Solution — 2017 Question 6 Normal Distribution

Let $X$ be the height, in m, of a randomly chosen adult male.

$X \sim N (μ, σ^{2})$

\begin{matrix} P (X < 1.6) & = 0.2 \\ P (Z < \frac{1.6 - μ}{σ}) & = 0.2 \\ \frac{1.6 - μ}{σ} & = - 0.841 62 \end{matrix}

apply standardization and inverse norm to find mu/sigma

\begin{matrix} P (X > 1.75) & = 0.3 \\ P (Z > \frac{1.75 - μ}{σ}) & = 0.3 \\ \frac{1.75 - μ}{σ} & = 0.524 40 \end{matrix}

apply standardization and inverse norm to find mu/sigma

Cross-multiplying and rearranging,

\begin{matrix} μ - 0.8416 σ & = 1.6 & (1) \end{matrix}

\begin{matrix} μ + 0.5244 σ & = 1.75 & (2) \end{matrix}

Solving (1) and (2) with a GC, $μ = 1.6924, σ = 0.109 81$

\begin{matrix} Mean & = 1.6924 \\ = 1.69 ∎ (to 3 s.f.) \end{matrix}

\begin{matrix} Variance & = 0.109 81^{2} \\ = 0.0121 ∎ (to 3 s.f.) \end{matrix}