2013 H2 Mathematics Paper 2 Question 6

The Normal Distribution

Answers

μ=1.29a{\mu = 1.29a}

Full solutions

X∼N(μ,σ2)Z=X−μσ∼N(0,1)\begin{gather*} X \sim \textrm{N}(\mu, \sigma^2) \\ Z = \frac{X-\mu}{\sigma} \sim \textrm{N}(0, 1) \end{gather*}
P(X<2a)=0.95P(Z<2a−μσ)=0.95\begin{align*} \textrm{P}(X < 2a) &= 0.95 \\ \textrm{P}\left(Z < \frac{2a-\mu}{\sigma}\right) &= 0.95 \\ \end{align*}
2a−μσ=1.6449\frac{2a-\mu}{\sigma} = 1.6449
σ=2a−μ1.6449\begin{equation} \sigma = \frac{2a-\mu}{1.6449} \end{equation}
P(X<a)=0.25P(Z<a−μσ)=0.25\begin{align*} \textrm{P}(X < a) &= 0.25 \\ \textrm{P}\left(Z < \frac{a-\mu}{\sigma}\right) &= 0.25 \\ \end{align*}
a−μσ=−0.67449\frac{a-\mu}{\sigma} = -0.67449
σ=a−μ−0.67449\begin{equation} \sigma = \frac{a-\mu}{-0.67449} \end{equation}
Equating (1){(1)} and (2){(2)},
2a−μ1.6449=a−μ−0.67449−1.3490a+0.67449μ=1.6449a−1.6449μ\begin{gather*} \frac{2a-\mu}{1.6449} = \frac{a-\mu}{-0.67449} \\ -1.3490a + 0.67449\mu = 1.6449a -1.6449\mu \\ \end{gather*}
2.3193μ=2.9938aμ=1.29a (3 sf)  ■\begin{gather*} 2.3193\mu = 2.9938a \\ \mu = 1.29a \textrm{ (3 sf)} \; \blacksquare \end{gather*}