Finding the Area on One Side
In Using normalcdf to Find a Normal Area, you found the area between two values by entering a lower bound and an upper bound. Many questions ask for a one-sided area instead: the proportion of values below a cutoff or above it. The same normalcdf command works when one bound is set to an extremely low or high value.
Suppose \(X\) follows a normal model with mean \(\mu\) and standard deviation \(\sigma\). A left-tail probability such as \(P(X<c)\) is the area to the left of \(c\). A right-tail probability such as \(P(X>c)\) is the area to the right of \(c\). The cutoff \(c\) separates the requested tail from the rest of the curve.
On a calculator, use \(-1E99\) as a very low lower bound and \(1E99\) as a very high upper bound. These are practical stand-ins for negative and positive infinity, respectively. They are not values of the variable that have substantive meaning; they let normalcdf capture essentially the entire tail.
The input order remains lower bound, upper bound, mean, and standard deviation. For a left tail, the cutoff is the upper bound; for a right tail, it is the lower bound. Keep the cutoff, mean, and standard deviation in the same original units.
Using a Complement for a Tail
The two tails at a cutoff divide the entire area under the curve. Since the total area is 1, the area on one side is 1 minus the area on the other side. In particular, a right-tail area can be found by subtracting the left-tail area from 1.
Thus, you can find a right-tail probability either directly with \(\operatorname{normalcdf}(c,1E99,\mu,\sigma)\), or by calculating \(1-\operatorname{normalcdf}(-1E99,c,\mu,\sigma)\). Both methods should agree apart from rounding. The complement can be helpful when you already have a left-tail area from normalcdf or Table A.
As in Sketching and Shading Normal Curves, sketch or picture the requested region before calculating. Shading the right side of a cutoff makes it clear that the answer should be the part left over after subtracting the left area from 1. A quick sketch can also catch the mistake of using the left-tail command for a question asking “above.”
Normal variables are continuous. Therefore, the probability of a value being exactly equal to one cutoff is 0. For a normal model, \(P(X<c)=P(X\leq c)\), and \(P(X>c)=P(X\geq c)\). So wording such as “at most” or “at least” does not change the area calculation at a single boundary.
A Routine for One-Sided Areas
Identify what \(X\) measures, its units, and the model \(N(\mu,\sigma)\).
Decide whether the requested values are below or above the cutoff, and write the inequality for \(X\).
Use \(-1E99\) as the lower bound for a left tail or \(1E99\) as the upper bound for a right tail. Alternatively, use a complement.
Check that the answer is between 0 and 1, then state what proportion of modeled values is in the requested tail.
You can check a calculator result by standardizing the cutoff with \(z=(c-\mu)/\sigma\) and using Table A, which gives the area to the left of a z-score. A left-tail probability is the table’s left area. A right-tail probability is 1 minus that left area. Table entries are rounded, so a small difference from the calculator’s result is expected when the table value is less precise.
Worked Example: Delivery Times Below a Cutoff
Worked Example: Delivery Times Below a Cutoff
A fictional courier service models a particular delivery route’s travel time as normal with mean 32 minutes and standard deviation 6 minutes. Find the proportion of modeled travel times below 24.5 minutes.
State. Let \(X\) be travel time on this route, in minutes. The model is \(X\sim N(32,6)\), and the event is \(X<24.5\).
Plan. This is a left-tail question because it asks for times below a cutoff. Enter a very low lower bound, then the cutoff, the mean, and the standard deviation.
Do. Using normalcdf:
Check the result by standardizing the cutoff: \(z=(24.5-32)/6=-1.25\). Table A gives a left area of about 0.1056 at \(z=-1.25\), agreeing with the calculator result to four decimal places.
Conclude. About 0.1056, or 10.56%, of travel times in this model are below 24.5 minutes.
Worked Example: Fill Volumes Above a Target
Worked Example: Fill Volumes Above a Target
A fictional bottling line models container fill volume as normal with mean 250 milliliters and standard deviation 12 milliliters. Find the proportion of containers with fill volumes above 268 milliliters.
State. Let \(X\) be a container’s fill volume, in milliliters. The model is \(X\sim N(250,12)\), and the event is \(X>268\).
Plan. This is a right-tail question. Find the area directly by making the cutoff the lower bound and using \(1E99\) as the upper bound. Then verify with the complement of the left-tail area.
Do. The direct calculator command gives:
For the complement check, the left-tail area to the cutoff is approximately 0.9332. Subtracting it from 1 gives \(1-0.9332=0.0668\). A z-score check gives \(z=(268-250)/12=1.5\). Table A gives a left area of about 0.9332 at \(z=1.5\), so the right area is \(1-0.9332=0.0668\). The calculator’s unrounded areas give the reported result to four decimal places.
Conclude. About 0.0668, or 6.68%, of container fill volumes in this model are above 268 milliliters.
Worked Example: Comparing Direct and Complement Methods
Worked Example: Comparing Direct and Complement Methods
A fictional greenhouse models the time a particular seedling takes to emerge as normal with mean 18 days and standard deviation 4 days. Find the proportion of seedlings that take at least 22 days.
State. Let \(X\) be emergence time, in days. The model is \(X\sim N(18,4)\). “At least 22 days” means \(X\geq22\), which has the same probability as \(X>22\) for this continuous model.
Plan. The event is a right tail. Calculate it directly with a very high upper bound, then calculate the complement of the left-tail area as a check.
Do. The direct method is:
For the complement method, first find the area to the left of 22: \(\operatorname{normalcdf}(-1E99,22,18,4)\approx0.8413\). Then subtract from 1: \(1-0.8413=0.1587\).
A z-score gives a second check: \(z=(22-18)/4=1\). Table A’s left area at \(z=1\) is about 0.8413, so the area to the right is \(1-0.8413=0.1587\). Both methods produce the same probability to four decimal places.
Conclude. About 15.87% of seedlings in this model take at least 22 days to emerge.
Common Mistakes and AP Exam Tips
A correct normalcdf command depends on translating the words into the correct side of the cutoff. The extreme bound does not decide which side to calculate; the event does.
- Putting the cutoff on the wrong side. For \(P(X<c)\), the cutoff is the upper bound: use \(\operatorname{normalcdf}(-1E99,c,\mu,\sigma)\). For \(P(X>c)\), it is the lower bound: use \(\operatorname{normalcdf}(c,1E99,\mu,\sigma)\).
- Using the left area when the question asks for the right tail. normalcdf with \(-1E99\) and the cutoff returns the area to the left. For an “above” or “greater than” question, subtract that result from 1 or use the direct right-tail command.
- Treating \(1E99\) as a model parameter. It is only a convenient calculator bound. The mean and standard deviation still come from the stated normal model.
- Entering the wrong input order. The order is lower bound, upper bound, mean, standard deviation. Do not switch the mean and standard deviation or enter the variance in place of the standard deviation.
- Mixing units. A cutoff in minutes must be used with a mean and standard deviation in minutes. Convert values first if the model and cutoff are given in different units.
- Reporting an area without its context. A result such as 0.1587 is a probability, not 0.1587 days. Interpret it as a proportion or percentage of values in the specified tail under the model.
For full-credit communication, state what \(X\) represents and its model, write the requested event, show the correct normalcdf command or complement calculation, and interpret the resulting probability in context. A bare decimal does not show which tail was calculated or what the area means.
Key Takeaway
For a one-sided normal probability, use an extreme bound to make normalcdf capture the desired tail. A left tail uses a very low lower bound; a right tail uses a very high upper bound. You can also find a right tail by subtracting the corresponding left-tail area from 1. Standardizing the cutoff and consulting Table A provides a useful check.
Check Your Understanding
For each question, identify whether the event is a left or right tail and write the calculator setup before calculating.
- A fictional greenhouse models flower height as \(N(36,5)\), in centimeters. Write a normalcdf command for the proportion of flowers shorter than 31 centimeters.
- A fictional sensor’s response time follows \(N(120,15)\), in milliseconds. Write a direct normalcdf command for the proportion of response times greater than 150 milliseconds.
- For \(X\sim N(80,10)\), explain how to find \(P(X\geq90)\) using a complement of a left-tail area.
- A student enters \(\operatorname{normalcdf}( -1E99,42,38,2)\) to find the proportion of values above 42. Identify the error and write the corrected setup.
- For a continuous normal variable, explain why \(P(X<25)\) and \(P(X\leq25)\) have the same value.