| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Symmetric properties of normal |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine z-score calculations for parts (a)-(b), then using symmetry properties and simple algebra to find shaded areas and values in parts (c)-(d). While multi-part, each step follows textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.04g Normal distribution properties: empirical rule (68-95-99.7), points of inflection |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 70) = P\left(Z < \frac{70-79}{12}\right)\) | M1 | standardise 79, 12 or 79, 144 |
| \(= P(Z < -0.75) = 0.2266\) | A1A1 | \(-0.75\), \(0.2266\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(64 < X < 96) = P\left(\frac{64-79}{12} < Z < \frac{96-79}{12}\right)\) | M1 | standardise both, 79 & 12 only |
| \(= P(-1.25 < Z < 1.42) = 0.8166\) | A1, A1 | \(-1.25\) & \(1.42\), \(0.8166\) |
| Answer | Marks |
|---|---|
| Shaded area \(= \frac{1}{3}(1 - 0.6463)\) | M1A1 |
| \(= 0.1179\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 79 + b) = 0.7642\) | B1 | \(0.7642\) |
| \(\Rightarrow \frac{b}{12} = 0.72\) | M1A1 | standardise LHS = probability, all correct |
| \(b = 8.64\) | A1 |
## Question 7:
**Part (a):**
$P(X < 70) = P\left(Z < \frac{70-79}{12}\right)$ | M1 | standardise 79, 12 or 79, 144
$= P(Z < -0.75) = 0.2266$ | A1A1 | $-0.75$, $0.2266$
**Part (b):**
$P(64 < X < 96) = P\left(\frac{64-79}{12} < Z < \frac{96-79}{12}\right)$ | M1 | standardise both, 79 & 12 only
$= P(-1.25 < Z < 1.42) = 0.8166$ | A1, A1 | $-1.25$ & $1.42$, $0.8166$
**Part (c):**
Shaded area $= \frac{1}{3}(1 - 0.6463)$ | M1A1 |
$= 0.1179$ | A1 |
**Part (d):**
$P(X \leq 79 + b) = 0.7642$ | B1 | $0.7642$
$\Rightarrow \frac{b}{12} = 0.72$ | M1A1 | standardise LHS = probability, all correct
$b = 8.64$ | A1 |7. The random variable $X$ is normally distributed with mean 79 and variance 144 .
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( X < 70 )$,
\item $\mathrm { P } ( 64 < X < 96 )$.
It is known that $\mathrm { P } ( 79 - a \leq X \leq 79 + b ) = 0.6463$. This information is shown in the figure below.\\
\includegraphics[max width=\textwidth, alt={}, center]{df898ff4-c3ef-400c-b4f7-f4df3757941d-6_581_983_818_590}
Given that $\mathrm { P } ( X \geq 79 + b ) = 2 \mathrm { P } ( X \leq 79 - a )$,
\item show that the area of the shaded region is 0.1179 .
\item Find the value of $b$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2005 Q7 [13]}}