The normal curve and the 68-95-99.7 rule for area under it, finding probabilities from the rule, interpreting real normally distributed data, converting values to z-scores, and reading probabilities from the standard normal table.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 11 — Data Analysis and Statistics
Use Normal Distributions
Objectives
Five outcomes. Two numbers, a mean and a standard deviation, describe the whole shape.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-761 — the lesson these objectives are drawn from
Warm-up
Lesson 11.1 gave you the mean and the standard deviation; Lesson 10.6 gave you probability distributions.
Discussion prompt
Heights, test scores and measurement errors all pile up around a central value and thin out symmetrically on both sides. What single shape describes all of them?
Hint: Sketch what such a histogram would look like.
Answer:
A bell: high in the middle, falling away symmetrically on both sides. That shape is the normal distribution, and it appears whenever many small independent effects add together.
Its whole shape is fixed by just two numbers — the mean, which says where the peak sits, and the standard deviation, which says how wide the bell is.
So the statistics of Lesson 11.1 become a complete description rather than a summary, and any probability question can be answered from them.
Concept
A normal distribution is modelled by a bell-shaped curve symmetric about the mean. The total area under it is 1, with about 68 percent within one standard deviation of the mean, 95 percent within two and 99.7 percent within three.
z-score — The number of standard deviations a value lies above or below the mean, found by subtracting the mean and dividing by the standard deviation.
\[ z = \frac{x-\bar{x}}{s} \]
An area under the curve can be read either as a percentage of the data or as a probability for one randomly chosen value.
Figure (svg): A normal curve with the percentage of area in each standard deviation band
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-758
Section
Section 1
Concept
The area under a normal curve totals 1. About 68 percent of it lies within one standard deviation of the mean, about 95 percent within two, and about 99.7 percent within three.
\[ 68\%, \; 95\%, \; 99.7\% \]
Splitting those figures gives the individual bands: 34 percent, 13.5 percent, 2.35 percent and 0.15 percent on each side of the mean.
Figure (svg): A normal curve with the percentage of area in each standard deviation band
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-757 — Areas Under a Normal Curve
Picture it
The normal curve with every band labelled.
Figure (svg): A normal curve with the percentage of area in each standard deviation band
The six bands and two tails total 100 percent. Reading a probability is a matter of adding whichever of them the region covers.
Worked example
Reading the key concept's diagram.
\[ \text{Add every band and tail of the normal curve and confirm the total.} \]
Add the two central bands
Why: Thirty-four percent on each side of the mean.
\[ 68 \% \]
Add the next pair
Why: Thirteen point five percent on each side.
\[ 27 \%,\text{ running total } 95 \]
Add the third pair
Why: Two point three five percent on each side.
\[ 4.7 \%,\text{ running total } 99.7 \]
Add the two tails
Why: Point one five percent on each side.
\[ 0.3 \%,\text{ total } 100 \]
Figure (svg): A normal curve with the percentage of area in each standard deviation band
\[ 68+27+4.7+0.3 = 100\% \]
Verify: check the cumulative figures
Why: The three headline numbers 68, 95 and 99.7 are the running totals after each pair is added, which is why they are quoted rather than the individual bands. Knowing both forms means a region can be built up either way.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-757
Fill the middle
Example 1.
Fill in the blanks
0.135+0.34 = 0.475
Why: The two bands from two standard deviations below the mean up to the mean total 0.475. That is half of 95 percent, as symmetry requires.
Worked example
Guided Practice 1 and 2.
\[ \text{Find } P(x \leq \bar{x}) \text{ and } P(x \geq \bar{x}). \]
Use the symmetry
Why: The curve is a mirror image about the mean.
The two halves total 1
Why: Together they cover everything.
\[ \text{each is } 0.5 \]
State the first
Why: Everything at or below the mean.
\[ 0.5 \]
State the second
Why: Everything at or above it.
\[ 0.5 \]
Figure (svg): The solution to Worked example read off two symmetric facts shown as a ladder of expressions, one row per algebraic move
\[ P(x \leq \bar{x}) = P(x \geq \bar{x}) = 0.5 \]
Verify: check against the bands
Why: Adding the left-hand bands gives 34 plus 13.5 plus 2.35 plus 0.15, which is exactly 50 percent. The symmetry argument and the band arithmetic agree, and the symmetry argument is faster.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758
Trap
\[ \text{the band between } +1s \text{ and } +2s \]
Call its area 95 percent
Why: The headline figure is attached to the nearest band.
\[ 95\% \quad \text{(wrong)} \]
That band holds 13.5 percent. The 95 percent figure covers everything from two standard deviations below the mean to two above.
\[ 95\% = 2.35+13.5+34+34+13.5+2.35 \]
Read the headline figures as cumulative totals
Why: Sixty-eight, 95 and 99.7 each measure a symmetric region around the mean.
\[ \text{one band from } +1s \text{ to } +2s \text{ is } 13.5\% \]
The individual bands are 34, 13.5, 2.35 and 0.15 percent. Keeping the two lists separate in your head is what prevents the mix-up.
Matching
Add the bands it covers.
Match the pairs
Why: The first three are symmetric regions around the mean and the fourth is a single band on one side. Confusing the two lists is the most common error with this rule.
Sorting
Read each region off the curve.
Sort into buckets
Sort each region by its area.
Sixteen percent above one standard deviation is worth remembering: it is 100 minus 84, and 84 is half of 68 plus 50.
Prediction
Commit before reasoning.
Predict first
Only 0.3 percent of a normal distribution lies beyond three standard deviations. What does that mean in practice?
Correct: A value that far out happens about three times in a thousand, so it is genuinely unusual.
\[ 100-99.7 = 0.3\% \text{ beyond } 3s \]
Why: Three in a thousand is rare but not impossible — in a sample of ten thousand you would expect about thirty such values. The curve extends forever in both directions and never touches the axis, so no value is strictly impossible. That is why manufacturing and quality control use three standard deviations as a threshold: it flags the genuinely odd without rejecting ordinary variation.
Section
Section 2
Concept
Shade the region the question describes, then add the areas of the bands it covers. An area under the curve can be read as a probability for one randomly selected value.
\[ P(\bar{x}-2s \leq x \leq \bar{x}) = 0.135+0.34 = 0.475 \]
For a region running out to a tail, subtracting from 1 or from 0.5 is usually quicker than adding several bands.
Figure (svg): A normal curve with the region from two standard deviations below the mean up to the mean shaded
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-757 — Find a normal probability
Picture it
Example 1: from two standard deviations below the mean up to the mean.
Figure (svg): A normal curve with the region from two standard deviations below the mean up to the mean shaded
The shaded region covers the 13.5 percent band and the 34 percent band, totalling 0.475 — exactly half of the 95 percent figure.
Worked example
Example 1.
\[ \text{Find } P(\bar{x}-2s \leq x \leq \bar{x}). \]
Sketch and shade
Why: From two standard deviations below the mean to the mean.
Identify the bands
Why: The 13.5 percent band and the 34 percent band.
\[ 0.135\text{ and } 0.34 \]
Add them
Why: Point one three five plus 0.34.
\[ 0.475 \]
Interpret
Why: Just under half the distribution lies in this region.
\[ \text{about } 47.5 \% \]
Figure (svg): A normal curve with the region from two standard deviations below the mean up to the mean shaded
\[ 0.135+0.34 = 0.475 \]
Verify: check against the 95 percent figure
Why: The full region from negative two to positive two standard deviations holds 95 percent, and symmetry splits that into two equal halves of 47.5 percent each. The band addition and the symmetry shortcut agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-757
Fill the middle
Guided Practice 6.
Fill in the blanks
P(x \geq \bar0.16+s) = 1-0.84 = ___
Why: Everything below one standard deviation above the mean is 0.84, so the region above it is 0.16. Subtracting from 1 is quicker than adding the three bands out to the tail.
Worked example
Guided Practice 3 to 6.
\[ \text{Find } P(\bar{x}\leq x\leq\bar{x}+2s), \; P(\bar{x}-s\leq x\leq\bar{x}), \; P(x\leq\bar{x}-3s), \; P(x\geq\bar{x}+s). \]
First: two bands above the mean
Why: Thirty-four plus 13.5 percent.
\[ 0.475 \]
Second: one band below the mean
Why: The 34 percent band alone.
\[ 0.34 \]
Third: the left tail
Why: Everything beyond three standard deviations below.
\[ 0.0015 \]
Fourth: above one standard deviation
Why: Thirteen point five plus 2.35 plus 0.15.
\[ 0.16 \]
Figure (svg): The solution to Worked example four more probabilities shown as a ladder of expressions, one row per algebraic move
\[ 0.475, \; 0.34, \; 0.0015, \; 0.16 \]
Verify: check the fourth by subtraction
Why: Everything below one standard deviation above the mean is 0.5 plus 0.34, which is 0.84, so above it is 1 minus 0.84, or 0.16 — matching. Subtracting from 1 uses two numbers rather than three and is the quicker route for a tail region.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758
Error analysis
A student finds the probability of lying within one standard deviation of the mean.
Annotate
On: \( 34\%+34\%+34\% = 102\% \)
A probability above 1 flags the error at once. The six bands and two tails account for everything, with no extra region at the centre.
Sorting
Which route is shorter for each region?
Sort into buckets
Sort each probability question.
The third of these is a single tail, which is short enough to add directly — 0.0015 on its own.
Matching
Add the bands it covers.
Match the pairs
Why: The four answers span three orders of magnitude, from nearly half the distribution down to fifteen in ten thousand — which is how quickly a normal curve thins out as you move away from the centre.
Prediction
Commit before reasoning.
Predict first
What is the probability that a value from a normal distribution is exactly the mean?
Correct: Essentially zero, since a single point has no area.
\[ P(x = \bar{x}) = 0; \quad P(x \leq \bar{x}) = P(x < \bar{x}) = 0.5 \]
Why: Areas under a curve measure intervals, and a single point is an interval of zero width. That is why probabilities here are always asked about ranges rather than exact values, and why it makes no difference whether an inequality is strict. A continuous distribution assigns probability to regions, never to individual points.
Section
Section 3
Concept
Given a mean and a standard deviation, label the axis at the mean and at each standard deviation either side. Any boundary landing on one of those marks can be answered by the band rule directly.
\[ 172 \pm 14: \; 158, 186; \quad 172+2(14) = 200 \]
Two labelled axes help: one in the data's own units and one counting standard deviations, which is the z-score scale of the next idea.
Figure (svg): A normal curve of cholesterol readings with the actual measurement scale beneath it
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758 — Interpret normally distributed data
Picture it
Example 2: cholesterol readings with mean 172 and standard deviation 14.
Figure (svg): A normal curve of cholesterol readings with the actual measurement scale beneath it
Readings of 158 and 186 are one standard deviation either side, and 200 is two above. Both questions land exactly on a mark.
Worked example
Example 2, both parts.
\[ \text{Readings have mean } 172 \text{ and } s = 14. \text{ Find the percent between } 158 \text{ and } 186, \text{ and the percent above } 200. \]
Label the axis
Why: Add and subtract 14 repeatedly from 172.
\[ 130, 144, 158, 172, 186, 200, 214 \]
Locate the first boundaries
Why: One fifty-eight and 186 are one s either side.
\[ \text{within } 1\text{ standard deviation} \]
Read the first answer
Why: The 68 percent figure.
\[ 68 \% \]
Locate 200 and read the tail
Why: Two hundred is two s above; beyond it is 2.35 plus 0.15.
\[ 2.5 \% \]
Figure (svg): A normal curve of cholesterol readings with the actual measurement scale beneath it
\[ 68\%; \qquad 2.35\%+0.15\% = 2.5\% \]
Verify: check the second by subtraction
Why: Everything below 200 is 0.5 plus 0.475, which is 0.975, so above it is 0.025 or 2.5 percent — matching the band addition. Two routes agreeing is a good sign that the boundary was placed at the right mark.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758
Fill the middle
Example 2.
Fill in the blanks
172+2(14) = 200
Why: One seventy-two plus 28 is 200, which is exactly two standard deviations above the mean. Marking the axis first turns every question into a band question.
Worked example
Guided Practice 7 and lesson exercises 11 to 16.
\[ \text{Find the percent of readings between } 172 \text{ and } 200; \text{ then, for mean } 33 \text{ with } s = 4, \text{ find } P(29 \leq x \leq 37) \text{ and } P(x \geq 29). \]
Cholesterol: locate the boundaries
Why: One seventy-two is the mean and 200 is two s above.
\[ 0\text{ to } +2 s \]
Cholesterol: add the bands
Why: Thirty-four plus 13.5 percent.
\[ 47.5 \% \]
Second data: 29 and 37
Why: One s either side of 33.
\[ 0.68 \]
Second data: at least 29
Why: Everything above one s below the mean.
\[ 0.84 \]
Figure (svg): The solution to Worked example two more from the same data shown as a ladder of expressions, one row per algebraic move
\[ 47.5\%; \; 0.68; \; 0.84 \]
Verify: check the last one two ways
Why: Adding gives 34 plus 34 plus 13.5 plus 2.35 plus 0.15, which is 84 percent. Subtracting gives 1 minus the 16 percent below negative one s, also 84. The second route needs one number rather than five.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-760
Trap
\[ \text{mean } 172, \; s = 14; \text{ find the percent above } 195 \]
Treat 195 as two standard deviations above
Why: It is near 200, so the nearest mark is used.
\[ 2.5\% \quad \text{(wrong)} \]
One ninety-five is 23 above the mean, which is about 1.64 standard deviations — not 2. The band rule does not reach it.
\[ z = \frac{195-172}{14} \approx 1.64 \]
Compute the z-score and use the table
Why: The band rule only applies at whole standard deviations.
\[ \text{look up } z \approx 1.6 \text{ in the table} \]
Checking whether the boundary is a whole number of standard deviations is the first thing to do, and it decides which method the question needs.
Sorting
For mean 33 with s equal to 4.
Sort into buckets
Sort each boundary.
The value 35 is half a standard deviation above the mean, which the band rule cannot reach — the table of the last idea handles it.
Comparison
Fill the blanks. The same axis, twice.
Comparison matrix
| Reading | Distance from the mean | z-score |
|---|---|---|
| 158 | 14 below | -1 |
| 172 | the mean itself | 0 |
| 200 | 28 above | 2 |
| 214 | 42 above | 3 |
The right-hand column is the same axis measured in standard deviations, which makes it identical for every normal distribution — the idea behind the z-score.
Prediction
Commit before reasoning.
Predict first
The cholesterol standard deviation is 14. If it were 28 instead, what would happen to the percent between 158 and 186?
Correct: It would fall, since 158 and 186 would now be only half a standard deviation either side.
\[ z = \frac{186-172}{28} = 0.5 \text{ rather than } 1 \]
Why: A larger standard deviation means a wider, flatter bell, so a fixed interval around the mean captures less of it. With s equal to 28 the interval from 158 to 186 spans only from negative one half to positive one half of a standard deviation, holding about 38 percent rather than 68. The standard deviation controls the width of the curve, which is exactly what the second of the two describing numbers does.
Section
Section 4
Concept
The z-score of a value is the number of standard deviations it lies above or below the mean. Subtracting the mean and dividing by the standard deviation converts any normal distribution into the standard one, with mean 0 and standard deviation 1.
\[ z = \frac{x-\bar{x}}{s} \]
A positive z-score means above the mean and a negative one below, and the size says how unusual the value is regardless of the original units.
Figure (svg): The z-score formula converting a data value into a number of standard deviations
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758 — Standard normal distribution
Picture it
Subtract, then divide.
Figure (svg): The z-score formula converting a data value into a number of standard deviations
Fifty seals against a mean of 73 with a standard deviation of 14.1 gives a z-score of about negative 1.6 — well below average but not extraordinary.
Worked example
Example 3, Step 1.
\[ \text{Seal counts have mean } 73 \text{ and } s = 14.1. \text{ Find the } z \text{-score of } 50. \]
Subtract the mean
Why: Fifty minus 73.
\[ -23 \]
Divide by the standard deviation
Why: Negative 23 over 14.1.
\[ \text{about } -1.63 \]
Round to one decimal
Why: The table works to tenths.
\[ \text{about } -1.6 \]
Interpret the sign
Why: Negative means below the mean.
\[ 1.6\text{ standard deviations below} \]
Figure (svg): The z-score formula converting a data value into a number of standard deviations
\[ z = \frac{50-73}{14.1} \approx -1.6 \]
Verify: sanity-check the size
Why: Fifty is 23 below a mean of 73, and one standard deviation is 14.1 — so 23 is a bit more than one and a half of them, matching 1.6. Estimating the z-score before dividing catches a sign error or a misplaced decimal.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 759-759
Fill the middle
Example 3.
Fill in the blanks
z = \frac-23___ = \frac___}___ \approx -1.6
Why: Fifty minus 73 is negative 23, and dividing by 14.1 gives about negative 1.6. The subtraction always comes first.
Worked example
What a z-score is for.
\[ \text{A student scores } 84 \text{ on a test with mean } 75 \text{ and } s = 6, \text{ and } 63 \text{ on another with mean } 55 \text{ and } s = 4. \text{ Which is better?} \]
First test: subtract and divide
Why: Eighty-four minus 75, over 6.
\[ z = 1.5 \]
Second test: subtract and divide
Why: Sixty-three minus 55, over 4.
\[ z = 2 \]
Compare the z-scores
Why: Two is further above the mean than 1.5.
Note why raw scores mislead
Why: Eighty-four is the larger number but the weaker performance.
Figure (svg): The solution to Worked example compare two different scales shown as a ladder of expressions, one row per algebraic move
\[ z = 1.5 \text{ against } z = 2 \]
Verify: read the two z-scores as percentages
Why: A z-score of 1.5 puts the student above roughly 93 percent of the class, and 2 above roughly 98 percent. The second performance really is the stronger one, even though 63 looks worse than 84 — which is exactly what a z-score is for.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 758-758
Error analysis
A student computes the z-score of 50 with mean 73 and standard deviation 14.1.
Annotate
On: \( z = 50-\frac{73}{14.1} \approx 44.8 \)
A z-score is nearly always between negative 3 and 3, so a value of 44.8 is impossible on its face and flags the error immediately.
Sorting
Read the size of each z-score.
Sort into buckets
Sort each z-score.
The sign says which side of the mean, and the size says how unusual — which is why z-scores can be compared across completely different measurements.
Matching
Mean 73, standard deviation 14.1.
Match the pairs
Why: The mean always has z-score 0, which is a useful anchor. Every other value's z-score measures its distance from that anchor in standard deviations rather than in seals.
Prediction
Commit before reasoning.
Predict first
Why convert values to z-scores rather than working with the original data?
Correct: Because one table then serves every normal distribution, whatever its mean and units.
\[ \text{any normal} \;\to\; \text{one standard normal} \]
Why: There are infinitely many normal distributions, one for each mean and standard deviation, and tabulating all of them is impossible. Converting to z-scores maps every one of them onto the single standard normal distribution, so a single table answers every question. It also makes different measurements comparable, which is why test scores and medical results are often reported as z-scores.
Section
Section 5
Concept
The standard normal table gives the probability that z is at most a given value. The row supplies the whole part of the z-score and the column its tenths.
\[ P(z \leq -1.6) = 0.0548 \]
The table always reports the area to the LEFT of the z-score, so a right-hand region is found by subtracting the entry from 1.
Figure (svg): Part of the standard normal table with one lookup highlighted
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 759-759 — Standard Normal Table
Picture it
Example 3: looking up a z-score of negative 1.6.
Figure (svg): Part of the standard normal table with one lookup highlighted
Row negative 1 and column point 6 meet at 0.0548, so about 5.5 percent of surveys observed at most 50 seals.
Worked example
Example 3.
\[ \text{Seal counts have mean } 73 \text{ and } s = 14.1. \text{ Find } P(x \leq 50). \]
Compute the z-score
Why: Fifty minus 73, over 14.1.
\[ \text{about } -1.6 \]
Find the row
Why: The whole part, negative 1.
\[ r o w - 1 \]
Find the column
Why: The tenths, point 6.
\[ \text{column } . 6 \]
Read the entry
Why: Where the row and column meet.
\[ 0.0548 \]
Figure (svg): Part of the standard normal table with one lookup highlighted
\[ P(x \leq 50) \approx 0.0548 \]
Verify: check against the band rule
Why: A z-score of negative 1.6 lies between negative 1 and negative 2, so the probability should sit between 0.0250 and 0.1587 — and 0.0548 does. The band rule brackets every table lookup, which is a quick guard against reading the wrong row.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 759-759
Fill the middle
Example 3, extended.
Fill in the blanks
P(x \geq 50) = 1-0.0548 = 0.9452
Why: The table gives the area to the left, so the area to the right is 1 minus it. About 94.5 percent of surveys observed at least 50 seals.
Worked example
Guided Practice 8 and lesson exercises 19 and 21.
\[ \text{Find } P(x\leq 90) \text{ for the seals; then } P(x\leq 68) \text{ and } P(x\leq 45) \text{ for mean } 64 \text{ with } s = 7. \]
Seals at 90
Why: Seventeen over 14.1 is about 1.2.
\[ P = 0.8849 \]
Second: 68 with mean 64
Why: Four over 7 is about 0.6.
\[ P = 0.7257 \]
Third: 45 with mean 64
Why: Negative 19 over 7 is about negative 2.7.
\[ P = 0.0035 \]
Note the pattern
Why: A positive z gives above one half; a negative one below.
\[ 0\text{ is the dividing line} \]
Figure (svg): The solution to Worked example three more lookups shown as a ladder of expressions, one row per algebraic move
\[ 0.8849, \; 0.7257, \; 0.0035 \]
Verify: check each against the halfway mark
Why: Two of the three z-scores are positive, and both gave probabilities above 0.5; the negative one gave a probability well below it. Every entry above z equal to 0 exceeds one half and every entry below it falls short, which is a one-glance check on the row.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 759-760
Trap
\[ z = -1.6, \; \text{table gives } 0.0548 \]
Report it as the probability of at least 50 seals
Why: The direction of the inequality is not checked.
\[ P(x \geq 50) = 0.0548 \quad \text{(wrong)} \]
Fifty is below the mean of 73, so far more than 5 percent of surveys should exceed it — about 95 percent, in fact.
\[ P(x \geq 50) = 1-0.0548 = 0.9452 \]
The table gives the LEFT area; subtract for the right
Why: Every entry is a cumulative probability from the far left.
\[ \text{table entry} + \text{right tail} = 1 \]
Sketching the curve and shading the wanted region before looking anything up settles the direction every time.
Ranking
Finding a probability with the table.
Put in order
Why: The sketch in step one is what tells you whether step five is needed, and skipping it is how a left area gets reported for a right-hand question. It costs ten seconds and prevents the commonest error in the idea.
Comparison
Fill the blanks. Two ways to get an area.
Comparison matrix
| Question | Band rule | Table |
|---|---|---|
| Which z-scores? | whole numbers only | any value to one decimal |
| What it gives | the area of a band | the area to the left of z |
| P(z <= -2) | 0.0250 from the bands | 0.0228 from the table |
| Effort | add a few numbers | one lookup |
The two disagree slightly in the third row because the band figures are rounded — 68, 95 and 99.7 are approximations to the exact areas the table reports.
Prediction
Commit before reasoning.
Predict first
Table entries beyond three standard deviations are given as tiny numbers like 0.00001. Why not extend the table?
Correct: Because the remaining area is negligible for practical purposes, though never actually zero.
\[ \text{beyond } 3s: \; 0.3\%; \quad \text{beyond } 4s: \; 0.006\% \]
Why: Only 0.3 percent of the area lies beyond three standard deviations, and beyond four it is under one in fifteen thousand. The curve extends forever and never touches the axis, so every entry is positive — the book's own note explains that 0.00001 means slightly more than zero. Extending the table further would add rows nobody needs, which is why practical tables stop where they do.
Comparison
Fill the blanks. Whole standard deviations, or any value.
Comparison matrix
| Question | Band rule | z-score and table |
|---|---|---|
| When it applies | boundaries at whole standard deviations | any boundary at all |
| First step | sketch and shade | compute the z-score |
| What you read | 68, 95, 99.7 or the individual bands | the area to the left of z |
| For a right-hand region | add the bands, or subtract from 1 | subtract the entry from 1 |
Both answer the same kind of question, and the band rule is simply the table's answers at whole standard deviations, rounded.
Pattern
Sketch first, then choose a method.
Every probability lies between 0 and 1, and the band rule brackets every table lookup — both are free checks.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-761
Check
Add the bands.
Check your understanding
A normal distribution has mean 84 and standard deviation 5. What is P(74 <= x <= 94)?
Answer: A
Why: 74 and 94 are two standard deviations either side of 84.
Check
A tail region.
Check your understanding
A normal distribution has mean 51 and standard deviation 3. What is P(x <= 48)?
Answer: A
Why: 48 is one standard deviation below the mean, and 13.5 plus 2.35 plus 0.15 is 16 percent.
Check
The table gives a left-hand area.
Check your understanding
Seal counts have mean 73 and standard deviation 14.1. What is P(x <= 50)?
Answer: A
Why: The z-score is about -1.6, and the table entry for that is 0.0548.
Real world
A factory fills bottles with a mean of 500 millilitres and a standard deviation of 4 millilitres, normally distributed. A bottle is rejected if it holds under 492 millilitres.
Discussion prompt
Find the rejection rate, then find what the mean would have to be to cut it to about one in a thousand.
Hint: Start with the z-score of 492.
Answer:
\[ z = \frac{492-500}{4} = -2 \;\Longrightarrow\; P \approx 0.0228 \]
About 2.3 percent of bottles are rejected — roughly one in forty-four, which on a line filling ten thousand bottles a day is 230 bottles wasted.
\[ P = 0.001 \;\Longrightarrow\; z \approx -3.1 \;\Longrightarrow\; \bar{x} = 492+3.1(4) \approx 504.4 \]
Raising the mean fill to about 504.4 millilitres would cut rejections to one in a thousand — at the cost of giving away 4.4 millilitres in every bottle, which over ten thousand bottles a day is 44 litres of free product. That trade-off is the whole of process control: tightening the standard deviation is usually cheaper than raising the mean, since halving s to 2 would achieve the same rejection rate with a mean of only 498.2. Reading the calculation backwards, from a target probability to a required setting, is what makes the normal distribution a design tool rather than only a description.
Commit first
Answer, then rate your confidence honestly.
Predict first
The table gives P(z at most -1.6) as 0.0548. Is that the probability of being ABOVE that value?
Correct: No — the table gives the area to the LEFT, so the right-hand area is 0.9452.
\[ P(z \leq -1.6)+P(z \geq -1.6) = 0.0548+0.9452 = 1 \]
Why: Every entry is a cumulative probability measured from the far left of the curve, which is why the entries increase as you move down the rows and across the columns. A sanity check settles it: a z-score of negative 1.6 is below the mean, so most values must be above it — and 0.9452 says so while 0.0548 does not. Sketching the curve and shading the wanted region before looking anything up is the habit that prevents this entirely.
Explain it
They know what a mean and a standard deviation are and have never seen a bell curve.
Discussion prompt
In four sentences or fewer, explain what the 68-95-99.7 rule says.
Hint: Describe where the data sits.
Answer:
Lots of measurements pile up around an average and thin out on both sides, making a bell shape. About 68 out of every 100 values land within one standard deviation of the average.
Stretch to two standard deviations and you capture 95 of them, and to three and you have 99.7. So anything beyond three standard deviations is genuinely rare — about three times in a thousand.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the bands, write 34, 13.5, 2.35 and 0.15 on a sketch once and keep it. For labelling, add and subtract the standard deviation repeatedly from the mean before reading the question. For z-scores, subtract first and divide second. For the table, sketch and shade before looking anything up.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a normal distribution page. Top: draw a large bell curve with all six bands and both tails labelled, and write the three cumulative figures beneath it. Middle left: shade three different regions on small copies of the curve and compute each probability by adding bands. Middle right: draw the cholesterol curve with both scales labelled, in milligrams and in z-scores, and answer both parts of Example 2. Bottom left: write the z-score formula and compute three z-scores of your own, including one negative. Bottom right: copy the row of the table for z between negative 1 and negative 1.9, look up one value, and write one sentence stating which side of the curve the entry measures.
If any of your probabilities came out above 1, recheck whether you added a band twice — there is no band at the mean itself.
Recap
Five things, all from a mean and a standard deviation.
| If you see | Then |
|---|---|
| A bell-shaped symmetric distribution | A normal distribution |
| Boundaries at whole standard deviations | Use the band rule |
| Any other boundary | Compute a z-score and use the table |
| A table entry | It is the area to the LEFT of that z-score |
| A right-hand region | Subtract the entry from 1 |
| A z-score beyond 3 | The value is genuinely unusual |
Lesson 11.4 turns from describing a population to sampling one, and asks how far a sample's statistics can be trusted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 11 Data Analysis and Statistics — Lesson 11.3 Use Normal Distributions §11.3, pp. 757-761 — everything on these slides traces back here
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