| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Standard +0.8 This S1 question requires understanding of conditional probability with normal distribution. Parts (a) and (b) are routine standardization and inverse normal problems. Part (c) is more sophisticated, requiring students to recognize that given the phone has lasted 127 hours, they need P(L > 133 | L > 127), which involves conditional probability reasoning with continuous distributions—a conceptually challenging step beyond standard S1 fare. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{127-100}{15}\) | M1 | Attempt to standardise with 127, 100 and 15; allow \(\pm\) |
| \(P(L>127)=P(Z>1.8)\) or \(1-P(Z\leq 1.8)\) | A1 | Allow a diagram but must have 1.8 and correct area |
| \(=1-0.9641=\mathbf{0.0359}\) (awrt \(0.0359\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{d-100}{15}=-1.2816\) | M1, B1 | Attempt to standardise with 100 and 15; \(z=\pm1.2816\) seen anywhere |
| \(d=80.776\), awrt \(\mathbf{80.8}\) | A1 | Can be scored for using \(1.28\) but then M1B0A1; \(80.8\) must follow from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Require \(P(L>133 \mid L>127)\) | M1 | Clear indication of correct conditional probability or attempt at correct ratio |
| \(=\frac{P(L>133)}{P(L>127)}=\frac{P(Z>2.2)}{P(L>127)}\) | dM1 | Dependent on 1st M1; \(P(L>133)\) leading to \(P(Z>2.2)\) |
| \(=\frac{1-0.9861}{1-0.9641}=\frac{0.0139}{0.0359}\) | A1 | For \(0.0139\) or better from \(P(Z>2.20)\); dependent on both Ms |
| \(=0.3871\ldots=\) awrt \(\mathbf{0.39}\) | A1 | Both Ms required |
# Question 4:
## Part (a)
| Working | Mark | Guidance |
|---------|------|---------|
| $\frac{127-100}{15}$ | M1 | Attempt to standardise with 127, 100 and 15; allow $\pm$ |
| $P(L>127)=P(Z>1.8)$ or $1-P(Z\leq 1.8)$ | A1 | Allow a diagram but must have 1.8 and correct area |
| $=1-0.9641=\mathbf{0.0359}$ (awrt $0.0359$) | A1 | |
## Part (b)
| Working | Mark | Guidance |
|---------|------|---------|
| $\frac{d-100}{15}=-1.2816$ | M1, B1 | Attempt to standardise with 100 and 15; $z=\pm1.2816$ seen anywhere |
| $d=80.776$, awrt $\mathbf{80.8}$ | A1 | Can be scored for using $1.28$ but then M1B0A1; $80.8$ must follow from correct working |
## Part (c)
| Working | Mark | Guidance |
|---------|------|---------|
| Require $P(L>133 \mid L>127)$ | M1 | Clear indication of correct conditional probability or attempt at correct ratio |
| $=\frac{P(L>133)}{P(L>127)}=\frac{P(Z>2.2)}{P(L>127)}$ | dM1 | Dependent on 1st M1; $P(L>133)$ leading to $P(Z>2.2)$ |
| $=\frac{1-0.9861}{1-0.9641}=\frac{0.0139}{0.0359}$ | A1 | For $0.0139$ or better from $P(Z>2.20)$; dependent on both Ms |
| $=0.3871\ldots=$ awrt $\mathbf{0.39}$ | A1 | Both Ms required |
---\begin{enumerate}
\item The length of time, $L$ hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours.\\
(a) Find $\mathrm { P } ( L > 127 )$.\\
(b) Find the value of $d$ such that $\mathrm { P } ( L < d ) = 0.10$
\end{enumerate}
Alice is about to go on a 6 hour journey.\\
Given that it is 127 hours since Alice last charged her phone,\\
(c) find the probability that her phone will not need charging before her journey is completed.
\hfill \mbox{\textit{Edexcel S1 2013 Q4 [10]}}