2020 H2 Mathematics Paper 2 Question 6

The Normal Distribution

Answers

(i)

(ii)

0.202{0.202}

(iii)

0.108{0.108}

(iv)

0.606{0.606}

Full solutions

(i)

(ii)

P(T>6)=0.202 (3 sf)  ■\begin{align*} & \textrm{P}(T > 6) \\ &= 0.202 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

T+W+…∼N(5+21,1.22+32)T+W∼N(26,10.44)\begin{align*} T + W + \ldots &\sim \textrm{N}(5 + 21, 1.2^2 + 3^2) \\ T + W &\sim \textrm{N}( 26, 10.44 ) \end{align*}
P(late∣walk)=P(T+W>30)=0.10786=0.108 (3 sf)  ■\begin{align*} & \textrm{P}(\textrm{late} \mid \textrm{walk}) \\ &= \textrm{P}(T+W > 30) \\ &= 0.10786 \\ &= 0.108 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

T+D∼N(5+19,1.22+62)T+D∼N(24,37.44)\begin{align*} T + D &\sim \textrm{N}(5 + 19, 1.2^2 + 6^2) \\ T + D &\sim \textrm{N}( 24, 37.44 ) \end{align*}
P(late∣drive)=P(T+D>30)=0.16340\begin{align*} & \textrm{P}(\textrm{late} \mid \textrm{drive}) \\ &= \textrm{P}(T+D > 30) \\ & = 0.16340 \end{align*}
P(weather fine∣late)=P(weather fine ∩ late)P(late)=0.7×0.107860.7×0.10786+0.3×0.16340=0.606 (3 sf)  ■\begin{align*} & \textrm{P}(\textrm{weather fine} \mid \textrm{late}) \\ &= \frac{\textrm{P}(\textrm{weather fine } \cap \textrm{ late})}{\textrm{P}(\textrm{late})} \\ &= \frac{0.7 \times 0.10786}{0.7 \times 0.10786 + 0.3 \times 0.16340} \\ &= 0.606 \textrm{ (3 sf)} \; \blacksquare \end{align*}