S2 3.2 - Normal approximations

S2 3.2 - Normal approximations

A smooth Normal curve can estimate probabilities for a count, but its mean, spread and boundaries must match the discrete model. You will approximate binomial and Poisson distributions, apply the continuity correction and judge when the approximation is sensible.

Matching binomial moments

How can a continuous curve estimate the probability of an integer count? A binomial histogram can look increasingly like a bell as its mass spreads over many values away from both endpoints. We first match the centre and spread; later we match the event's boundaries.

For X∼B(n,p)X\sim\mathrm B(n,p), the mean is npnp and variance is np(1−p)np(1-p). A matching Normal variable therefore has the same two moments.

Normal approximation to a binomial

When the binomial distribution is sufficiently spread out and not strongly skewed,

X∼B(n,p)is approximated byY∼N(np,np(1−p)).X\sim\mathrm B(n,p)\quad\text{is approximated by}\quad Y\sim\mathrm N\bigl(np,np(1-p)\bigr).

The second Normal parameter is the variance. Standardise using its square root. The binomial moments are in the formula booklet; moment matching gives this approximation.

The approximation does not make XX continuous. XX remains the original count and YY is a separate continuous model used to estimate its probabilities.

Worked example 1: building the approximating model

Let X∼B(200,0.4)X\sim\mathrm B(200,0.4). State a matching Normal distribution and its standard deviation.

Plan. Calculate the binomial mean and variance before writing Normal notation.

μ=200(0.4)=80,σ2=200(0.4)(0.6)=48.\mu=200(0.4)=80,\qquad \sigma^2=200(0.4)(0.6)=48.

Answer. Y∼N(80,48)Y\sim\mathrm N(80,48), with standard deviation 48≈6.93\sqrt{48}\approx6.93.

Check. Both expected successes, 80, and expected failures, 120, are substantial, and the standard deviation is much smaller than either distance to an endpoint.

Change only the probability from 0.4 to 0.6. The centre moves, while the variance stays the same because the two factors exchange places.

Worked example 2: exchanging success and failure

Let X∼B(200,0.6)X\sim\mathrm B(200,0.6). State a matching Normal distribution and its standard deviation.

Plan. Recalculate npnp and np(1−p)np(1-p); do not carry over the old mean.

μ=200(0.6)=120,σ2=200(0.6)(0.4)=48.\mu=200(0.6)=120,\qquad \sigma^2=200(0.6)(0.4)=48.

Answer. Y∼N(120,48)Y\sim\mathrm N(120,48), with standard deviation 48≈6.93\sqrt{48}\approx6.93.

Check. If successes in the first example become failures, the count changes from XX to 200−X200-X. That changes the mean to 120 but leaves the spread unchanged.

Writing N(120,48)\mathrm N(120,\sqrt{48}) would mistake standard deviation for the second parameter.

Matching Poisson moments

A Poisson count has equal mean and variance. As its mean grows, the distribution spreads over more integer values and becomes less skewed, allowing a Normal approximation.

Normal approximation to a Poisson

For a sufficiently large mean,

X∼Po⁡(λ)is approximated byY∼N(λ,λ),σ=λ.X\sim\operatorname{Po}(\lambda)\quad\text{is approximated by}\quad Y\sim\mathrm N(\lambda,\lambda), \qquad \sigma=\sqrt{\lambda}.

The Poisson moments are in the formula booklet. This approximation follows by matching them; it is not a claim that the distributions are identical.

If a rate is given per minute but the count covers several minutes, scale the Poisson mean first. Both the Normal mean and variance then use that scaled value.

Worked example 3: approximating a count over time

Calls arrive independently at a constant mean rate of 16 per minute. State a matching Normal model for the number in four minutes.

Plan. Find the Poisson mean for the whole observation interval, then match its moments.

X∼Po⁡(4×16)=Po⁡(64),E(X)=Var⁡(X)=64.X\sim\operatorname{Po}(4\times16)=\operatorname{Po}(64), \qquad \mathrm E(X)=\operatorname{Var}(X)=64.

Answer. Y∼N(64,64)Y\sim\mathrm N(64,64), with standard deviation 8.

Check. The four-minute mean is four times the one-minute mean; standard deviation is only twice the one-minute standard deviation, since 64=216\sqrt{64}=2\sqrt{16}.

Change only the observation interval from four minutes to nine. Variance scales with time, but standard deviation does not scale directly with time.

Worked example 4: scaling the interval

For the same call process, state a matching Normal model for the number in nine minutes.

Plan. Use the nine-minute Poisson mean in both Normal parameters.

X∼Po⁡(9×16)=Po⁡(144),Y∼N(144,144).X\sim\operatorname{Po}(9\times16)=\operatorname{Po}(144), \qquad Y\sim\mathrm N(144,144).

Answer. Mean 144, variance 144 and standard deviation 12.

Check. Relative to four minutes, the variance scales by 9/49/4, while the standard deviation scales by 9/4=3/2\sqrt{9/4}=3/2.

The Poisson model still requires a reasonable constant-rate, independent-event description. A Normal curve cannot repair an unsuitable original count model.

Continuity correction

An integer count of 64 represents the whole histogram bar from 63.5 to 64.5. A continuity correction asks the Normal curve to cover that bar, rather than stopping through its centre.

For example, X≤70X\le70 includes the bar centred at 70 and needs the boundary 70.5. In contrast, X<70X<70 means X≤69X\le69 and needs 69.5. Translate the integer event first, then move to half-integers.

Discrete eventMatching continuous event
X≤kX\le kY<k+0.5Y<k+0.5
X<kX<kY<k−0.5Y<k-0.5
X≥kX\ge kY>k−0.5Y>k-0.5
X>kX>kY>k+0.5Y>k+0.5
X=kX=kk−0.5<Y<k+0.5k-0.5<Y<k+0.5

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