| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Percentages or proportions given |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring inverse normal lookups to form simultaneous equations, then conditional probability calculations. All steps are routine textbook procedures with no novel insight required, though the multi-part structure and conditional probability context in parts (d)-(e) elevate it slightly above the most basic normal distribution exercises. |
| Spec | 2.03d Calculate conditional probability: from first principles2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(P(L < 45) = 0.4 \Rightarrow \frac{45 - \mu}{\sigma} = -0.2533\) or \(\Rightarrow 45 - \mu = -0.2533\sigma\) (o.e.) | M1 |
| \(45 + 0.2533\sigma = \mu\) (\(\ast\)) | A1cso | A1cso for sight of \(P(L < 45) = 0.4\) (o.e.) and 0.2533 leading to given ans. [\(0.2533471...\) from calc] |
| (b) | \(P(L < 35) = 0.15 \Rightarrow \frac{35 - \mu}{\sigma} = -1.0364\) | M1 |
| e.g. \(35 + 1.0364\sigma = \mu\) | A1 | A1 for any correct equation, \(z = 1.04\) or better and correct signs |
| (c) | Solving: \(10 - 0.7831\sigma = 0\) | M1 |
| \(\sigma = 12.7697...\) awrt \(\mathbf{12.8}\) | A1 | 1st A1 for \(\sigma =\) awrt 12.8 (NB use of 1.04 gives 12.7113... so we penalise that here) |
| \(\mu = \) awrt \(\mathbf{48.2}\) | A1 | 2nd A1 for \(\mu =\) awrt 48.2 [allow 48.3 if 12.8 used in a correct eqn e.g. \(35 + 1.04 \times 12.8\) or better] |
| (d)(i) | \(P(L > 35 \mid L < 45) = \frac{P(35 < L < 45)}{P(L < 45)} = \frac{0.25}{0.15 + 0.25} = \frac{5}{8}\) (o.e.) | A1, M1, A1 |
| (d)(ii) | \(P(L < 45 \mid L > 35) = \frac{P(35 < L < 45)}{P(L > 35)} = \frac{0.25}{0.60 + 0.25} = \frac{5}{17}\) or awrt 0.294 | M1, A1 |
| (e) | Prob. of a yellow stick from Hei is \(\frac{3}{5}\) which is > prob. of \(\frac{2}{17}\) for Tang So more likely to be Hei | B1ft, dB1ft |
| **(a)** | $P(L < 45) = 0.4 \Rightarrow \frac{45 - \mu}{\sigma} = -0.2533$ or $\Rightarrow 45 - \mu = -0.2533\sigma$ (o.e.) | M1 | M1 for attempting to standardise with 45, $\mu$ and $\sigma$ Allow $\pm$ and allow $z =$ awrt 0.25 |
| | $45 + 0.2533\sigma = \mu$ ($\ast$) | A1cso | A1cso for sight of $P(L < 45) = 0.4$ (o.e.) and 0.2533 leading to given ans. [$0.2533471...$ from calc] |
| **(b)** | $P(L < 35) = 0.15 \Rightarrow \frac{35 - \mu}{\sigma} = -1.0364$ | M1 | M1 for standardising with 35 $\mu$ and $\sigma$ and setting equal to a $z$ value with $1 < |z| < 1.05$ |
| | e.g. $35 + 1.0364\sigma = \mu$ | A1 | A1 for any correct equation, $z = 1.04$ or better and correct signs |
| **(c)** | Solving: $10 - 0.7831\sigma = 0$ | M1 | M1 for solving their 2 linear equations in $\mu$ and $\sigma$ – reducing to an equation in 1 variable |
| | $\sigma = 12.7697...$ awrt $\mathbf{12.8}$ | A1 | 1st A1 for $\sigma =$ awrt 12.8 (NB use of 1.04 gives 12.7113... so we penalise that here) |
| | $\mu = $ awrt $\mathbf{48.2}$ | A1 | 2nd A1 for $\mu =$ awrt 48.2 [allow 48.3 if 12.8 used in a correct eqn e.g. $35 + 1.04 \times 12.8$ or better] |
| **(d)(i)** | $P(L > 35 \mid L < 45) = \frac{P(35 < L < 45)}{P(L < 45)} = \frac{0.25}{0.15 + 0.25} = \frac{5}{8}$ (o.e.) | A1, M1, A1 | M1 for a correct expression [num = $P(35 < L < 45)$] with some correct values substituted This M1 may be implied by one of the correct probabilities for (i) or (ii) |
| **(d)(ii)** | $P(L < 45 \mid L > 35) = \frac{P(35 < L < 45)}{P(L > 35)} = \frac{0.25}{0.60 + 0.25} = \frac{5}{17}$ or awrt 0.294 | M1, A1 | 2nd A1 for $\frac{5}{17}$ or awrt 0.294 |
| **(e)** | Prob. of a yellow stick from Hei is $\frac{3}{5}$ which is > prob. of $\frac{2}{17}$ for Tang **So more likely to be Hei** | B1ft, dB1ft | 1st B1ft for a correct comparison of their probabilities from (d) "probs" $\in [0, 1]$ is B0 2nd dB1ft for choosing Hei (dependent on a suitable reason that it is more likely to be hers) Allow e.g. "Hei, because her prob is greater" to score B1B1 provided (d)(i) > (d)(ii) and a correct comparison stated. Allow "Tang" if their (d)(i) < their (d)(ii) and a correct comparison stated. |
**[Total 12]**
---3. Hei and Tang are designing some pieces of art. They collected a large number of sticks. The random variable $L$ represents the length of a stick in centimetres and has a normal distribution with mean $\mu$ and standard deviation $\sigma$.
They sorted the sticks into lengths and painted them.\\
They found that $60 \%$ of the sticks were longer than 45 cm and these were painted red, whilst $15 \%$ of the sticks were shorter than 35 cm and these were painted blue. The remaining sticks were painted yellow.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mu$ and $\sigma$ satisfy
$$45 + 0.2533 \sigma = \mu$$
\item Find a second equation in $\mu$ and $\sigma$.
\item Hence find the value of $\mu$ and the value of $\sigma$.
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( L > 35 \mid L < 45 )$
\item $\mathrm { P } ( L < 45 \mid L > 35 )$
Hei created her piece of art using a random selection of blue and yellow sticks.\\
Tang created his piece of art using a random selection of red and yellow sticks.\\
Hei and Tang each used the same number of sticks to create their piece of art.\\
George is viewing Hei's and Tang's pieces of art. He finds a yellow stick on the floor that has fallen from one of these pieces.
\end{enumerate}\item With reference to your answers to part (d), state, giving a reason, whether the stick is more likely to have fallen from Hei's or Tang's piece of art.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2017 Q3 [12]}}