Solution — 2024 Question 6 Normal Distribution
Let be the r.v. of the height of a randomly chosen tree
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$̲ X \sim \mathrm{N} \left( \mu , \sigma ^2 \right)
\begin{align*} \mathrm{P}\left(\mu - 1.5 < X < \mu + 1.5\right) &= 0.7 \\ \mathrm{P}\left(\frac{\mu - 1.5 - \mu }{\sigma } < Z < \frac{\mu + 1.5 - \mu }{\sigma }\right) &= 0.7 \\ \mathrm{P}\left(-\frac{1.5}{\sigma } < Z < \frac{1.5}{\sigma }\right) &= 0.7 \\ \frac{1.5}{\sigma } &= 1.0364 \\ 1.0364\sigma &= 1.5 \\ \sigma &= 1.4473 \\ &= 1.45\; \QED \text{ (to 3 s.f.)} \end{align*}$$ ParseError: {align*} can be used only in display mode.
[apply standardization and inverse norm to find mu/sigma]
\begin{align*}
\mathrm{P}\left(X \leq 17.3\right) &= 0.9
\\ \mathrm{P}\left(Z \leq \frac{17.3 - \mu }{1.4473}\right) &= 0.9
\\ \frac{17.3 - \mu }{1.4473} &= 1.2816
\\ 17.3 - \mu &= 1.8548
\\ \mu &= 15.445
\\ &= 15.4\; \QED \text{ (to 3 s.f.)}
\end{align*}$$
[apply standardization and inverse norm to find mu/sigma]
Using invNorm (center) for the GC,