| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Moderate -0.3 This is a standard S1 normal distribution question requiring table lookups and basic probability rules. Parts (a)(i)-(iii) are routine table reading, (a)(iv) needs simple union rule application, and part (b) requires conditional probability with standardization—all textbook techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.04h Select appropriate distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A)=P(Z>1.1)=1-0.8643=\mathbf{0.1357}\) (accept awrt 0.136) | B1 | Mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B)=P(Z>-1.9)=\mathbf{0.9713}\) (accept awrt 0.971) | B1 | 0.9713 followed by \(1-0.9713\) is B0; for rounding errors e.g. 29.245 followed by 29.3 apply ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
\(P(C)=[P(-1.5| M1, A1 |
M1 for correct expression with probability values; correct answer implies M1A1 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A\cup C)=P(Z>-1.5)\) or \(P(Z<1.5)\) or \(=P(A)+P(C)-P(A\cap C)="0.1357"+"0.8664"-(0.9332-0.8643)=\mathbf{0.9332}\) (accept awrt 0.933) | M1, A1 | M1 for correct addition formula with some correct substitution (or correct ft) or \(P(Z>-1.5)\) (o.e.) or fully correct expression with correct probabilities; A1 for 0.9332; correct answer only is M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[P(X>w\ | X>28)\right]=\frac{P(X>w)}{P(X>28)}=[0.625]\) | M1 |
| \(P(X>28)=P\left(Z>\frac{28-21}{5}\right)=P(Z>1.4)=[0.0808\ \text{calc: }0.80756..]\) | M1 | M1 for standardising 28 with 21 and 5; allow \(\pm\); may be implied by 0.0808 [or awrt 0.081] in correct position |
\(P(X>w)=0.0808\times0.625\ (=0.0505)\) or \(P(X| A1 |
A1 for \(P(X>w)=0.0808\times0.625\) or \(P(X>w)=0.0505\) or \(P(X | |
| \(\frac{w-21}{5}=1.64\) | M1 B1 | M1 for standardising \(w\) with 21 and 5 and setting equal to a \(z\)-value \(\ |
| \(w=\) awrt \(\mathbf{29.2}\) | A1 | allow awrt 29.2 |
# Question 6:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A)=P(Z>1.1)=1-0.8643=\mathbf{0.1357}$ (accept awrt 0.136) | B1 | Mark final answer |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B)=P(Z>-1.9)=\mathbf{0.9713}$ (accept awrt 0.971) | B1 | 0.9713 followed by $1-0.9713$ is B0; for rounding errors e.g. 29.245 followed by 29.3 apply ISW |
## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(C)=[P(-1.5<Z<1.5)]=0.9332-(1-0.9332)$ or $(0.9332-0.5)\times2=\mathbf{0.8664}$ (accept awrt 0.866) | M1, A1 | M1 for correct expression with probability values; correct answer implies M1A1 |
## Part (a)(iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A\cup C)=P(Z>-1.5)$ or $P(Z<1.5)$ or $=P(A)+P(C)-P(A\cap C)="0.1357"+"0.8664"-(0.9332-0.8643)=\mathbf{0.9332}$ (accept awrt 0.933) | M1, A1 | M1 for correct addition formula with some correct substitution (or correct ft) or $P(Z>-1.5)$ (o.e.) or fully correct expression with correct probabilities; A1 for 0.9332; correct answer only is M1A1 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[P(X>w\|X>28)\right]=\frac{P(X>w)}{P(X>28)}=[0.625]$ | M1 | M1 for correct expression for conditional probability; must have $P(X>w)$ as numerator; may be implied by $P(X>w)=0.625\times(\text{any probability})$ |
| $P(X>28)=P\left(Z>\frac{28-21}{5}\right)=P(Z>1.4)=[0.0808\ \text{calc: }0.80756..]$ | M1 | M1 for standardising 28 with 21 and 5; allow $\pm$; may be implied by 0.0808 [or awrt 0.081] in correct position |
| $P(X>w)=0.0808\times0.625\ (=0.0505)$ or $P(X<w)=0.9495$ | A1 | A1 for $P(X>w)=0.0808\times0.625$ or $P(X>w)=0.0505$ or $P(X<w)=0.9495$; depends on both Ms |
| $\frac{w-21}{5}=1.64$ | M1 B1 | M1 for standardising $w$ with 21 and 5 and setting equal to a $z$-value $\|z\|>1$; B1 for 1.64 (or better) used correctly |
| $w=$ awrt $\mathbf{29.2}$ | A1 | allow awrt 29.2 |
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Could you please share the actual mark scheme pages that contain the questions, answers, mark allocations, and guidance notes? This appears to be either a blank page or a back cover page from the document.\begin{enumerate}
\item The random variable $Z \sim \mathrm {~N} ( 0,1 )$\\
$A$ is the event $Z > 1.1$\\
$B$ is the event $Z > - 1.9$\\
$C$ is the event $- 1.5 < Z < 1.5$\\
(a) Find\\
(i) $\mathrm { P } ( A )$\\
(ii) $\mathrm { P } ( B )$\\
(iii) $\mathrm { P } ( C )$\\
(iv) $\mathrm { P } ( A \cup C )$
\end{enumerate}
The random variable $X$ has a normal distribution with mean 21 and standard deviation 5\\
(b) Find the value of $w$ such that $\mathrm { P } ( X > w \mid X > 28 ) = 0.625$
\hfill \mbox{\textit{Edexcel S1 2015 Q6 [12]}}