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Worked solution

Locate the mean midway between equal normal tails

X∼N(μ,σ2)X\sim N(\mu,\sigma^2) and P(X<85)=P(X>101)=0.0548P(X<85)=P(X>101)=0.0548. Find μ and σ.

[2009/MI/BT/II/9(a)(modified)]

CJC · Discussion · Question 5

Answer

μ=93\mu=93, σ≈5.00\sigma\approx5.00 (3 s.f.).

Use equal-tail symmetry

The normal curve is symmetric about its mean. Equal lower and upper tail areas therefore give μ=(85+101)/2=93\mu=(85+101)/2=93.

How did this part go?

Standardise the lower cutoff

P(X<85)=0.0548P(X<85)=0.0548 gives P(Z<(85−93)/σ)=0.0548P(Z<(85-93)/\sigma)=0.0548. Hence (85−93)/σ=−1.599993613(85-93)/\sigma=-1.599993613, so σ≈5.00 to 3 significant figures.

How did this part go?

Compare your sketch

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A normal curve symmetric about μ=93 has equal tail areas 0.0548 below 85 and above 101. Both cutoffs are 8 units from the mean.
A normal curve symmetric about μ=93 has equal tail areas 0.0548 below 85 and above 101. Both cutoffs are 8 units from the mean.
A normal curve symmetric about μ=93 has equal tail areas 0.0548 below 85 and above 101. Both cutoffs are 8 units from the mean.

Common mistakes

The labelled tail areas are probabilities, not density heights.

Solution notes

Complete ask, all source symmetry/quantile steps and rounding retained. Equal-tail curve, shaded tails, both 0.0548 labels,85/μ/101 axis labels and mean symmetry line reconstructed in accessible SVG. No marks printed.

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