.

(i) Worked example — Maths test ranking

Question. Mean in a maths test was $61\%$ with standard deviation $3\%$. I got $59\%$. Where was I in the class of $100$?

$\mu = 61 \qquad \sigma = 3 \qquad x = 59$

$z \,=\, \dfrac{59 - 61}{3} \,=\, \dfrac{-2}{3} \,\approx\, -0.66$

$P(z \leq -0.66) \,=\, 1 - P(z \leq 0.66) \,=\, 1 - 0.7454 \,=\, 0.2546$

That's the fraction of the class who scored below me. Out of 100 students:

$0.2546 \times 100 \,=\, 25.46$ (students below me)

I am the $26^{\text{th}}$ from the bottom — i.e. $74^{\text{th}}$ from the top out of 100

Even a score $2\%$ below the mean is well into the bottom quarter — because $\sigma$ is only $3\%$, so $2\%$ is nearly a full $\sigma$.

(ii) Worked example — Cows producing milk

Question. A cow produces $26$ litres per day on average. The standard deviation is $3$ litres. How many of my $350$ cows produce between $24$ and $27$ litres?

$\mu = 26 \qquad \sigma = 3 \qquad 24 \leq x \leq 27$

At $x = 24$: $z \,=\, \dfrac{24 - 26}{3} \,=\, \dfrac{-2}{3} \,\approx\, -0.67$

At $x = 27$: $z \,=\, \dfrac{27 - 26}{3} \,=\, \dfrac{1}{3} \,\approx\, 0.33$

$P(-0.67 \leq z \leq 0.33) \,=\, P(z \leq 0.33) \,-\, P(z \leq -0.67)$

$\hphantom{=} \,=\, 0.6293 \,-\, [\,1 - 0.7486\,]$

$\hphantom{=} \,=\, 0.6293 \,-\, 0.2514$

$\hphantom{=} \,=\, 0.3779$

Number of cows $\,=\, 0.3779 \times 350 \,=\, 132.265$

$\approx 133$ cows

You try

A bag of apples has $\mu = 180\text{ g}$ and $\sigma = 20\text{ g}$. What percentage of apples weigh more than $200\text{ g}$?

Find $z$ for $x=200$, then use the complement.

$z \,=\, \dfrac{200 - 180}{20} \,=\, 1$.

$P(z \geq 1) \,=\, 1 - 0.8413 \,=\, 0.1587$.

$\approx 15.87\%$ of apples weigh more than $200\text{ g}$.

$\approx 15.87\%$

You try

A test has $\mu = 65$ and $\sigma = 8$. Out of $2000$ students, how many scored between $60$ and $75$?

Convert each endpoint to a z-score, then find the interval area, then multiply by 2000.

At $x=60$: $z = \dfrac{60-65}{8} = -\dfrac{5}{8} \approx -0.63$.

At $x=75$: $z = \dfrac{75-65}{8} = \dfrac{10}{8} = 1.25$.

$P(-0.63 \leq z \leq 1.25) \,=\, P(z \leq 1.25) \,-\, P(z \leq -0.63)$

$\hphantom{=} \,=\, 0.8944 \,-\, [\,1 - 0.7357\,] \,=\, 0.8944 \,-\, 0.2643 \,=\, 0.6301$

$0.6301 \times 2000 \,=\, 1260.2$.

$\approx 1261$ students (round up — discrete count)

$\approx 1261$ students

You try

In a normal test with $\mu = 50$ and $\sigma = 10$, a student scored $42$. Where would they finish in a class of $200$?

Negative $z$. The area to the left tells you how many scored worse.

$z \,=\, \dfrac{42 - 50}{10} \,=\, -0.8$.

$P(z \leq -0.8) \,=\, 1 - P(z \leq 0.8) \,=\, 1 - 0.7881 \,=\, 0.2119$.

Number scoring below this student $\,=\, 0.2119 \times 200 \,=\, 42.38$.

$\approx 43^{\text{rd}}$ from the bottom (out of $200$).

$\approx 43^{\text{rd}}$ from the bottom

Section 9 of 9

Threshold Problems: Top % and Quartiles

Last flavour. The question gives you a percentile ("top 10%", "upper quartile", "bottom 5%") and asks for the raw score that hits it. This is Section 6 (inverse lookup) combined with Section 7 (the formula).

Threshold recipe — must learn

1.Convert the percentile into a left-tail probability. e.g. “Top $10\%$” means $P(z \leq k) = 0.9$. “Bottom $5\%$” means $P(z \leq k) = 0.05$. “Upper quartile” means $P(z \leq k) = 0.75$.

2.Look up $k$ in the z-table (inverse lookup — find the area in the body, read the z-value).

3.Solve $\dfrac{x - \mu}{\sigma} = k$ for $x$. That gives $\,x = \mu + k\sigma$.

(i) Worked example — Top 10% in a test

Question. A test has mean $56\%$ with standard deviation $4\%$. What result will put you in the top $10\%$ of students?

$\mu = 56 \qquad \sigma = 4$

Top $10\%$ means $10\%$ of students score higher than this cut-off. So $90\%$ score lower:

$P(z \leq k) \,=\, 0.9$

From the z-table: $0.8997$ is at row $1.2$, column $0.08$.

$k \,=\, 1.28$

Now solve back for $x$:

$\dfrac{x - 56}{4} \,=\, 1.28$

$x - 56 \,=\, 1.28 \times 4 \,=\, 5.12$

$x \,=\, 61.12$

$\approx 62\%$

Round up for a threshold — if you score $61\%$ exactly you don't quite make the top $10\%$, so the safe answer is $62\%$.

(ii) Worked example — Upper quartile

Question. Test scores have mean $\mu = 156$ with standard deviation $\sigma = 3$. What score puts you in the upper quartile?

$\mu = 156 \qquad \sigma = 3$

"Upper quartile" means the top $25\%$, so $75\%$ score below the cut-off:

$P(z \leq k) \,=\, 0.75 \;\Rightarrow\; k \,=\, 0.67$ (from $0.7486$ in the table)

$\dfrac{x - 156}{3} \,=\, 0.67$

$x - 156 \,=\, 0.67 \times 3 \,=\, 2.01$

$x \,=\, 158.01$

$\approx 158$

You try

IQ scores have $\mu = 100$ and $\sigma = 15$. What IQ puts you in the top $5\%$?

Top $5\%$ means $P(z \leq k) = 0.95$. Find $k$, then solve for $x$.

Top $5\%$ $\Rightarrow$ $P(z \leq k) \,=\, 0.95$.

The closest entries are $P(z \leq 1.64) = 0.9495$ and $P(z \leq 1.65) = 0.9505$. Take $k = 1.65$ (the safer pick — gives $\geq 95\%$).

$\dfrac{x - 100}{15} \,=\, 1.65 \;\Rightarrow\; x \,=\, 100 + 1.65(15) \,=\, 124.75$.

$\text{IQ} \approx 125$

You try

Crisp packets have $\mu = 30\text{ g}$ and $\sigma = 2\text{ g}$. What weight is in the bottom $10\%$?

Negative $k$. Bottom 10% means $P(z \leq k) = 0.1$.

Bottom $10\%$ $\Rightarrow$ $P(z \leq k) \,=\, 0.1$.

$0.1 < 0.5$, so $k$ is negative. Use $1 - 0.1 \,=\, 0.9$ $\Rightarrow$ $k_+ \,=\, 1.28$, so $k = -1.28$.

$\dfrac{x - 30}{2} \,=\, -1.28$

$x \,=\, 30 + 2(-1.28) \,=\, 30 - 2.56 \,=\, 27.44$.

$x \approx 27.4\text{ g}$

$\approx 27.4\text{ g}$

You try

Salaries in a company have $\mu =$ €$45{,}000$ and $\sigma =$ €$6{,}000$. What salary puts you in the lower quartile?

Lower quartile = bottom 25%, so $P(z \leq k) = 0.25$, giving $k = -0.67$.

Lower quartile $\Rightarrow$ $P(z \leq k) \,=\, 0.25 \,<\, 0.5$, so $k$ is negative.

$1 - 0.25 \,=\, 0.75 \;\Rightarrow\; k_+ \,=\, 0.67$, so $k = -0.67$.

$\dfrac{x - 45000}{6000} \,=\, -0.67$

$x \,=\, 45000 + 6000(-0.67) \,=\, 45000 - 4020 \,=\, 40{,}980$.

$x \approx$ €$40{,}980$

$\approx$ €$40{,}980$

You're done.

Z-Scores & the Normal Distribution · Higher Level · Mathslive.ie