Decide whether an interval is central

Let $H\sim N(169,8^2)$. The lower bound 160 is fixed. First find

$P(H<160)\approx 0.1302945171.$

Because $d>160$ and $P(160<H<d)=0.2$,

$\begin{aligned}&P(H<d)\\ & =P(H<160)\\ &{}+0.2\approx 0.3302945171.\end{aligned}$

Use normalcdf(−10^99,160,169,8) to obtain the first area, keep that full result and add 0.2. Then use invNorm(0.3302945171,169,8,LEFT) to obtain $d\approx 165.4871995$ cm, or $\boxed{165\text{ cm}}$ (3 s.f.). The decimal 165.487 cm is included to show the interval’s position; the source does not prescribe a rounding instruction for this prompt.

No. A central interval with lower bound 160 and mean 169 would have upper bound 178 by symmetry. That interval’s area is not 0.2. In the required solution, both 160 and d≈165.487 lie below the mean 169, so the interval is asymmetric.

Use the sketch to check the given area, rather than assuming that any interval probability is central. The source prints “169 am”; this is a unit typo, interpreted as 169 cm in its height context. The printed prompt has no worked answer; the cumulative-area solution here is authored to complete the question.