| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Moderate -0.3 This S1 question tests basic properties of normal distribution symmetry and conditional probability. Parts (a) and (b) require only understanding that probabilities sum to 1 and the normal distribution is symmetric about μ. Part (c) applies the definition of conditional probability P(A|B) = P(A∩B)/P(B), which is straightforward once symmetry is recognized. While it requires conceptual understanding rather than pure calculation, these are standard S1 concepts with no complex problem-solving needed. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(Y > 17) = 1 - P(Y \le 17) = 0.4\) | B1 (1) | For 0.4. Note: do not isw if 0.6554 is given as answer after 0.4 has been seen. |
| (b) \(P(Y < \mu) = 0.5\) or \([P(\mu < Y < 17) =] 0.6 - 0.5\) \(= 0.1\) | M1, A1 (2) | |
| (c) \([P(Y < \mu | Y < 17) =]\) | M1, A1 (2) |
| \(\frac{P(Y < \mu)}{P(Y < 17)}\) or \(\frac{0.5}{0.6}\) | ||
| \(= \frac{5}{6}\) awrt \(0.833\) | [5] |
**(a)** $P(Y > 17) = 1 - P(Y \le 17) = 0.4$ | B1 (1) | For 0.4. Note: do not isw if 0.6554 is given as answer after 0.4 has been seen.
**(b)** $P(Y < \mu) = 0.5$ or $[P(\mu < Y < 17) =] 0.6 - 0.5$ $= 0.1$ | M1, A1 (2) |
**(c)** $[P(Y < \mu | Y < 17) =]$ | M1, A1 (2) | May be implied by $\frac{P(Y < \mu)}{0.6}$ or $\frac{0.5}{P(Y < 17)}$
$\frac{P(Y < \mu)}{P(Y < 17)}$ or $\frac{0.5}{0.6}$ | |
$= \frac{5}{6}$ awrt $0.833$ | | [5]
---4. The random variable $Y \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$
Given that $\mathrm { P } ( Y < 17 ) = 0.6$ find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( Y > 17 )$
\item $\mathrm { P } ( \mu < Y < 17 )$
\item $\mathrm { P } ( Y < \mu \mid Y < 17 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2014 Q4 [5]}}