Calculate z-scores and use them to compare values from different normal distributions and find probabilities

Compute each z-score: $z = \frac{x - \mu}{\sigma}$.

Anika: $z = \frac{76 - 68}{6} = \frac{8}{6} \approx 1.33$.

Ben: $z = \frac{82 - 76}{8} = \frac{6}{8} = 0.75$.

Anika's z-score is higher, so she performed better relative to her cohort (further above the mean in standard-deviation units), even though Ben's raw score is higher.

Markers reward both z-scores, comparison, and the final conclusion with the reason ("higher z-score means further above the mean").

(a) For the 90th percentile, $z = 1.28$. Use $x = \mu + z\sigma$.

$x = 164 + 1.28 \times 6 = 164 + 7.68 = 171.68$ cm. Round to $171.7$ cm.

(b) By symmetry, the 10th percentile has $z = -1.28$.

$x = 164 + (-1.28) \times 6 = 164 - 7.68 = 156.32$ cm. Round to $156.3$ cm.

Markers reward the formula rearrangement, the symmetric z-score for the lower tail, and answers to one decimal place.

Apply the z-score formula. Substitute $x = 11.2$, $\mu = 12.0$ and $\sigma = 0.4$ into $z = \frac{x - \mu}{\sigma}$:

Keep the sign. Her time is below the mean, so the z-score is negative. (For a race, a faster time below the mean is a good result.)

Answer: $z = -2.00$.

Apply the z-score formula (a). Substitute $x = 620$, $\mu = 550$ and $\sigma = 50$ into $z = \frac{x - \mu}{\sigma}$:

Interpret the size (b). A z-score of $1.40$ means the score sits $1.40$ standard deviations above the mean.

Answer: (a) $z = 1.40$; (b) $1.40$ standard deviations above the mean.

Choose the reverse formula. A z-score is given and the raw mark is wanted, so de-standardise with $x = \mu + z\sigma$.

Substitute the values. Use $\mu = 72$, $z = 1.5$ and $\sigma = 8$:

Answer: the mark is $84$.

Recall the 50th percentile z-score (a). The 50th percentile is the middle of a symmetric normal distribution, which sits exactly at the mean, so $z = 0$.

De-standardise (b). Substitute $\mu = 100$, $z = 0$ and $\sigma = 15$ into $x = \mu + z\sigma$:

Answer: (a) $z = 0$; (b) the 50th percentile IQ score is $100$ (the mean).

Standardise Mia's time. Use $z = \frac{x - \mu}{\sigma}$ with her own event's mean and standard deviation:

Standardise Zoe's time. Use Zoe's event:

Compare, remembering lower is better. Both times are below their means (good for a race). Zoe's z-score of $-1.80$ is further below the mean than Mia's $-1.50$, so Zoe is more standard deviations faster than her field.

Answer: Zoe's swim ($z = -1.80$) was better relative to her event than Mia's ($z = -1.50$).

Identify the percentile (a). The top $25\%$ begins at the 75th percentile, which has $z = 0.67$.

De-standardise with $x = \mu + z\sigma$. Substitute $\mu = 68$, $z = 0.67$ and $\sigma = 10$:

Interpret. A mark of $74.7$ marks the boundary, so a student needs about $75$ to be safely in the top $25\%$.

Answer: the cutoff for the top $25\%$ is a mark of about $74.7$.

Standardise the Paper 1 mark (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 46$, $\mu = 40$, $\sigma = 4$:

De-standardise onto Paper 2 (b). The equivalent mark keeps the same z-score, so use $x = \mu + z\sigma$ with Paper 2's $\mu = 60$, $\sigma = 10$ and $z = 1.50$:

Answer: (a) $z = 1.50$; (b) the equivalent Paper 2 mark is $75$.

Set up the symmetric pair. The middle $90\%$ runs from $z = -1.65$ to $z = +1.65$, so apply $x = \mu + z\sigma$ at each end with $\mu = 72$ and $\sigma = 4$.

Lower bound ($z = -1.65$).

Upper bound ($z = +1.65$).

Answer: the middle $90\%$ of athletes have resting heart rates between $65.4$ and $78.6$ beats per minute.

Standardise the Mathematics mark (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 82$, $\mu = 70$, $\sigma = 9$:

Standardise the English mark (a). Use $x = 75$, $\mu = 66$, $\sigma = 6$:

Compare and justify (b). English has the higher z-score ($1.50$ against $1.33$), so Priya sits more standard deviations above the mean in English. She performed better in English relative to her cohort, even though her raw Mathematics mark is higher.

Answer: (a) Maths $z \approx 1.33$, English $z = 1.50$; (b) English, because its higher z-score puts her further above her cohort's mean.

Standardise Sam's salary (a). Use $z = \frac{x - \mu}{\sigma}$ for finance:

Standardise Lee's salary (a). Use technology:

Compare (b)

Lee's z-score ($1.64$) is higher than Sam's ($1.50$), so Lee earns more relative to his own industry despite the smaller dollar salary.

Use the table for the tail (c)

The area below Lee is $\Phi(1.64) = 0.9495$, so the proportion above is $1 - 0.9495 = 0.0505$.

Answer

(a) Sam $z = 1.50$, Lee $z \approx 1.64$; (b) Lee; (c) about $5.05\%$ of technology workers earn more than Lee.

Standardise the height (a). Use $z = \frac{x - \mu}{\sigma}$ with $x = 192$, $\mu = 178$, $\sigma = 7$:

Find the top $5\%$ cutoff (b). The tallest $5\%$ start at the 95th percentile, $z = 1.65$. De-standardise with $x = \mu + z\sigma$:

Find the right-tail probability (c). First standardise $185$ cm: $z = \frac{185 - 178}{7} = \frac{7}{7} = 1.00$. The area below is $\Phi(1.00) = 0.8413$, so the area above is $1 - 0.8413 = 0.1587$.

Answer: (a) $z = 2.00$; (b) a man must exceed about $189.55$ cm; (c) $P(X > 185) = 0.1587$, about $15.87\%$.

Standardise both endpoints (a). Apply $z = \frac{x - \mu}{\sigma}$ with $\mu = 18$ and $\sigma = 2.5$ at each value:

Convert the lower tail with the symmetry rule (b). The interval $13 \le X \le 20$ becomes $-2.00 \le z \le 0.80$. The area below the upper z is $\Phi(0.80) = 0.7881$. For the lower z, use $\Phi(-2.00) = 1 - \Phi(2.00) = 1 - 0.9772 = 0.0228$.

Subtract to get the between-area (b). The probability between the two values is the difference of the two areas:

Take the complement for the two outer tails (c). Lasting under $13$ or over $20$ hours is everything outside the interval, so subtract from $1$:

Scale to the delivery (d). Multiply the between-probability by the number of batteries: $500 \times 0.7653 = 382.65$, so about $383$ batteries.

Answer: (a) $z_{13} = -2.00$, $z_{20} = 0.80$; (b) $P(13 \le X \le 20) = 0.7653$ (about $76.53\%$); (c) $0.2347$ (about $23.47\%$); (d) about $383$ batteries.

Write an equation at each percentile (a). Substitute each known thickness and z-score into $x = \mu + z\sigma$:

Eliminate $\mu$ by subtracting (a). Subtracting the second equation from the first cancels $\mu$:

Back-substitute for the mean (b). Put $\sigma = 10$ into the first equation:

A quick check with the second equation agrees: $486 + 0.50 \times 10 = 491$.

Build the symmetric middle-$95\%$ interval (c). The central $95\%$ runs from $z = -1.96$ to $z = +1.96$, so apply $x = \mu + z\sigma$ at each end with $\mu = 491$ and $\sigma = 10$:

Answer: (a) $\sigma = 10$ microns; (b) $\mu = 491$ microns; (c) the middle $95\%$ of washers have thicknesses between $471.4$ and $510.6$ microns.

Standardise the Major Project mark (a). Apply $z = \frac{x - \mu}{\sigma}$ with $x = 89$, $\mu = 72$, $\sigma = 8$:

Standardise the Written Exam mark (a). Apply the formula with $x = 86$, $\mu = 65$, $\sigma = 12$:

Compare the relative standings (b)

The Major Project has the higher z-score ($2.13$ against $1.75$), so the student sits further above the mean, in standard-deviation units, on the Major Project. That was the stronger result relative to the cohort.

Find the right-tail probability (c)

Scoring above $86$ on the Written Exam means $z > 1.75$. The area below is $\Phi(1.75) = 0.9599$, so the area above is $1 - 0.9599 = 0.0401$. In a cohort of $240$, the expected number above is $240 \times 0.0401 = 9.624$, so about $10$ students.

Explain the flaw (d)

Raw marks from two different tasks are not comparable, because the tasks have different means and standard deviations. Equal-looking raw marks can sit at different distances above their own means once standardised, so the z-scores ($2.13$ and $1.75$) show the performances were not equally good.

Answer

(a) Project $z \approx 2.13$, Exam $z = 1.75$; (b) the Major Project, because its higher z-score places the student further above the cohort mean; (c) $P(X > 86) = 0.0401$ (about $4.01\%$), so roughly $10$ of the $240$ students; (d) raw marks from different distributions are not directly comparable, and the z-scores show the Major Project was the better result.