| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Quadratic relationship μ = kσ² |
| Difficulty | Challenging +1.2 This question requires finding μ and σ from two probability conditions (involving inverse normal calculations), then computing expected value with three different outcomes. While it involves multiple steps and careful probability work, the techniques are standard S1 material with no novel insight required—students follow a clear procedure of standardizing, using tables/inverse normal, solving simultaneous equations, then applying expected value formula. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | \([P(A < 388) = 0.001 \Rightarrow] \frac{388 - \mu}{\sigma} = \pm z\) (where \( | z |
| \([P(A > 410) = 0.0197 \Rightarrow] \frac{410 - \mu}{\sigma} = \pm z\) (where \(1 < | z | < 2.5\)); So \(410 - \mu = 2.06\sigma\) (o.e.) |
| So \(22 = 5.1502 \sigma\) (\(\text{calc gives } 5.1502323\ldots\)) \(\sigma = 4.2716\ldots\) (awrt 4.27) and \(\mu = 401.20\ldots\) (awrt 401) | M1, A1, A1 | 3rd M1 for solving their 2 linear equations in \(\mu\) and \(\sigma\) (i.e. reduce to equation in one var') Correct processes used but allow 1 sign or numerical slip. Must see the equation in one variable unless correct answers obtained in which case can be implied.; 3rd A1 for \(\sigma =\) awrt 4.27 (Allow 4.26 if 1st A0 for awrt −3.1); 4th A1 for \(\mu =\) awrt 401. Use of \(\sigma^2\) instead of \(\sigma\) in (a) will score 0/7 |
| 5(b) | [Let \(X =\) profit in £]; \(\begin{array}{c | cccc} x & -100 & -0.30 & 0.25 \\ P(X = x) & 0.001 & 0.0197 & 1 - (0.001 + 0.0197) = 0.9793 \end{array}\) |
| \(E(X) = -100 \times 0.001 - 0.30 \times 0.0197 + 0.25 \times 0.9793 = 0.138915\ldots\) (£) 0.14 (allow any answer in range 0.138…~0.141…) | M1, A1 | 3rd M1 dep on 3 values of X and 3 probs. For an expression for E(X) using their values; A1 for answer of 0.14 or in range (0.138…~0.141…). Not awrt 0.14. Accept 0.13 if correct expression seen beforehand |
**5(a)** | $[P(A < 388) = 0.001 \Rightarrow] \frac{388 - \mu}{\sigma} = \pm z$ (where $|z| > 2.25$); So $388 - \mu = -3.0902\sigma$ (o.e.) | M1, A1 | 1st M1 for attempt to standardise with 388 and set equal to $\pm$ a z value where $|z| > 2.25$; 1st A1 for fully correct equation with $z = -3.0902$ or better (calc gives 3.0902320…)
| $[P(A > 410) = 0.0197 \Rightarrow] \frac{410 - \mu}{\sigma} = \pm z$ (where $1 < |z| < 2.5$); So $410 - \mu = 2.06\sigma$ (o.e.) | M1, A1 | 2nd M1 for attempt to standardise with 410 and set equal to $\pm$ a z value where $1 < |z| < 2.5$; 2nd A1 for fully correct equation with $z =$ awrt 2.06 (calc gives 2.059984…)
| So $22 = 5.1502 \sigma$ ($\text{calc gives } 5.1502323\ldots$) $\sigma = 4.2716\ldots$ (awrt 4.27) and $\mu = 401.20\ldots$ (awrt 401) | M1, A1, A1 | 3rd M1 for solving their 2 linear equations in $\mu$ and $\sigma$ (i.e. reduce to equation in one var') Correct processes used but allow 1 sign or numerical slip. Must see the equation in one variable unless correct answers obtained in which case can be implied.; 3rd A1 for $\sigma =$ awrt 4.27 (Allow 4.26 if 1st A0 for awrt −3.1); 4th A1 for $\mu =$ awrt 401. Use of $\sigma^2$ instead of $\sigma$ in (a) will score 0/7
**5(b)** | [Let $X =$ profit in £]; $\begin{array}{c|cccc} x & -100 & -0.30 & 0.25 \\ P(X = x) & 0.001 & 0.0197 & 1 - (0.001 + 0.0197) = 0.9793 \end{array}$ | M1, M1 | 1st M1 for 3 correct values of x (i.e. −100, −0.30, +0.25); 2nd M1 for attempt at all 3 probabilities (correct expression for 0.9793)
| $E(X) = -100 \times 0.001 - 0.30 \times 0.0197 + 0.25 \times 0.9793 = 0.138915\ldots$ (£) 0.14 (allow any answer in range 0.138…~0.141…) | M1, A1 | 3rd M1 dep on 3 values of X and 3 probs. For an expression for E(X) using their values; A1 for answer of 0.14 or in range (0.138…~0.141…). Not awrt 0.14. Accept 0.13 if correct expression seen beforehand
---\begin{enumerate}
\item The label on a jar of Amy's jam states that the jar contains about 400 grams of jam. For each jar that contains less than 388 grams of jam, Amy will be fined $\pounds 100$. If a jar contains more than 410 grams of jam then Amy makes a loss of $\pounds 0.30$ on that jar.
\end{enumerate}
The weight of jam, $A$ grams, in a jar of Amy's jam has a normal distribution with mean $\mu$ grams and standard deviation $\sigma$ grams. Amy chooses $\mu$ and $\sigma$ so that $\mathrm { P } ( A < 388 ) = 0.001$ and $\mathrm { P } ( A > 410 ) = 0.0197$\\
(a) Find the value of $\mu$ and the value of $\sigma$.
Amy can sell jars of jam containing between 388 grams and 410 grams for a profit of $\pounds 0.25$\\
(b) Calculate the expected amount, in £s, that Amy receives for each jar of jam.\\
\hfill \mbox{\textit{Edexcel S1 2016 Q5 [11]}}