N2020 P2 Q6

N2020 P2 Q6

Junior College 2
10 marks
Free

In this question you should assume that \(T\), \(W\) and \(D\) follow independent normal distributions.

James leaves home to go to work at \(T\) minutes past 8 am each day, where \(T\) follows the distribution \(N(5,1.2^2)\).

  1. Sketch this distribution for the period from 8 am to 8.10 am.[2]
  2. Find the probability that, on a randomly chosen day, James leaves for work later than 8.06 am.[1]
  3. When the weather is fine, James walks to work. The time, \(W\) minutes, he takes to walk to work follows the distribution \(N(21,3^2)\). James is supposed to start work at 8.30 am. Find the probability that, on a randomly chosen day when James walks, he is late for work.[2]
  4. When the weather is not fine, James drives to work. He still leaves at \(T\) minutes past 8 am each day; the time, \(D\) minutes, he takes to drive to work follows the distribution \(N(19,6^2)\). On average, the weather is fine on \(70\%\) of mornings. One day, James is late for work. Find the probability that the weather is fine that day.[5]

Default solution

(i) Sketch a symmetric bell-shaped normal density centred at 8.05 am over 8.00–8.10 am, with spread \(1.2\) minutes.

(ii) \(P(T>6)=1-\Phi((6-5)/1.2)=0.202\).

(iii) \(T+W\sim N(26,1.2^2+3^2)\). Hence \(P(T+W>30)=1-\Phi(4/\sqrt{10.44})=0.108\).

(iv) \(T+D\sim N(24,1.2^2+6^2)\), so \(P(\text{late}\mid\text{not fine})=1-\Phi(6/\sqrt{37.44})=0.163\).

By Bayes' rule, \(P(\text{fine}\mid\text{late})=\frac{0.7(0.107864)}{0.7(0.107864)+0.3(0.163400)}=0.606\).

Answer:(ii) \(0.202\); (iii) \(0.108\); (iv) \(0.606\)

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