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The normal distribution and the empirical rule for HSC Maths Standard 2

Syllabus dot point

“Recognise the features of the normal distribution and apply the empirical 6868-9595-99.799.7 rule”

HSCMaths Standard 2Year 12: Statistical Analysis12 min read

Quick answer

For a normal distribution, the empirical rule says about 68%68\% of values lie within 11 standard deviation of the mean, 95%95\% within 22, and 99.7%99.7\% within 33 standard deviations; by symmetry, the right-tail percentages above μ+σ\mu + \sigma, μ+2σ\mu + 2\sigma and μ+3σ\mu + 3\sigma are 16%16\%, 2.5%2.5\% and 0.15%0.15\%.

Jump to a section
  1. What this dot point is asking
  2. The answer
  3. Exam-style questions
  4. Practice questions

What this dot point is asking

NESA wants three things. First, recognise the features of the normal distribution: it is symmetric, bell-shaped, and set by its mean and standard deviation. Second, apply the empirical 6868-9595-99.799.7 rule to find the percentage of data inside standard-deviation bands. Third, use this to solve practical problems. This is one of the most reliably examined ideas in the Statistical Analysis module. The marks are easy once you stop thinking in raw numbers and start thinking in standard deviations.

The answer

Note

Think about the heights of everyone in your year. A few people are very short, a few are very tall, and most are bunched up in the middle. If you draw a graph of this, you get a smooth hill shape called a bell curve. The standard deviation is just a measure of how spread out the data is. The empirical rule is a handy shortcut that tells you what slice of the group sits close to the middle and what slice is way out at the edges. So you can answer questions like "how many people are unusually tall" without counting every single person.

The normal distribution

The normal distribution (or bell curve) is a smooth curve that shows how data is spread out. It turns up naturally whenever a quantity is built from many small, separate effects added together. Common examples are adult heights, exam scores across a large group, repeated measurement errors, and the fill weight of mass-produced bottles. Its shape is set entirely by two numbers:

  • μ\mu (mu): the mean, which sits under the peak and the centre of symmetry.
  • σ\sigma (sigma): the standard deviation, which controls the width. A small σ\sigma gives a tall narrow curve; a large σ\sigma gives a short wide one.

Here is the insight that makes every question easy to handle. Once you know μ\mu and σ\sigma, you know the whole curve. Every normal curve is the same shape, just stretched or shifted. So a question never really cares about the raw value xx. It only cares about how many standard deviations xx sits from the mean. That count is the only thing that decides a percentage.

Key properties to be able to state:

  • Symmetric about the mean, so the left and right halves mirror each other.
  • Mean == median == mode =μ= \mu (a consequence of the symmetry and single peak).
  • Highest at x=μx = \mu, falling away smoothly on both sides and never quite touching the axis.
  • The total area under the curve is 11, because it is a probability density: an area is a proportion of the data.

The empirical rule, band by band

The empirical rule is the main result here. Do not just memorise three separate numbers. Instead, build the rule up one band at a time, because that is exactly how the harder questions are set out step by step. The diagrams below shade one more standard-deviation band at each stage.

Stage 1, the central 68%68\%. Go one standard deviation either side of the mean, from μ−σ\mu - \sigma to μ+σ\mu + \sigma. About 68%68\% of all values fall in this central band. This is the bulk of the data, clustered near the mean.

Empirical rule stage 1 Within 1 SD of the mean (mu minus sigma to mu plus sigma) lies about 68 percent of the data. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% Stage 1 Within 1 SD of the mean (mu minus sigma to mu plus sigma) lies about 68 percent of the data.

Stage 2, out to 95%95\%. Widen the band to two standard deviations either side, μ−2σ\mu - 2\sigma to μ+2σ\mu + 2\sigma. Now about 95%95\% of values are captured. The extra strip on each side (from 1σ1\sigma to 2σ2\sigma out) adds 13.5%13.5\% per side, which is how 68%68\% grows to 95%95\%.

Empirical rule stage 2 Extending to 2 SD (mu minus 2 sigma to mu plus 2 sigma) captures about 95 percent. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% 95% Stage 2 Extending to 2 SD captures about 95% of the data.

Stage 3, out to 99.7%99.7\%. Go three standard deviations either side, μ−3σ\mu - 3\sigma to μ+3σ\mu + 3\sigma, and you have about 99.7%99.7\% of the data. Almost nothing lies beyond 33 standard deviations: only 0.3%0.3\% in total, split between the two tails.

Empirical rule stage 3 Out to 3 SD (mu minus 3 sigma to mu plus 3 sigma) captures about 99.7 percent; almost everything. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% 95% 99.7% Stage 3 Out to 3 SD (mu minus 3 sigma to mu plus 3 sigma) captures about 99.7 percent; almost everything.

Stage 4, the half-band percentages. Because the curve is symmetric, each band splits evenly about the mean, so it is worth knowing the percentage of each individual strip. From the mean outward each side reads 34%34\%, then 13.5%13.5\%, then 2.35%2.35\%, then 0.15%0.15\% in the far tail. These are the pieces you add and subtract to answer any "between these two values" question.

Empirical rule complete The complete empirical rule: 68 percent within 1 SD, 95 percent within 2 SD, 99.7 percent within 3 SD, with each half band labelled 34 percent, 13.5 percent and 2.35 percent. μ−3σ μ−2σ μ−σ μ μ+σ μ+2σ μ+3σ 68% 95% 99.7% 34% 34% 13.5% 13.5% 2.35% 2.35% About 68% within 1 SD, 95% within 2 SD, 99.7% within 3 SD.

Key fact

For any normal distribution, about 68%68\% of values lie within 11 standard deviation of the mean, 95%95\% within 22, and 99.7%99.7\% within 33. Reading from the mean outward, the half-bands are 34%34\%, 13.5%13.5\%, 2.35%2.35\% and 0.15%0.15\% per side. Because the curve is symmetric, the percentage above μ+kσ\mu + k\sigma equals the percentage below μ−kσ\mu - k\sigma: the tail beyond 1σ1\sigma is 16%16\%, beyond 2σ2\sigma is 2.5%2.5\%, and beyond 3σ3\sigma is 0.15%0.15\%.

The tails, by symmetry

The single most useful follow-up move is splitting "inside the band" into "outside the band", then halving for one tail. If 68%68\% is inside μ±σ\mu \pm \sigma, then 100−68=32%100 - 68 = 32\% is outside, and by symmetry each tail carries half:

  • Above μ+σ\mu + \sigma (or below μ−σ\mu - \sigma): 100−682=16%\dfrac{100 - 68}{2} = 16\%.
  • Above μ+2σ\mu + 2\sigma (or below μ−2σ\mu - 2\sigma): 100−952=2.5%\dfrac{100 - 95}{2} = 2.5\%.
  • Above μ+3σ\mu + 3\sigma (or below μ−3σ\mu - 3\sigma): 100−99.72=0.15%\dfrac{100 - 99.7}{2} = 0.15\%.

Common standard-deviation regions

This table is the empirical rule in its most usable form. Every "what percentage" question is built by adding or subtracting these strips.

Region Percentage
Within μ±1σ\mu \pm 1\sigma 68%68\%
Within μ±2σ\mu \pm 2\sigma 95%95\%
Within μ±3σ\mu \pm 3\sigma 99.7%99.7\%
Between μ\mu and μ+1σ\mu + 1\sigma 34%34\%
Between μ+1σ\mu + 1\sigma and μ+2σ\mu + 2\sigma 13.5%13.5\%
Between μ+2σ\mu + 2\sigma and μ+3σ\mu + 3\sigma 2.35%2.35\%
Above μ+3σ\mu + 3\sigma 0.15%0.15\%

As a check, the strips on one side of the mean add to a half: 34+13.5+2.35+0.15=50%34 + 13.5 + 2.35 + 0.15 = 50\%.

Applying the rule: the three-step method

To find the percentage of data in a range:

  1. Express each endpoint as a number of standard deviations from the mean, using value−μσ\dfrac{\text{value} - \mu}{\sigma}. You are turning raw values into the only currency that matters.
  2. Sketch the curve, mark the mean and the two endpoints, and shade the region you want.
  3. Add or subtract the strips from the region table to total the shaded area.

The sketch is not optional flourish: it stops the most common error, which is forgetting which strips lie inside the region you were actually asked about.

If the endpoints are not whole numbers of standard deviations from the mean, the empirical rule cannot give an exact answer. Standard 2 then expects z-scores and the standard normal table (the next dot point); the empirical rule is the special case where the endpoints land exactly on ±1\pm 1, ±2\pm 2 or ±3\pm 3.

When the normal distribution applies

The rule only works when the data is actually (approximately) normal. The model fits when:

  • Natural variation is at play: adult heights and weights, exam scores in a large cohort, IQ scores.
  • Measurement error accumulates: repeated readings of the same fixed quantity.
  • Manufacturing tolerances apply: fill weights, component dimensions on a production line.

It does not fit obviously skewed data (household incomes, house prices, reaction times), where a few extreme values pull one tail out. Applying the 6868-9595-99.799.7 percentages to skewed data is a conceptual error that costs marks.

How exam questions ask about the normal distribution

The wording changes but the task is always "turn endpoints into standard deviations, then add strips". Learn the translations:

  • "What percentage lie between aa and bb?" Convert both endpoints to standard deviations from the mean, then sum the strips between them.
  • "What percentage are more than / greater than aa?" A tail question. Find how many SDs aa is above the mean, take the inside-band percentage, subtract from 100100, then halve for the single tail (or read the tail directly: 16%16\%, 2.5%2.5\%, 0.15%0.15\%).
  • "What percentage are less than aa?" Everything to the left: 50%50\% for the whole left half, plus or minus the strips between the mean and aa.
  • "How many students / bottles / people ...?" Find the percentage, then multiply by the total. Give a whole number for a count of people.
  • "Between what two values do the middle 95%95\% lie?" Run the rule backwards: the middle 95%95\% is μ±2σ\mu \pm 2\sigma, so compute those two values.
  • "Is it unusual to score above xx?" Code for "how far into the tail is it": a value beyond 2σ2\sigma (top 2.5%2.5\%) is uncommon; beyond 3σ3\sigma (top 0.15%0.15\%) is rare.
Exam technique

Always sketch the bell curve, mark μ\mu in the middle and the ±1,2,3σ\pm 1, 2, 3\sigma gridlines, and shade the region asked for before doing any arithmetic; markers reward a labelled diagram and it stops you adding the wrong strips. State each endpoint as a number of standard deviations explicitly ("8484 is 11 SD above the mean"). Watch the wording: "within" is the central band, "more than" is one tail (remember to halve), and "between" is a sum of strips. If the question asks for a count, multiply the percentage by the total and round to a whole number, with units.

Worked example

Exam marks: a tail and a between

Exam marks in HSC Mathematics Standard 2 are approximately normally distributed with mean 7070 and standard deviation 1414 (illustrative numbers; the real values vary by year).

Part A: What percentage of students score above 8484?

Express the boundary in standard deviations:

84−7014=1 SD above the mean.\frac{84 - 70}{14} = 1 \text{ SD above the mean.}

Take the tail by symmetry: The empirical rule puts 68%68\% within 11 SD of the mean, so 100−68=32%100 - 68 = 32\% lies outside. By symmetry that 32%32\% splits into two equal tails, giving 16%16\% above 8484.

So for Part A, about 16%16\% of students score above 8484.

Part B
What percentage score between 5656 and 9898?
Locate each endpoint on the scale
56=70−14=μ−1σ56 = 70 - 14 = \mu - 1\sigma, and 98=70+28=μ+2σ98 = 70 + 28 = \mu + 2\sigma.
Sum the strips between the boundaries
From μ−1σ\mu - 1\sigma to μ+2σ\mu + 2\sigma is three strips: the left half-band (34%34\%), the right half-band (34%34\%), and the next strip out on the right (13.5%13.5\%):

34%+34%+13.5%=81.5%.34\% + 34\% + 13.5\% = 81.5\%.

So for Part B, about 81.5%81.5\% of students score between 5656 and 9898.

Heights: turning a percentage into a count

The heights of 200200 Year 12 boys at a Sydney school are approximately normally distributed with mean 175175 cm and standard deviation 77 cm. How many are taller than 189189 cm?

Express the cutoff in standard deviations:

189−1757=2 SD above the mean.\frac{189 - 175}{7} = 2 \text{ SD above the mean.}

Find the tail percentage, then the count: Beyond 22 SD above the mean is the top 2.5%2.5\%. Apply it to the 200200 boys:

2.5%×200=0.025×200=5 boys.2.5\% \times 200 = 0.025 \times 200 = 5 \text{ boys}.

Answer: about 55 boys are taller than 189189 cm.

Manufacturing quality control

A bottling factory fills 600600 mL drink bottles with mean 602602 mL and standard deviation 1.51.5 mL. Bottles below 597.5597.5 mL fail quality control. What proportion fail?

Express the cutoff in standard deviations:

597.5−6021.5=−4.51.5=−3 SD below the mean.\frac{597.5 - 602}{1.5} = \frac{-4.5}{1.5} = -3 \text{ SD below the mean.}

Read the tail: Below μ−3σ\mu - 3\sigma is the bottom 0.15%0.15\%. So about 0.15%0.15\% of bottles fail, roughly 33 in every 20002000 filled.

Answer: about 0.15%0.15\% of bottles fail quality control.

Running the rule backwards

Heights of Year 12 girls at an Australian school are approximately normally distributed with mean 164164 cm and standard deviation 66 cm. Between what two heights do the middle 95%95\% lie?

Identify the standard-deviation band for the middle 95%95\%: The middle 95%95\% of a normal distribution is exactly the band μ±2σ\mu \pm 2\sigma.

Convert the band back to heights:

μ−2σ=164−2(6)=152 cm,μ+2σ=164+2(6)=176 cm.\mu - 2\sigma = 164 - 2(6) = 152 \text{ cm}, \qquad \mu + 2\sigma = 164 + 2(6) = 176 \text{ cm}.

Answer: the middle 95%95\% of girls are between 152152 cm and 176176 cm tall.

Common traps
Confusing within and outside the band
68%68\% inside means 32%32\% outside, split as 16%16\% in each tail. Read whether the question wants inside or a tail.
Forgetting to halve
"Above 11 SD" is 16%16\%, not 32%32\%. The 32%32\% is both tails combined.
Using the rule on non-normal data
The empirical rule applies only to (approximately) normal distributions, not to skewed data like incomes or house prices.
Standard deviation versus variance
The rule uses the standard deviation σ\sigma, not the variance σ2\sigma^2. If you are given the variance, square-root it first.
Treating the endpoints as exact when they are not whole SDs
The rule gives exact strips only at ±1,2,3σ\pm 1, 2, 3\sigma. For other endpoints you need z-scores and the table.
Reporting a count as a decimal
A number of people or bottles is a whole number; round at the end.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style3 marks
The heights of Year 12 boys at a Sydney school are normally distributed with mean 175175 cm and standard deviation 77 cm. What percentage are taller than 189189 cm?
Show worked answer →

189189 cm is 189−1757=2\frac{189 - 175}{7} = 2 standard deviations above the mean.

By the empirical rule, 95%95\% of values lie within 22 standard deviations of the mean, so 5%5\% lie outside this band.

By symmetry, half of that (2.5%2.5\%) is above 189189 cm.

So about 2.5%2.5\% of Year 12 boys are taller than 189189 cm.

Markers reward the number of standard deviations from the mean, the application of 95%95\% inside, and the halving by symmetry.

2021 HSC-style4 marks
A factory produces bags of rice with a mean weight of 11 kg and standard deviation 2020 g. The weights are normally distributed. (a) What percentage of bags weigh between 980980 g and 10201020 g? (b) What percentage weigh between 940940 g and 10601060 g?
Show worked answer →

Convert standard deviation to grams: σ=20\sigma = 20 g.

(a) 980980 g is 11 SD below the mean (10001000 g), and 10201020 g is 11 SD above. By the empirical rule, 68%68\% of bags lie in this range.

(b) 940940 g is 33 SD below, and 10601060 g is 33 SD above the mean. By the empirical rule, 99.7%99.7\% of bags lie in this range.

Markers reward both: the identification of how many SDs each endpoint is from the mean, and the empirical rule percentage.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation1 mark
A set of data is normally distributed. State the approximate percentage of values that lie within (a) 11 standard deviation of the mean, (b) 22 standard deviations of the mean, (c) 33 standard deviations of the mean.
Show worked solution →

Recall the empirical rule. For any normal distribution the central bands carry fixed percentages, measured outward from the mean μ\mu.

State each band.

(a) Within 11 standard deviation, μ−σ\mu - \sigma to μ+σ\mu + \sigma: about 68%68\%.

(b) Within 22 standard deviations, μ−2σ\mu - 2\sigma to μ+2σ\mu + 2\sigma: about 95%95\%.

(c) Within 33 standard deviations, μ−3σ\mu - 3\sigma to μ+3σ\mu + 3\sigma: about 99.7%99.7\%.

Answer: 68%68\%, 95%95\% and 99.7%99.7\%.

foundation1 mark
The lifetimes of a brand of light globe are normally distributed with mean μ=1000\mu = 1000 hours and standard deviation σ=100\sigma = 100 hours. How many standard deviations above the mean is a lifetime of 12001200 hours?
Show worked solution →

Set up the standard-deviation count. Find how far the value sits above the mean, then divide by one standard deviation: value−μσ\dfrac{\text{value} - \mu}{\sigma}.

Substitute the numbers.

1200−1000100=200100=2.\frac{1200 - 1000}{100} = \frac{200}{100} = 2.

Answer: 12001200 hours is 22 standard deviations above the mean.

foundation2 marks
The time a fully charged phone battery lasts is normally distributed with mean μ=30\mu = 30 hours and standard deviation σ=4\sigma = 4 hours. What percentage of batteries last more than 3434 hours?
Show worked solution →

Express the cutoff in standard deviations.

34−304=1 standard deviation above the mean.\frac{34 - 30}{4} = 1 \text{ standard deviation above the mean.}

Take the tail by symmetry. The empirical rule puts 68%68\% within 11 standard deviation, so 100−68=32%100 - 68 = 32\% lies outside the band. By symmetry that splits into two equal tails, so the upper tail above μ+σ\mu + \sigma is 322=16%\dfrac{32}{2} = 16\%.

Answer: about 16%16\% of batteries last more than 3434 hours.

foundation2 marks
Annual rainfall in a town is normally distributed with mean μ=800\mu = 800 mm and standard deviation σ=50\sigma = 50 mm. What percentage of years have rainfall between 850850 mm and 900900 mm?
Show worked solution →

Locate each endpoint on the standard-deviation scale.

850=800+50=μ+1σ,900=800+2(50)=μ+2σ.850 = 800 + 50 = \mu + 1\sigma, \qquad 900 = 800 + 2(50) = \mu + 2\sigma.

Read the single strip between them. The band from μ+1σ\mu + 1\sigma to μ+2σ\mu + 2\sigma is one strip of the curve, worth 13.5%13.5\%.

Answer: about 13.5%13.5\% of years have rainfall between 850850 mm and 900900 mm.

core2 marks
A mill packs flour into bags whose weights are normally distributed with mean μ=2000\mu = 2000 g and standard deviation σ=20\sigma = 20 g. What percentage of bags weigh between 19801980 g and 20202020 g?
Show worked solution →

Convert each endpoint to standard deviations.

1980=2000−20=μ−1σ,2020=2000+20=μ+1σ.1980 = 2000 - 20 = \mu - 1\sigma, \qquad 2020 = 2000 + 20 = \mu + 1\sigma.

Identify the band
From μ−1σ\mu - 1\sigma to μ+1σ\mu + 1\sigma is the central band of the empirical rule.
Read the percentage
The central band within 11 standard deviation carries about 68%68\% of the data.
Answer
about 68%68\% of bags weigh between 19801980 g and 20202020 g.
core3 marks
Marks in a class test are normally distributed with mean μ=60\mu = 60 and standard deviation σ=12\sigma = 12. What percentage of students score above 8484?
Show worked solution →

Express the boundary in standard deviations.

84−6012=2 standard deviations above the mean.\frac{84 - 60}{12} = 2 \text{ standard deviations above the mean.}

Find the outside-band percentage
The empirical rule puts 95%95\% within 22 standard deviations, so 100−95=5%100 - 95 = 5\% lies outside the band.
Halve for the single tail
By symmetry the 5%5\% splits evenly between the two tails, so above μ+2σ\mu + 2\sigma is 52=2.5%\dfrac{5}{2} = 2.5\%.
Answer
about 2.5%2.5\% of students score above 8484.
core3 marks
The heights of Year 12 students at a school are normally distributed with mean μ=168\mu = 168 cm and standard deviation σ=8\sigma = 8 cm. What percentage of students are shorter than 152152 cm?
Show worked solution →

Express the cutoff in standard deviations.

152−1688=−168=−2 standard deviations, so 152 cm=μ−2σ.\frac{152 - 168}{8} = \frac{-16}{8} = -2 \text{ standard deviations, so } 152 \text{ cm} = \mu - 2\sigma.

Find the outside-band percentage
With 95%95\% inside 22 standard deviations, 100−95=5%100 - 95 = 5\% lies outside the band, split equally between the two tails.
Take the lower tail
Below μ−2σ\mu - 2\sigma is one tail, so 52=2.5%\dfrac{5}{2} = 2.5\%.
Answer
about 2.5%2.5\% of students are shorter than 152152 cm.
core4 marks
In a cohort of 500500 students, marks in an examination are normally distributed with mean μ=65\mu = 65 and standard deviation σ=10\sigma = 10. How many students score between 5555 and 7575?
Show worked solution →

Convert each endpoint to standard deviations.

55=65−10=μ−1σ,75=65+10=μ+1σ.55 = 65 - 10 = \mu - 1\sigma, \qquad 75 = 65 + 10 = \mu + 1\sigma.

Read the band percentage. From μ−1σ\mu - 1\sigma to μ+1σ\mu + 1\sigma is the central band, about 68%68\% of students.

Turn the percentage into a count. Multiply by the cohort size:

68%×500=0.68×500=340.68\% \times 500 = 0.68 \times 500 = 340.

Answer: about 340340 students score between 5555 and 7575.

exam4 marks
A factory packs rice into bags whose weights are normally distributed with mean μ=1000\mu = 1000 g and standard deviation σ=15\sigma = 15 g. A batch contains 800800 bags. (a) What percentage of bags weigh less than 970970 g? (b) How many bags in the batch is this?
Show worked solution →

Part (a): express the cutoff in standard deviations.

970−100015=−3015=−2, so 970 g=μ−2σ.\frac{970 - 1000}{15} = \frac{-30}{15} = -2, \text{ so } 970 \text{ g} = \mu - 2\sigma.

Take the lower tail. With 95%95\% within 22 standard deviations, 100−95=5%100 - 95 = 5\% is outside, split evenly, so below μ−2σ\mu - 2\sigma is 52=2.5%\dfrac{5}{2} = 2.5\%.

So for part (a), about 2.5%2.5\% of bags weigh less than 970970 g.

Part (b): convert the percentage to a count.

2.5%×800=0.025×800=20.2.5\% \times 800 = 0.025 \times 800 = 20.

Answer: (a) about 2.5%2.5\%; (b) about 2020 bags.

exam4 marks
A manufacturer tests 20002000 light globes whose lifetimes are normally distributed with mean μ=1200\mu = 1200 hours and standard deviation σ=80\sigma = 80 hours. (a) What percentage of globes last longer than 13601360 hours? (b) Estimate how many of the 20002000 globes this represents.
Show worked solution →

Part (a): express the boundary in standard deviations.

1360−120080=16080=2 standard deviations above the mean.\frac{1360 - 1200}{80} = \frac{160}{80} = 2 \text{ standard deviations above the mean.}

Take the upper tail. Since 95%95\% lies within 22 standard deviations, 100−95=5%100 - 95 = 5\% is outside, and halving for the single tail gives 52=2.5%\dfrac{5}{2} = 2.5\% above μ+2σ\mu + 2\sigma.

So for part (a), about 2.5%2.5\% of globes last longer than 13601360 hours.

Part (b): convert the percentage to a count.

2.5%×2000=0.025×2000=50.2.5\% \times 2000 = 0.025 \times 2000 = 50.

Answer: (a) about 2.5%2.5\%; (b) about 5050 globes.

exam5 marks
Marks in an HSC trial examination are approximately normally distributed with mean μ=70\mu = 70 and standard deviation σ=8\sigma = 8. A school enters 600600 students. (a) What percentage of students score between 6262 and 8686? (b) How many students would be expected to score above 8686?
Show worked solution →

Part (a): convert each endpoint to standard deviations.

62=70−8=μ−1σ,86=70+2(8)=μ+2σ.62 = 70 - 8 = \mu - 1\sigma, \qquad 86 = 70 + 2(8) = \mu + 2\sigma.

Sum the strips between the boundaries. From μ−1σ\mu - 1\sigma to μ+2σ\mu + 2\sigma is three strips: the left half-band 34%34\%, the right half-band 34%34\%, and the next strip out on the right 13.5%13.5\%:

34%+34%+13.5%=81.5%.34\% + 34\% + 13.5\% = 81.5\%.

So for part (a), about 81.5%81.5\% of students score between 6262 and 8686.

Part (b): read the upper tail, then count. Above μ+2σ\mu + 2\sigma is the top 2.5%2.5\%. Applying this to the 600600 students:

2.5%×600=0.025×600=15.2.5\% \times 600 = 0.025 \times 600 = 15.

Answer: (a) about 81.5%81.5\%; (b) about 1515 students.

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