Finding the Area Between Two Specification Limits
In Finding Areas Above and Below a Value, you used a cutoff to find a one-sided area under a normal curve. Manufacturing questions often have two cutoffs: a lower and an upper specification limit. The desired proportion is the area between those limits.
For example, a part may be considered acceptable if its measured diameter is neither too small nor too large. If \(L\) is the lower tolerance limit and \(U\) is the upper tolerance limit, acceptable measurements lie between \(L\) and \(U\). A normal model lets us calculate the proportion expected to fall in that interval.
The area between the limits can be found by subtracting the area to the left of the lower limit from the area to the left of the upper limit. This works because the cumulative area to the left of \(U\) includes both the area below \(L\) and the area between \(L\) and \(U\).
The calculator command \(\operatorname{normalcdf}(L,U,\mu,\sigma)\) finds the area directly. Keep the input order straight: lower bound, upper bound, mean, standard deviation. The two limits, the mean, and the standard deviation must all use the same units.
A second way to think about the calculation is to standardize both limits. As in Calculating a z-Score, each z-score measures how many standard deviations its value is from the mean. After converting both limits to z-scores, subtract the lower cumulative area from the upper cumulative area using Table A.
This standardization is a useful check, but you do not need to convert to z-scores before using normalcdf. The command can calculate an area using the original measurement units when you enter the model’s mean and standard deviation.
Interpreting a Tolerance Probability
A probability between two tolerance limits is also a proportion under the stated model. If the result is 0.90, the model predicts that about 90% of measurements will fall within the limits. It does not mean that every item will meet the specification, nor does it guarantee the exact percentage in a particular production run.
The proportion outside the tolerance interval is the complement of the proportion inside it. This can be helpful when the question asks about the share of parts that fail to meet specifications.
For a continuous normal variable, the probability of landing on exactly one specified measurement is 0. Therefore, using inclusive limits instead of strict limits does not change the probability: \(P(L\leq X\leq U)=P(L<X<U)\). In a manufacturing context, it is usually natural to say that measurements “from \(L\) to \(U\), inclusive” meet the stated tolerance; the model calculation is the same.
A Routine for Tolerance Intervals
Name the random variable \(X\), its units, and its normal model \(N(\mu,\sigma)\).
Write the lower limit \(L\) and upper limit \(U\), making sure \(L<U\). Translate the requested range into an event for \(X\).
Use \(\operatorname{normalcdf}(L,U,\mu,\sigma)\), or subtract the cumulative area at \(L\) from the cumulative area at \(U\).
Optionally standardize each limit and check with Table A. State the result as a proportion or percentage of measurements in the model.
Worked Example: Bearing Pins Within Diameter Tolerances
Worked Example: Bearing Pins Within Diameter Tolerances
A fictional machine produces bearing pins whose diameters are modeled as normal with mean 12.00 millimeters and standard deviation 0.04 millimeters. Pins meet the diameter specification if they measure from 11.94 to 12.06 millimeters. Find the model-based proportion that meets the specification.
State. Let \(X\) be the diameter of a pin, in millimeters. The model is \(X\sim N(12.00,0.04)\). The event of interest is \(11.94\leq X\leq12.06\).
Plan. This is an area between two values. Use the lower and upper specification limits as the first two normalcdf inputs, followed by the mean and standard deviation. All inputs are in millimeters.
Do. The calculator command gives:
Check by standardizing the limits: \(z_L=(11.94-12.00)/0.04=-1.5\) and \(z_U=(12.06-12.00)/0.04=1.5\). Table A gives a left area of about 0.9332 at \(z=1.5\) and about 0.0668 at \(z=-1.5\). Their difference is \(0.9332-0.0668=0.8664\), agreeing with the calculator result.
Conclude. About 0.8664, or 86.64%, of pin diameters in this model meet the specification. This is the model-based proportion expected within the stated tolerance limits.
Worked Example: Gasket Thickness in an Asymmetric Interval
Worked Example: Gasket Thickness in an Asymmetric Interval
A fictional production line models gasket thickness as normal with mean 3.20 millimeters and standard deviation 0.10 millimeters. The acceptable range is 3.05 to 3.38 millimeters. Find the proportion of modeled gasket thicknesses in this range.
State. Let \(X\) be a gasket’s thickness, in millimeters. Then \(X\sim N(3.20,0.10)\), and the event is \(3.05\leq X\leq3.38\).
Plan. The limits are not equally far from the mean, so do not assume the area is symmetric. Use normalcdf with the two original limits and the given model parameters.
Do. The direct calculation is:
For a check, the standardized limits are \(z_L=(3.05-3.20)/0.10=-1.5\) and \(z_U=(3.38-3.20)/0.10=1.8\). Table A gives approximate left areas of 0.0668 and 0.9641. Subtracting gives \(0.9641-0.0668=0.8973\), rounded to four decimal places.
Conclude. About 89.73% of gasket thicknesses in this normal model fall within the acceptable range of 3.05 to 3.38 millimeters.
Worked Example: Finding the Proportion of Wire Within Tolerance
Worked Example: Finding the Proportion of Wire Within Tolerance
A fictional wire manufacturer models wire diameter as normal with mean 0.500 inches and standard deviation 0.004 inches. A customer’s specifications require diameters from 0.494 to 0.507 inches. Find the proportion within tolerance, then find the proportion outside tolerance.
State. Let \(X\) be a wire’s diameter, in inches. The model is \(X\sim N(0.500,0.004)\). The in-tolerance event is \(0.494\leq X\leq0.507\).
Plan. First find the area between the two limits with normalcdf. Then use the complement to find the proportion outside the specification range.
Do. The in-tolerance proportion is:
As a check, \(z_L=(0.494-0.500)/0.004=-1.5\) and \(z_U=(0.507-0.500)/0.004=1.75\). Table A gives approximate cumulative areas of 0.0668 and 0.9599. Their difference is \(0.9599-0.0668=0.8931\). The outside proportion is \(1-0.8931=0.1069\).
Conclude. About 89.31% of wire diameters in the model meet the specifications, and about 10.69% fall outside them. These are complementary model-based proportions.
Common Mistakes and AP Exam Tips
Most errors in between-values problems come from using the wrong bounds or interpreting a cumulative area as though it were the requested interval area. Make the event clear before entering values.
- Reporting the area to the left of the upper limit. That cumulative area includes values below the lower limit too. Subtract the area below the lower limit to isolate the interval.
- Reversing the limits. Enter the smaller value first and the larger value second. In \(\operatorname{normalcdf}(L,U,\mu,\sigma)\), \(L\) is the lower bound and \(U\) is the upper bound.
- Using the variance instead of the standard deviation. The fourth normalcdf input is \(\sigma\), not \(\sigma^2\). For a model with standard deviation 0.04 millimeters, enter 0.04.
- Assuming a tolerance interval is symmetric. The limits may not be equally far from the mean. Calculate both sides from their actual values instead of doubling one area.
- Mixing units. A diameter limit in inches cannot be combined directly with a mean in millimeters. Convert all values to the same units before calculating.
- Confusing inside and outside proportions. The interval area is the proportion meeting both limits. The complement is the proportion below the lower limit or above the upper limit.
- Giving only a calculator decimal. A complete interpretation names the measurement, the tolerance interval, and what proportion the result represents.
For a strong AP response, define \(X\) in context, state its normal model and units, identify the interval event, show the normalcdf command or the difference of cumulative areas, and interpret the probability as a proportion of modeled measurements. If you also report the outside proportion, show that it is the complement.
Key Takeaway
A manufacturing tolerance interval has a lower and an upper limit. For a normal model, calculate the proportion within the limits directly with normalcdf, or subtract the cumulative left area at the lower limit from the cumulative left area at the upper limit. Standardizing both limits offers a useful check, and the final probability should be interpreted as a model-based proportion in context.
Check Your Understanding
For each question, identify the interval event and write the normalcdf setup before calculating.
- A fictional pump component has length modeled as \(N(48,0.3)\), in millimeters. What command finds the proportion with lengths from 47.5 to 48.4 millimeters?
- A fictional packaging line models packet mass as \(N(250,8)\), in grams. Which cumulative areas would you subtract to find the proportion between 238 and 262 grams?
- A fictional metal rod’s diameter follows \(N(6.00,0.05)\), in millimeters. If the acceptable range is 5.92 to 6.08 millimeters, how would you find the proportion outside the range after finding the proportion inside?
- For a normal model, explain why the probability from \(L\) to \(U\) is the left area at \(U\) minus the left area at \(L\).
- A student enters \(\operatorname{normalcdf}(4.2,4.0,4.1,0.1)\) for the proportion between 4.0 and 4.2 centimeters. Identify the input error and give the corrected setup.