5.5 - Approximating a binomial distribution
A binomial probability may involve many terms. Learn when a normal curve is a suitable approximation, use half-integer boundaries to match a discrete count to a continuous area, and check that the final approximation describes the intended event.
When approximation is suitable
Suppose . How can you estimate without adding binomial terms? A normal curve can approximate the distribution, but matching just the mean is not enough: its spread and the event boundary matter too.
When both the expected numbers of successes and failures are sufficiently large, the binomial bars form a roughly symmetric bell shape. Very small expected counts on either side can produce pronounced skew, so a normal curve is less suitable.
Normal approximation to a binomial distribution
For , put . Use a normal approximation when
Match the mean and variance with
The binomial mean and variance are in the formula booklet. The approximation conditions and the choice to use this normal model are not: learn them.
These are suitability conditions, not a promise of an exact result or a fixed maximum error. A large alone is insufficient if is very close to or . The underlying experiment must first satisfy the binomial assumptions.
Worked example 1: a suitable approximation
Question. For , justify a normal approximation and state its parameters.
Plan. Check both expected counts, then use the binomial mean and variance.
Answer. Both conditions hold, so use , with standard deviation .
Check. The proposed normal mean equals , and its variance equals . The second parameter is not the standard deviation.
Keep trials and change only the success probability. A large trial count does not rescue a small expected number of failures.
Worked example 2: one condition fails
Question. Is a normal approximation suitable for ?
Plan. Check both and , rather than just the larger expected count.
Answer. No: the expected number of failures is too small for the stated normal-approximation conditions.
Check. The mean is close to the maximum possible count , so a symmetric bell curve would poorly reflect the short upper side and longer lower side. This agrees with the failed numerical condition.
At equality, such as , the stated condition is not met. Give both numerical checks when asked to justify the approximation.
Watch out
Do not write . Normal notation takes the variance as its second parameter. Use only when standardising.
Test both conditions in this choice.
Continuity correction
A binomial variable puts probability at integers; a normal variable spreads it continuously. Represent each integer count by a unit-width bar centred on that integer. The bar for occupies to , so including all of count means going to the right edge, .
Continuity correction
A continuity correction replaces an integer event boundary by the appropriate half-integer edge when using a continuous normal model for a discrete count.
Write the allowed integer values first. This is safer than memorising “add ” because the direction depends on the event.
Worked example 3: including the boundary count
Question. For , write the continuity-corrected normal event approximating .
Plan. Include the whole unit bar centred at , so use its upper edge.
The allowed counts end at . Their bars extend up to , giving
Answer. Use the normal area below .
Check. The bar from to is included in full. Stopping at would include only half of the boundary bar.
Change only to . The largest allowed integer becomes , so the corrected boundary moves by a whole unit.
Worked example 4: excluding the boundary count
Question. For the same binomial model, write the continuity-corrected normal event approximating .
Plan. Translate to , then use the upper edge of the bar for .
Answer. Use the normal area below .
Check. The bar for , from to , is excluded. This matches the strict inequality in the original discrete event.
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