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Worked solution

Find a sample size for a central normal interval

For an independent random sample X1,…,XnX_1,\ldots,X_n from X∼N(μ,1)X\sim N(\mu,1):

(i) State the distribution of Xˉ\bar X.

(ii) Find the sample size required to ensure that the probability that Xˉ\bar X is within 0.1 of μ\mu is greater than 0.95.

CJC · Lecture · Example 6

(i) Standardise the sample mean

Xˉ∼N(μ,1/n)\bar X\sim N(\mu,1/n) exactly and Z=(Xˉ−μ)/(1/n)∼N(0,1)Z=(\bar X-\mu)/(1/\sqrt n)\sim N(0,1). Thus the event is −0.1n<Z<0.1n-0.1\sqrt n<Z<0.1\sqrt n.

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(ii) Method 1: inverse normal

A central area 0.95 leaves 0.025 in each tail. The source’s DISTR → invNorm, area .95, μ 0, σ 1, CENTER gives ±1.959963986\pm1.959963986. To make the probability strictly larger, 0.1n>1.9599639860.1\sqrt n>1.959963986, hence n>384.1458821n>384.1458821. The least integer is 385. The source rounds the intermediate bound to 384.14; it leads to the same integer.

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(ii) Method 2: calculator table

Enter Y1=normalcdf⁡(−0.1X,0.1X,0,1)Y_1=\operatorname{normalcdf}(-0.1\sqrt X,0.1\sqrt X,0,1) and inspect integer X=nX=n. The printed table is:

nCentral probability as printed
3840.95
3850.9503
3860.9505

At 384 the unrounded value is 0.9499564788, below .95 despite its display rounding. At 385 it is 0.9502540093. Monotonic widening of the central interval proves 385 is minimal.

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Compare your sketch

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Standard normal curve centred at zero. At the boundary, cutoffs −0.1√n and 0.1√n enclose area 0.95, leaving (1−0.95)/2=0.025 in each tail. The requested probability greater than 0.95 requires wider cutoffs; n385 is the least integer.
Standard normal curve centred at zero. At the boundary, cutoffs −0.1√n and 0.1√n enclose area 0.95, leaving (1−0.95)/2=0.025 in each tail. The requested probability greater than 0.95 requires wider cutoffs; n385 is the least integer.
Standard normal curve centred at zero. At the boundary, cutoffs −0.1√n and 0.1√n enclose area 0.95, leaving (1−0.95)/2=0.025 in each tail. The requested probability greater than 0.95 requires wider cutoffs; n385 is the least integer.

Common mistakes

The source sketch shows the equality boundary: central area .95 and tails .025. The requested strict inequality requires an interval wider than that boundary, not acceptance of the rounded .95 at n384.

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