Sample means of discrete distributions | CAIE 9709 Probability & Statistics 2 Notes

Probability & Statistics 2

Mathematics9709

Translate between the total and the mean

For independent observations $X_1,\ldots,X_n$ , define the total $S=X_1+\cdots+X_n$ and $\bar X=S/n$ .

E(S)=n\mu,\qquad\operatorname{Var}(S)=n\sigma^2

E(\bar X)=\mu,\qquad\operatorname{Var}(\bar X)=\frac{\sigma^2}{n}

A total changes by 1 when one extra event is counted.
Dividing by n means the sample mean changes by $1/n$ .

Half a total-step becomes 1 over 2n

Correct the total, then divide by n

integer total S

95 | 95.5 | 96

half a total-step = 0.5

divide every value by 20 ↓

sample mean X̄

4.75 | 4.775 | 4.80

half a mean-step = 1/(2n) = 0.025

95.5 ÷ 20 = 4.775

Correct the integer total by one half, or the sample mean by one over two n. These are the same boundary.

For a discrete sample mean, the correction is

\pm\frac{1}{2n}

The safest method is to translate the event to the integer total, apply the usual $\pm1/2$ , then divide the corrected boundary by n.

Worked example

Correct a mean of Poisson counts

Let $X\sim\mathrm{Po}(5)$ and take a sample of 20. Approximate $P(\bar X<4.8)$ .

Since $S=20\bar X$ , the event is $S<96$ , or $S\le95$ . The corrected boundary is 95.5.

z=\frac{95.5-100}{\sqrt{100}}=-0.45

Use the lower-tail probability $\Phi(-0.45)$ .

Answer

P(\bar X<4.8)\approx0.326

Use the correction only when the lattice matters

Use it when a normal distribution approximates a mean of discrete counts, unless the question says not to.
Do not use it for a mean of genuinely continuous measurements.
If the question explicitly says a correction is not expected, follow that instruction.

Your turnIndependent discrete-mean check [5 marks]

Let $X\sim\mathrm B(10,0.4)$ . A random sample of 50 values is taken. Use a normal approximation with continuity correction to find $P(\bar X\ge4.3)$ .

Show worked answer

The total of the 50 binomial values is $S\sim\mathrm B(500,0.4)$ , with mean 200 and variance 120. The event $\bar X\ge4.3$ is $S\ge215$ , so use 214.5.

z=\frac{214.5-200}{\sqrt{120}}=1.324

Use the upper-tail probability $1-\Phi(1.324)$ .

Answer

P(\bar X\ge4.3)\approx0.0928

Common mistake

Do not use

\pm1/2

directly on a sample mean. That is half a total-step. For a mean of n counts, the equivalent correction is

\pm1/(2n)

Translate between the total and the mean

For independent observations $X_1,\ldots,X_n$ , define the total $S=X_1+\cdots+X_n$ and $\bar X=S/n$ .

E(S)=n\mu,\qquad\operatorname{Var}(S)=n\sigma^2

E(\bar X)=\mu,\qquad\operatorname{Var}(\bar X)=\frac{\sigma^2}{n}

A total changes by 1 when one extra event is counted.
Dividing by n means the sample mean changes by $1/n$ .

Half a total-step becomes 1 over 2n

Correct the total, then divide by n

integer total S

95 | 95.5 | 96

half a total-step = 0.5

divide every value by 20 ↓

sample mean X̄

4.75 | 4.775 | 4.80

half a mean-step = 1/(2n) = 0.025

95.5 ÷ 20 = 4.775

Correct the integer total by one half, or the sample mean by one over two n. These are the same boundary.

For a discrete sample mean, the correction is

\pm\frac{1}{2n}

The safest method is to translate the event to the integer total, apply the usual $\pm1/2$ , then divide the corrected boundary by n.

Worked example

Correct a mean of Poisson counts

Let $X\sim\mathrm{Po}(5)$ and take a sample of 20. Approximate $P(\bar X<4.8)$ .

Since $S=20\bar X$ , the event is $S<96$ , or $S\le95$ . The corrected boundary is 95.5.

z=\frac{95.5-100}{\sqrt{100}}=-0.45

Use the lower-tail probability $\Phi(-0.45)$ .

Answer

P(\bar X<4.8)\approx0.326

Use the correction only when the lattice matters

Use it when a normal distribution approximates a mean of discrete counts, unless the question says not to.
Do not use it for a mean of genuinely continuous measurements.
If the question explicitly says a correction is not expected, follow that instruction.

Your turnIndependent discrete-mean check [5 marks]

Let $X\sim\mathrm B(10,0.4)$ . A random sample of 50 values is taken. Use a normal approximation with continuity correction to find $P(\bar X\ge4.3)$ .

Show worked answer

The total of the 50 binomial values is $S\sim\mathrm B(500,0.4)$ , with mean 200 and variance 120. The event $\bar X\ge4.3$ is $S\ge215$ , so use 214.5.

z=\frac{214.5-200}{\sqrt{120}}=1.324

Use the upper-tail probability $1-\Phi(1.324)$ .

Answer

P(\bar X\ge4.3)\approx0.0928

Common mistake

Do not use

\pm1/2

directly on a sample mean. That is half a total-step. For a mean of n counts, the equivalent correction is

\pm1/(2n)