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Worked solution

Practise standard-normal intervals and two tails

For Z∼N(0,1)Z\sim N(0,1), find (i) P(−2.56<Z≤0.134)P(-2.56<Z\le 0.134), (ii) P(Z>1.863)P(Z>1.863) and (iii) P(∣Z∣>0.754)P(|Z|>0.754).

CJC · Lecture · Self-Practice 3

(i) An interval across zero

Use normalcdf(−2.56,0.134,0,1) to obtain approximately 0.5480651046, so the answer is 0.548\boxed{0.548} (3 s.f.). The ≤ endpoint has no effect on this continuous probability.

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(ii) A right tail

Use normalcdf(1.863,10^99,0,1) to obtain approximately 0.0312311331, giving 0.0312\boxed{0.0312} (3 s.f.).

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(iii) Outside a central interval

The printed prompt has absolute-value bars: ∣Z∣>0.754|Z|>0.754 means Z<−0.754Z<-0.754 or Z>0.754Z>0.754. By symmetry,

P(∣Z∣>0.754)=2P(Z>0.754)=2[1−Φ(0.754)]≈0.4508492217.\begin{aligned}&P(|Z|>0.754)\\ & =2 P(Z>0.754)\\ & =2\left[1-\Phi(0.754)\right]\approx 0.4508492217.\end{aligned}

Thus the answer is 0.451\boxed{0.451} (3 s.f.). Equivalently subtract the interval probability between −0.754 and 0.754 from 1. The printed source gives the three short answers; the steps here explain them.

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Common mistakes

Retain the absolute-value bars in (iii): the question uses both tails. A single upper tail at 0.754 is about 0.225, which is only half the required probability.

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