| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question with three routine parts: (a) basic z-score calculation, (b) inverse normal to find standard deviation, and (c) conditional probability using symmetry. All techniques are textbook exercises requiring no novel insight, though part (c) requires recognizing symmetry properties of the normal distribution which elevates it slightly above pure recall. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\left[P(M < 145)\right] = P\left(Z < \frac{145-150}{10}\right) = P(Z < -0.5)\) or \(P(Z > 0.5) = \mathbf{\text{awrt } 0.309}\) | M1, A1, A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \([P(B > 115) = 0.15 \Rightarrow] \frac{115-100}{d} = 1.0364\) (Calc gives 1.036433...), \(d = \mathbf{14.5}\) (Calc gives 14.4727...) | M1B1A1, A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| (c) \([P(X > \mu + 15 \mid X > \mu - 15)] = \frac{P(X > \mu + 15)}{P(X > \mu - 15)} = \frac{0.35}{1-0.35} = \frac{7}{13}\) or awrt 0.538 | M1, A1, A1 | (3) [10] |
**(a)** $\left[P(M < 145)\right] = P\left(Z < \frac{145-150}{10}\right) = P(Z < -0.5)$ or $P(Z > 0.5) = \mathbf{\text{awrt } 0.309}$ | M1, A1, A1 | (3) | M1 for standardising with 145, 150 and 10; 1st A1 for P(Z < –0.5) or P(Z > 0.5) i.e. a $z$ value of ±0.5 and correct region indicated; 2nd A1 for awrt 0.309. Answer only 3/3
**Condone** poor use of notation if correct line appears later.
**(b)** $[P(B > 115) = 0.15 \Rightarrow] \frac{115-100}{d} = 1.0364$ (Calc gives 1.036433...), $d = \mathbf{14.5}$ (Calc gives 14.4727...) | M1B1A1, A1 | (4) | M1 for $\pm \frac{115-100}{d} = z$ where $\|z\| > 1$. Condone MR of $\mu = 150$ instead of 100 for M1B1only; B1 for standardised expression = ±1.0364 (do not allow for use of $1 - 1.0364$); 1st A1 for $z = \text{awrt } 1.04$ and compatible signs i.e. correct equation with $z = \text{awrt } 1.04$ is seen; 2nd A1 for awrt 14.5 (allow awrt 14.4 if $z = \text{awrt } 1.04$ is seen)
**Calc** Answer only of awrt 14.473 scores M1B1A1A1; Answer only of awrt 14.48 scores M1B0A1A1
**(c)** $[P(X > \mu + 15 \mid X > \mu - 15)] = \frac{P(X > \mu + 15)}{P(X > \mu - 15)} = \frac{0.35}{1-0.35} = \frac{7}{13}$ or awrt 0.538 | M1, A1, A1 | (3) [10] | M1 for correct ratio expression need P(X > $\mu$ + 15) on numerator. Allow use of value for $\mu$. May be implied by next line. NB $\frac{0.35 \times 0.65}{0.65} = \frac{0.2275}{0.65}$ is M0; 1st A1 for correct ratio of probabilities; 2nd A1 for awrt 0.538 or $\frac{7}{13}$ (o.e.). Allow 0.5385 provided 2nd A1 is scored.
---\begin{enumerate}
\item The time, in minutes, taken to fly from London to Malaga has a normal distribution with mean 150 minutes and standard deviation 10 minutes.\\
(a) Find the probability that the next flight from London to Malaga takes less than 145 minutes.
\end{enumerate}
The time taken to fly from London to Berlin has a normal distribution with mean 100 minutes and standard deviation $d$ minutes.
Given that $15 \%$ of the flights from London to Berlin take longer than 115 minutes,\\
(b) find the value of the standard deviation $d$.
The time, $X$ minutes, taken to fly from London to another city has a normal distribution with mean $\mu$ minutes.
Given that $\mathrm { P } ( X < \mu - 15 ) = 0.35$\\
(c) find $\mathrm { P } ( X > \mu + 15 \mid X > \mu - 15 )$.
\hfill \mbox{\textit{Edexcel S1 2013 Q4 [10]}}