From an Upper-Tail Percentage to a Sample-Mean Cutoff
In “Using normalcdf for Probabilities About \(\bar{x}\),” you used a cutoff to find the probability of getting sample means above it. Now we reverse that question: given an upper-tail percentage, what sample-mean value separates that percentage from the rest of the sampling distribution?
For example, “find the value cutting off the top 10%” asks for a cutoff with 10% of the sampling distribution above it and 90% below it. The cutoff is the 90th percentile. Because invNorm uses the area to the left of a value, the calculator input is 0.90, not 0.10.
The random variable is \(\bar{x}\), the mean of a sample of a specified size. So the mean and standard deviation entered into invNorm must describe the sampling distribution of sample means. Its center is \(\mu\), and, when the independence conditions are satisfied, its standard deviation is \(\sigma/\sqrt{n}\), as covered in “Standard Deviation of the Sample Mean.”
On a TI-84, invNorm takes the cumulative area to the left, the distribution mean, and the distribution standard deviation, in that order. Here, \(c\) is the cutoff for sample means. The second input is \(\mu\), and the third is \(\sigma/\sqrt{n}\)—not the standard deviation \(\sigma\) of individual observations.
The cutoff is a value of \(\bar{x}\), so it has the same units as the original observations. For example, if the observations measure minutes, the cutoff is a sample mean measured in minutes. It is not the value that 10% of individual observations exceed; it describes the distribution of means from repeated samples of size \(n\).
A Four-Step Method
Use this method whenever a question gives a percentage and asks for the corresponding sample-mean cutoff. First translate the percentage into a cumulative area to the left, then justify the sampling distribution and its calculator inputs.
Define \(\bar{x}\) in context and identify the requested cutoff. For the top 10%, the event above the cutoff has probability 0.10.
Check the sampling assumptions and whether the sampling distribution is exactly or approximately Normal. Find its mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\).
Use invNorm with area 0.90, mean \(\mu\), and standard deviation \(\sigma/\sqrt{n}\). Report the resulting cutoff with appropriate units and rounding.
Interpret the cutoff in context: approximately 10% of sample means from samples of the stated size exceed it, under the stated model and assumptions.
For a random sample without replacement, check the 10% condition, as explained in “The 10% Condition for Sample Means.” If observations are independent by design, state that. For the shape, a Normal population makes the sampling distribution of \(\bar{x}\) exactly Normal; for a non-Normal population, use CLT reasoning only when the population shape and sample size support an approximately Normal model, as discussed in “Central Limit Theorem Explained” and “Is \(n=30\) Large Enough for the CLT.”
Worked Example: Top 10% of Battery Sample Means
Suppose battery lifetimes follow a Normal population with mean \(\mu=52\) hours and standard deviation \(\sigma=12\) hours. Independent random samples of \(n=36\) batteries are taken. Find the sample-mean lifetime that cuts off the top 10% of the sampling distribution.
State. Let \(\bar{x}\) be the mean lifetime, in hours, for a random sample of 36 batteries. We want the cutoff \(c\) such that \(P(\bar{x}>c)=0.10\).
Plan. The population is Normal and observations are independent, so the sampling distribution of \(\bar{x}\) is exactly Normal. Its mean is 52 hours. Its standard deviation is:
Check the spread using the variance: \(\sigma^2/n=12^2/36=144/36=4\) square hours, and \(\sqrt{4}=2\) hours. Since the top 10% is to the right, the area to the left of the cutoff is 0.90.
Do. Enter the left-tail area, sampling-distribution mean, and sampling-distribution standard deviation into invNorm:
The standardized cutoff is approximately \((54.5631-52)/2=1.282\). A standard Normal value of about 1.282 has approximately 90% of its area to the left and 10% to the right, confirming the direction and size of the cutoff. Equivalently, entering the cutoff as the lower bound in normalcdf gives an upper-tail area of approximately 0.1000.
Conclude. For repeated independent samples of 36 batteries, about 10% of the sample means are expected to exceed 54.5631 hours, under the stated Normal model. This cutoff concerns sample means, not the lifetimes of individual batteries.
Why invNorm Uses 0.90, Not 0.10
The phrase “top 10%” describes a right-tail area. But invNorm asks for the cumulative area to the left of the value it returns. Since the total area is 1, the left-tail area is \(1-0.10=0.90\). This is why the correct input is 0.90.
A useful sketch can prevent a tail error: draw a bell-shaped curve, mark the cutoff to the right of the center, shade the small region to its right, and label that region 0.10. The larger region to the left is 0.90, which is the invNorm area. A cutoff for the top 10% should be above the center of a symmetric Normal distribution; a result below the center is a warning to check the area input.
The standardized 90th-percentile value is about 1.282. This can help check an answer: the cutoff should be approximately \(\mu+1.282(\sigma/\sqrt{n})\). This is a check, not a reason to use the population standard deviation in place of the standard deviation of sample means.
Worked Example: Top 10% of Mean Package Weights
In a hypothetical packaging process, package weights have population mean \(\mu=250\) grams and population standard deviation \(\sigma=16\) grams. A simple random sample of \(n=64\) packages is selected from 50,000 packages. Estimate the sample-mean weight that cuts off the top 10%. Assume the population is mildly skewed with no extreme outliers.
State. Let \(\bar{x}\) be the mean weight, in grams, of 64 sampled packages. We seek \(c\) such that \(P(\bar{x}>c)\) is approximately 0.10.
Plan. The sample is random, and \(64<0.10(50{,}000)=5{,}000\), so the 10% condition is satisfied. The population is mildly skewed without extreme outliers, and \(n=64\) is large enough for CLT reasoning to support an approximately Normal sampling distribution. The center is 250 grams. Its standard deviation is:
As a check, the variance of the sample mean is \(16^2/64=256/64=4\) square grams, whose square root is 2 grams. The left-side cumulative area for the requested cutoff is 0.90.
Do. Use invNorm with the parameters of the sampling distribution:
The standardized cutoff is approximately \((252.5631-250)/2=1.282\), consistent with the 90th percentile. Checking with normalcdf, the area to the right of 252.5631 under a Normal model with mean 250 and standard deviation 2 is approximately 0.1000.
Conclude. Using the CLT-based Normal approximation, about 10% of random samples of 64 packages are expected to have a mean weight above approximately 252.5631 grams. This is an approximate cutoff because the population is not stated to be Normal.
The Sample Size Changes the Cutoff
The requested percentile is determined not only by the population mean and standard deviation, but also by the sample size. A larger \(n\) makes the sampling distribution less spread out because \(\sigma_{\bar{x}}=\sigma/\sqrt{n}\) decreases. The top-10% cutoff therefore moves closer to \(\mu\), even though the relevant standardized location remains about 1.282 standard deviations above the center.
Here is a comparison using the same population parameters as the first example. Each row gives the standard deviation of sample means and the approximate 90th percentile. The population is Normal and sampling is independent, so the Normal model is exact for each sample size.
| Sample size \(n\) | \(\sigma_{\bar{x}}=12/\sqrt{n}\) | Top-10% cutoff |
|---|---|---|
| 9 | 4 hours | 57.1262 hours |
| 36 | 2 hours | 54.5631 hours |
| 144 | 1 hour | 53.2816 hours |
For example, when \(n=9\), the cutoff is \(52+1.28155(4)\), approximately 57.1262 hours. When \(n=144\), it is \(52+1.28155(1)\), approximately 53.2816 hours. The smaller spread for the larger sample means sample averages cluster more tightly around 52 hours.
Worked Example: Top 10% of Mean Assessment Times
Suppose assessment completion times follow a Normal population with mean \(\mu=82\) minutes and standard deviation \(\sigma=15\) minutes. Independent random samples of \(n=25\) students are taken. Find the sample-mean time that cuts off the top 10%.
State. Let \(\bar{x}\) be the mean completion time, in minutes, for a sample of 25 students. We seek the cutoff \(c\) for which \(P(\bar{x}>c)=0.10\).
Plan. The population is Normal and observations are independent, so the sampling distribution is exactly Normal. Its center is 82 minutes, and its standard deviation is:
Verify using variance: \(15^2/25=225/25=9\) square minutes, and \(\sqrt{9}=3\) minutes. Since the request is for the top 10%, use cumulative area 0.90.
Do. Apply invNorm to the sampling distribution:
As a check, the standardized cutoff is approximately \((85.8447-82)/3=1.282\). The cutoff is above the mean, as expected for a value with 90% of the distribution below it. A normalcdf check gives an upper-tail probability of approximately 0.1000.
Conclude. About 10% of independent samples of 25 students are expected to have a mean completion time above approximately 85.8447 minutes, under the stated Normal model.
Common Mistakes and AP Exam Tip
- Entering 0.10 into invNorm: invNorm needs area to the left. For a top-10% cutoff, use 0.90. An input of 0.10 finds a cutoff near the bottom of the distribution instead.
- Using \(\sigma\) instead of \(\sigma/\sqrt{n}\): The random variable is the sample mean. Its spread is the standard deviation of sample means, not the spread among individual observations.
- Skipping the Normal-model justification: State that the population is Normal for an exact model, or explain why CLT reasoning supports an approximate one. Check random selection, independence, and the 10% condition when appropriate.
- Calling the result an individual-value cutoff: Say that about 10% of sample means from samples of size \(n\) exceed the result. Do not say that 10% of individual batteries, packages, or students exceed it.
- Reporting a number without units or context: Include the quantity being averaged, the sample size, and the units of the cutoff. Make clear whether the answer is exact or approximate.
For full-credit communication, identify the random variable, convert the right-tail percentage to the left-tail area, show the calculation of \(\sigma_{\bar{x}}\), justify the Normal model, and interpret the result in context. A calculator value alone does not show that the area, mean, and standard deviation were entered correctly.
Check Your Understanding
For each situation, identify the left-tail area, check the sampling-distribution model, and find or describe the requested cutoff in context.
- A Normal population has mean 30 centimeters and standard deviation 9 centimeters. Independent samples of size 9 are taken. Find the sample-mean value cutting off the top 10%, and state its units.
- A simple random sample of 100 items is selected from 20,000. Individual measurements have mean 18 units and standard deviation 5 units, and the population is mildly skewed without extreme outliers. Estimate the cutoff for the top 10% of sample means.
- A student enters invNorm with area 0.10 to find the value cutting off the top 10%. Explain the error and give the correct area input.
- A Normal population has mean 70 minutes and standard deviation 12 minutes. Compare the top-10% cutoff for samples of size 4 with the cutoff for samples of size 36. Which is closer to the population mean, and why?
- Explain why an invNorm result for the sampling distribution of \(\bar{x}\) is not the same as the value cutting off the top 10% of individual observations.