Approximating a Binomial Distribution with a Normal Curve
A binomial random variable counts successes in a fixed number of trials. As you saw in Shape of Binomial Distributions, the shape of its distribution depends on both \(n\) and \(p\). When the binomial distribution is sufficiently balanced, a normal curve can provide a useful approximation to its probabilities. This can make calculations easier than adding many separate binomial probabilities.
The approximation is not appropriate for every binomial setting. First check that the expected numbers of successes and failures are both large enough. Then use a normal distribution with the same mean and standard deviation as the binomial distribution. Because binomial counts are whole numbers and the normal curve is continuous, adjust the event’s boundaries by half a unit before calculating its area.
Check the Large Counts Condition First
The Large Counts condition checks whether the binomial model has enough expected successes and expected failures for a normal approximation to be reasonable. If \(X\sim B(n,p)\), the expected number of successes is \(np\), and the expected number of failures is \(n(1-p)\).
The Large Counts condition is a check on shape: it helps ensure that the binomial distribution is not too skewed for a normal curve to approximate it well. It is separate from the BINS conditions, which concern whether the chance process can be modeled as binomial in the first place. Passing the Large Counts condition does not mean an approximation will be exact.
For a binomial variable that passes the check, use the mean and standard deviation established in the earlier tutorials on binomial means and standard deviations:
The normal model is centered at the binomial mean and has the same standard deviation. In notation, we can write \(Y\sim N(np,\sqrt{np(1-p)})\), where \(Y\) represents the continuous normal variable used for the approximation. Here the second parameter is the standard deviation.
Why a Continuity Correction Is Needed
A binomial variable can only take whole-number values, such as 0, 1, 2, and so on. A normal variable can take any value along a continuous scale. To estimate a binomial probability with a normal area, include the half-unit interval around each whole-number count. This adjustment is called a continuity correction.
For example, the binomial count \(X=8\) is represented by the normal interval from 7.5 to 8.5. The count 8 is the whole-number value at the center of that interval. For a tail or a range, adjust the boundary in the direction that includes the requested integer counts.
| Binomial event | Continuity-corrected normal event |
|---|---|
| \(X\leq k\) | \(Y<k+0.5\) |
| \(X<k\) | \(Y<k-0.5\) |
| \(X\geq k\) | \(Y>k-0.5\) |
| \(X>k\) | \(Y>k+0.5\) |
| \(a\leq X\leq b\) | \(a-0.5<Y<b+0.5\) |
| \(X=k\) | \(k-0.5<Y<k+0.5\) |
Since a normal distribution is continuous, including or excluding a single endpoint does not change its area. The important part is choosing the correct half-unit boundaries. In particular, “at least \(k\)” includes \(k\), so its normal approximation begins at \(k-0.5\), not \(k+0.5\).
A Reliable Calculation Process
State what \(X\) counts and identify \(n\) and \(p\). Check the BINS conditions for the chance process.
Calculate \(np\) and \(n(1-p)\). Continue only if both are at least 10.
Use \(\mu=np\) and \(\sigma=\sqrt{np(1-p)}\). Apply the continuity correction to the event’s integer boundary or boundaries.
Use a normal cumulative probability calculation and explain that the result is an approximate probability for the specified binomial event.
Worked Example: At Least 60 Successful Shots
Worked Example: At Least 60 Successful Shots
A player makes each practice shot with probability 0.5. Suppose the player takes 100 independent shots. Let \(X\) be the number of shots made. Estimate the probability of making at least 60 shots.
State. The binomial model is \(X\sim B(100,0.5)\), where one trial is one shot and success means making the shot. The event of interest is \(X\geq60\).
Plan. Each shot has two outcomes, the number of shots is fixed at 100, the shots are independent under the stated assumption, and the probability of success is constant at 0.5. These are the BINS conditions. For the normal approximation, check \(np=100(0.5)=50\) and \(n(1-p)=100(0.5)=50\). Both are at least 10, so the Large Counts condition is satisfied.
Do. The normal model has mean and standard deviation:
The event \(X\geq60\) includes the count 60 and all higher counts. With continuity correction, use the area to the right of \(59.5\):
The corrected boundary is \(1.9\) standard deviations above the mean because \((59.5-50)/5=1.9\). The upper-tail normal area is approximately 0.0287, rounded to four decimal places.
Conclude. Under the binomial model, the estimated probability that the player makes at least 60 of 100 shots is about 0.0287. This is an approximation based on a normal curve, not the exact binomial probability.
Worked Example: At Most 12 Successful Checks
Worked Example: At Most 12 Successful Checks
A device passes an individual check with probability 0.2. Suppose it is checked 80 times independently, with the same pass probability each time. Let \(X\) count the number of passes. Estimate the probability of at most 12 passes.
State. The model is \(X\sim B(80,0.2)\), and the event is \(X\leq12\).
Plan. A check has two outcomes, there are a fixed 80 checks, the outcomes are assumed independent, and the pass probability stays at 0.2. Thus BINS is satisfied under the stated assumptions. For the Large Counts check, \(np=80(0.2)=16\) and \(n(1-p)=80(0.8)=64\); both are at least 10.
Do. The matching normal model has mean 16 and standard deviation:
“At most 12” includes 12, so the continuity-corrected upper boundary is \(12.5\). Standardizing that boundary gives:
Therefore, using the normal cumulative distribution:
The same calculation can be checked by evaluating the standard normal area to the left of \(z\approx-0.9783\), which is approximately 0.1640, rounded to four decimal places.
Conclude. If the device’s pass outcomes follow the stated binomial model, the estimated probability of at most 12 passes in 80 checks is about 0.1640.
Worked Example: A Probability Between Two Counts
Worked Example: A Probability Between Two Counts
A sensor correctly identifies a test signal with probability 0.5 on each independent attempt. It is tried 100 times. Let \(X\) be the number of correct identifications. Estimate the probability of getting from 46 through 54 correct identifications, inclusive.
State. The model is \(X\sim B(100,0.5)\), and the requested event is \(46\leq X\leq54\).
Plan. Each attempt has two outcomes, the fixed number of attempts is 100, outcomes are assumed independent, and the probability of a correct identification is constant at 0.5. The BINS conditions are satisfied under these assumptions. Also, \(np=50\) and \(n(1-p)=50\), so the Large Counts condition is satisfied.
Do. The mean is \(100(0.5)=50\), and the standard deviation is \(\sqrt{100(0.5)(0.5)}=5\). Since both endpoints of the requested range are included, the continuity-corrected range is from \(45.5\) to \(54.5\):
The standardized boundaries are:
Thus the estimated probability is the area between \(-0.9\) and \(0.9\):
As a check, the standard normal area is \(\Phi(0.9)-\Phi(-0.9)\), approximately \(0.8159-0.1841=0.6319\), rounded to four decimal places.
Conclude. Under the stated model, the probability of 46 through 54 correct identifications is approximately 0.6319.
Approximating the Probability of Exactly One Count
The continuity correction also applies when the question asks for one exact binomial count. The event \(X=k\) is represented by the normal area from \(k-0.5\) to \(k+0.5\). For example, for \(X\sim B(100,0.5)\), the mean is 50 and the standard deviation is 5. The probability of exactly 50 successes can be approximated by:
The interval is only one unit wide, but its area estimates the probability attached to the single integer count 50. Without the half-unit adjustment, a continuous distribution assigns probability 0 to the single point \(Y=50\), which would not approximate the binomial probability.
Common Mistakes and AP Exam Tips
- Skipping one of the Large Counts checks. A complete check states both \(np\) and \(n(1-p)\) and confirms that each is at least 10. If either is below 10, do not proceed with this normal approximation.
- Using the wrong standard deviation. For a binomial count, use \(\sqrt{np(1-p)}\), not \(np\), and not \(\sqrt{np}\) unless the remaining factor happens to equal 1.
- Forgetting the continuity correction. For “at least 60,” the normal boundary is 59.5. Starting at 60.5 would leave out the count 60 and approximate a different event.
- Confusing “at most” and “less than.” “At most 12” means \(X\leq12\), so the corrected boundary is 12.5. “Less than 12” means \(X\leq11\), so its corrected boundary is 11.5.
- Adjusting range endpoints in the wrong direction. For an inclusive range \(a\leq X\leq b\), use \(a-0.5\) and \(b+0.5\). This includes the full half-unit intervals around both endpoint counts.
- Calling the result exact. A normal-curve calculation is an approximation to the binomial probability. Say “approximately” and interpret the result as an estimated probability under the model.
For full-credit communication, define \(X\), state the binomial model, check both Large Counts values, show the mean and standard deviation, apply continuity correction, and report the resulting area in context. If a Large Counts value is too small, explain that the normal approximation is not recommended; use the binomial methods from the earlier tutorials instead.
Key Takeaway
A normal curve can approximate a binomial probability when the expected counts of both successes and failures are at least 10. Match the normal model’s mean and standard deviation to the binomial values, use a continuity correction to account for whole-number counts, and describe the result as an approximation in context.
Check Your Understanding
For each question, decide whether the Large Counts condition is satisfied before setting up a normal approximation.
- For \(X\sim B(60,0.3)\), calculate \(np\) and \(n(1-p)\). Is a normal approximation appropriate by the Large Counts condition?
- If \(X\sim B(100,0.4)\), what continuity-corrected boundary represents \(P(X\geq45)\)?
- For a binomial variable \(X\), write the continuity-corrected normal event for \(P(X\leq18)\).
- If \(X\sim B(100,0.5)\), write the continuity-corrected normal interval for estimating \(P(48\leq X\leq55)\).
- For \(X\sim B(30,0.1)\), check both Large Counts values. Should you use a normal approximation? Explain.