2.11 Normal Models: Areas, Cutoffs and Relative Position — Answers and Rubrics
These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
Describe and Calculate Normal-Model Parameters: worked answer
Model Q is a normal model for repeated temperature readings in °C. Its center is marked 75, and its one-SD marks are 71 and 79. Model R is also normal, centered 75 with one-SD marks 67 and 83. A separate model U is uniform on 65–85°C, with constant density over that interval and zero elsewhere. The markings describe supplied model parameters, not observed sample summaries.
Part a
Describe Q and R as continuous distributions, including unimodality, bell shape, symmetry and the effect of their differing spread on curve width/height. Explain how a normal curve can model both a histogram of many readings and the probability distribution of a new reading. Contrast U: explain its continuous domain and why symmetry alone does not make it normal.
Q and R are continuous, unimodal, bell-shaped and symmetric about 75. Q has a narrower one-SD span and is taller/more concentrated; R is wider and shorter. A histogram of many readings may be approximated by the curve, while the curve’s interval areas model probabilities for a new numerical temperature reading. U can take any value in 65–85 and assigns probabilities to intervals, but its constant flat density is not bell-shaped; continuous symmetry does not imply normality. Many continuous variables can be normal-modeled, not all.
| Evidence for credit | Points |
|---|---|
| Gives continuous/unimodal/bell/symmetric normal descriptors and correct width/height comparison. | 1 |
| Explains both data-histogram and random-reading probability roles. | 1 |
| Describes U’s interval domain/probabilities and rejects symmetry-as-sufficient-for-normality. | 1 |
Part b
Calculate the mean and SD of Q and R from the marked intervals. For Q define Z=(Q−75)/4 and calculate its mean and SD. Name this resulting normal distribution and state whether its standardized readings retain °C units.
Both means are 75°C. Q’s SD is 79−75=4°C; R’s is 83−75=8°C. Z has mean(75−75)/4=0 and SD 4/4=1, so it is standard normal. The standardized readings are dimensionless because the numerator’s temperature difference is divided by a temperature SD.
| Evidence for credit | Points |
|---|---|
| Calculates mean 75 and SD 4/8 from the supplied markings. | 1 |
| Calculates standardized mean 0 and SD 1 and names standard normal. | 1 |
| Explains the absence of temperature units for Z. | 1 |
Use the Empirical Rule for Commute Time: worked answer
A supplied normal commute-time model has mean 50 minutes and SD 6 minutes. Use empirical-rule approximations rather than exact normal-CDF values. One-, two-, and three-SD boundaries must be calculated from the model.
Part a
Calculate the three central intervals within one, two and three SDs and their approximate percentages. Identify both names of the rule you use.
The intervals are 44–56 minutes (about 68%),38–62 (about 95%), and 32–68 (about 99.7%). This is the empirical rule, also called the 68–95–99.7 rule.
| Evidence for credit | Points |
|---|---|
| Calculates all three correct intervals. | 1 |
| Associates 68/95/99.7% with one/two/threeSD and names the rule. | 1 |
Part b
Calculate the approximate percentages between 50 and 56, between 56 and 62, above 56, below 32, and outside 38–62. Explain the symmetry and subtraction for each; label the regions on a sketch or give their bound/area descriptions.
50–56 contains 68/2=34%. 56–62 contains(95−68)/2=13.5%. Above 56 contains(100−68)/2=16%. Below 32 contains(100−99.7)/2=.15%. Outside 38–62 contains 100−95=5%, split 2.5% per tail. The descriptions or sketch must use the stated time boundaries and approximate area labels.
| Evidence for credit | Points |
|---|---|
| Calculates 34,13.5,16,.15 and 5% correctly. | 1 |
| Explains symmetry/complement/subtraction and identifies each region. | 1 |
Calculate Normal Battery-Life Areas: worked answer
A supplied approximately normal model for a battery population has mean 12 hours and SD 1.5 hours. Let X be one battery’s life. You may use normal technology or the following standard-normal left-tail values: Φ(−1)=.158655, Φ(2)=.977250, Φ(−2/3)=.252493 and Φ(4/3)=.908789.
Part a
Calculate P(X<10.5), P(X>15), and P(11<X<14). For each, standardize the bounds, identify the shaded region or describe it with labeled bounds, and show the area calculation.
For 10.5,z=(10.5−12)/1.5=−1, so P(X<10.5)=.158655. For 15,z=2, so P(X>15)=1−.977250=.022750. The interval 11–14 has z=−2/3 and 4/3, giving .908789−.252493=.656296. A correct sketch/description shades the left tail below 10.5, right tail above 15 and region between 11 and 14 respectively.
| Evidence for credit | Points |
|---|---|
| Calculates all four standardized bounds correctly. | 1 |
| Uses left area, right complement and interval subtraction to obtain the three values. | 1 |
| Labels the three regions consistently. | 1 |
Part b
Interpret the interval probability for the stated battery population. State the full curve’s probability/area and explain whether including the exact endpoints changes these continuous-model probabilities.
About 65.63% of batteries under this model have lives between 11 and 14 hours. The total normal-curve area/probability is 1. A single exact continuous value has probability 0, so including 11 or 14 does not change the modeled interval area.
| Evidence for credit | Points |
|---|---|
| Interprets .656296 for battery lives in 11–14 hours. | 1 |
| States total area 1 and correctly explains continuous endpoints. | 1 |
Find Normal Cutoffs and Extreme Tails: worked answer
A supplied normal phone-use model has mean 10 hours and SD 2 hours. For an area expressed as a percentage number p, use p/100 as a probability. Supplied standard-normal quantiles are Φ⁻¹(.05)=−1.6448536, Φ⁻¹(.10)=−1.2815516, Φ⁻¹(.90)=1.2815516 and Φ⁻¹(.95)=1.6448536. Use these values as a table, or equivalent inverse-normal technology; explain which left-tail area you select.
Part a
Find the upper boundary x_a of the lowest 10% and write the corresponding probability inequality using p/100. Show the selected standard-normal area and the conversion to hours.
p=10. Select left-tail area .10,z=−1.2815516. hours. , so the lowest 10% lies below about 7.437 hours.
| Evidence for credit | Points |
|---|---|
| Selects left area .10 and z−1.28155. | 1 |
| Calculates cutoff about 7.437 hours. | 1 |
| Writes the lowest 10% strict inequality with 10/100. | 1 |
Part b
Find the lower boundary x_b of the highest 10% and write its strict probability inequality. Explain why the inverse input is .90 rather than .10.
The upper tail .10 leaves .90 to its left, so z=1.2815516 and hours. .
| Evidence for credit | Points |
|---|---|
| Selects left area .90 and correct z. | 1 |
| Calculates 12.563 hours and writes P(X>x_b)=.10. | 1 |
| Explains the complementary area. | 1 |
Part c
Find both bounds of the central 80% interval. Identify p, lower bound and upper bound, and write P(x_a<X<x_b)=p/100. Verify the area by subtracting the two left-tail areas.
p=80. The central 80% leaves 10% in each tail, so lower/upper left areas are .10/.90. Bounds are 7.4368968 and 12.5631032 hours. , verified by .90−.10=.80.
| Evidence for credit | Points |
|---|---|
| Finds both correct bounds from .10/.90. | 1 |
| Writes the bounded probability statement with 80/100. | 1 |
| Verifies the .80 interval area. | 1 |
Part d
Find the most extreme 10% on both sides. State each tail area as one-half times p/100, calculate both cutoffs and write both strict tail inequalities. Explain why using 10% in each tail would answer a different question.
Each tail area is . Use left areas .05 and.95, giving , hours. and ; together the two tails total .10. Ten percent in each tail would total 20%.
| Evidence for credit | Points |
|---|---|
| Splits total 10% into 5% per tail. | 1 |
| Calculates 6.710 and 13.290 hours from .05/.95. | 1 |
| Writes both half-times-percentage inequalities and explains the 20% error. | 1 |
Compare Relative Positions Within and Between Models: worked answer
For the within-model comparison, Model A is normal for fill mass with mean 250 g and SD 5 g; its two readings are 254 g and 247.5 g. For the separate between-model comparison, Model C is normal with mean 300 g and SD 8 g, with a reading of 309.6 g; Model D is normal with mean 400 g and SD 10 g, with a reading of 413 g. All are supplied population models. Standard-normal left-tail values: Φ(.8)=.788145, Φ(−.5)=.308538, Φ(1.2)=.884930 and Φ(1.3)=.903200.
Part a
Calculate the percentiles of both Model A readings and compare their relative positions within that distribution. Also calculate the proportions above each reading and connect the two comparisons.
The z-scores are(254−250)/5=.8 and(247.5−250)/5=−.5. Their percentiles are about 78.81 and 30.85. The proportions above are 1−.788145=.211855 and 1−.308538=.691462. Thus 254 g is higher relative to Model A, with a smaller share of bags above it.
| Evidence for credit | Points |
|---|---|
| Computes both percentiles correctly. | 1 |
| Computes both above proportions. | 1 |
| Uses them to compare within-Model A position in context. | 1 |
Part b
Compare 309.6 g in Model C with 413 g in Model D using percentiles and proportions above the readings. Show the standardized positions and area calculations. Explain why comparing only raw grams or reporting only z-scores would not complete the requested comparison.
For Model C, , giving percentile about 88.49 and above proportion . For Model D, , giving percentile about 90.32 and above proportion . The 413 g Model D reading is higher relative to its own distribution: a larger share lies below it and a smaller share lies above it. Raw grams ignore the different centers and scales; z-scores support the calculation, while the percentiles/proportions express the corresponding population shares.
| Evidence for credit | Points |
|---|---|
| Calculates Model C z=1.2, percentile about 88.49 and above proportion 0.115070. | 1 |
| Calculates Model D z=1.3, percentile about 90.32 and above proportion 0.096800. | 1 |
| Concludes Model D’s reading is relatively higher using both percentile and above-proportion evidence, with each share tied to its population. | 1 |
| Explains why different population centers/scales prevent raw-mass comparison from completing the relative-position question. | 1 |