Solution — 2024 Question 6 Normal Distribution

(a)

Let $X$ be the r.v. of the height of a randomly chosen tree

$X \sim N (μ, σ^{2})$

\begin{matrix} P (μ - 1.5 < X < μ + 1.5) & = 0.7 \\ P (\frac{μ - 1.5 - μ}{σ} < Z < \frac{μ + 1.5 - μ}{σ}) & = 0.7 \\ P (- \frac{1.5}{σ} < Z < \frac{1.5}{σ}) & = 0.7 \\ \frac{1.5}{σ} & = 1.0364 \\ 1.0364 σ & = 1.5 \\ σ & = 1.4473 \\ = 1.45 ∎ (to 3 s.f.) \end{matrix}

apply standardization and inverse norm to find mu/sigma

\begin{matrix} P (X \leq 17.3) & = 0.9 \\ P (Z \leq \frac{17.3 - μ}{1.4473}) & = 0.9 \\ \frac{17.3 - μ}{1.4473} & = 1.2816 \\ 17.3 - μ & = 1.8548 \\ μ & = 15.445 \\ = 15.4 ∎ (to 3 s.f.) \end{matrix}

apply standardization and inverse norm to find mu/sigma

(b)

\begin{matrix} P (a < X < b) & = 0.98 \end{matrix}

Using invNorm (center) for the GC,

$a = 12.1 ∎ b = 18.8 ∎$