Conditional probability and the Normal distribution – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Mark scheme: 1 mark for each of parts a) and b), 2 marks for part c) (1 for a correct comparison, 1 for the conclusion).

   1. Use the gym column total out of 80. \(\dfrac{42}{80}=0.525\). So \(P(\text{gym})=0.525\).
   2. Given gym, divide by the 42 gym users. \(\dfrac{24}{42}=\dfrac47=0.571\). So \(P(\text{Year 13}\mid\text{gym})\approx0.571\).
   3. Compare the joint probability with the product of the two probabilities. \(P(\text{gym})P(\text{Y12})=0.525\times0.5=0.2625\) but \(P(\text{gym}\cap\text{Y12})=\dfrac{18}{80}=0.225\). These differ, so the events are not independent. (Alternatively \(P(\text{gym}\mid\text{Y12})=0.45\ne0.525\).) Year 12 students are less likely to go to the gym.

2. Answer to Question 2 (5 marks)

   Mark scheme: a) 1 mark for the method (two paths added), 1 mark for 0.08; b) 1 mark for the conditional probability method, 1 mark for 0.75; c) 1 mark.

   1. Add the two paths bus then late and bicycle then late. \(P(L)=0.6\times0.1+0.4\times0.05=0.06+0.02=0.08\). So Tom is late on 8% of days.
   2. Divide the path bus then late by the total probability of being late. \(P(\text{bus}\mid L)=\dfrac{0.06}{0.08}=0.75\). So when Tom is late, he took the bus three times out of four.
   3. He is not late with probability \(1-0.08=0.92\) each day, and the days are independent. \(0.92^5=0.659\). So he is on time all week with probability about 0.659.

3. Answer to Question 3 (4 marks)

   Mark scheme: a) 1 mark, b) 1 mark, c) 1 mark for \(z=-0.8416\) and 1 mark for \(k\).

   1. Standardise: \(z=\dfrac{70-64}{9}=0.667\), then take the upper tail. \(P(X\gt 70)=P(Z\gt 0.667)=0.2525\). So about 25% of tomatoes weigh more than 70 g.
   2. Standardise: \(\dfrac{55-64}9=-1\) and \(\dfrac{73-64}9=1\). \(55\) and \(73\) are one standard deviation either side of the mean, so \(P=P(-1\lt Z\lt 1)=0.683\). So about 68% of tomatoes weigh between 55 g and 73 g.
   3. Use the inverse Normal with area 0.2 and rearrange \(k=\mu+z\sigma\). \(z=-0.8416\), so \(k=64-0.8416\times9=56.4\) g. So 20% of tomatoes weigh less than 56.4 g.

4. Answer to Question 4 (4 marks)

   Mark scheme: a) 1 mark per equation; b) 1 mark for \(\sigma\), 1 mark for \(\mu\).

   1. Change each statement into an area to the left: \(P(X\gt 30)=0.4\) means \(P(X\lt 30)=0.6\). Then use the inverse Normal. \(P(X\lt 20)=0.15\): \(\dfrac{20-\mu}{\sigma}=-1.0364\). \(P(X\lt 30)=0.6\): \(\dfrac{30-\mu}{\sigma}=0.2533\). These are two linear equations in \(\mu\) and \(\sigma\).
   2. Subtract the equations to eliminate \(\mu\), then substitute back. Subtracting: \(\dfrac{10}{\sigma}=1.2897\), so \(\sigma=7.75\). Then \(\mu=20+1.0364\times7.75=28.0\). So \(\mu\approx28.0\) and \(\sigma\approx7.75\).

5. Answer to Question 5 (3 marks)

   Method: a binomial \(B(n,p)\) with large \(n\) and \(p\) not too close to 0 or 1 can be approximated by \(N(np,\,np(1-p))\). Because \(X\) takes whole numbers, the value 40 corresponds to the interval from 39.5 to 40.5, so \(X\le40\) becomes \(Y\lt 40.5\).

   \(np=45\) and \(np(1-p)=31.5\), so \(Y\sim N(45,31.5)\) (1). \(P(X\le40)\approx P(Y\lt 40.5)\) (1) \(=P\left(Z\lt \dfrac{40.5-45}{\sqrt{31.5}}\right)=P(Z\lt -0.802)=0.211\) (1). (The exact value is 0.2126.)

   Answer: \(P(X\le40)\approx0.211\), which is very close to the exact binomial value.

Total : 20 marks