| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Moderate -0.8 This is a straightforward normal distribution question testing standard techniques: probability calculations using tables/calculator, finding percentiles, and working backwards from a probability to find a parameter. All parts are routine A-level statistics applications with no novel problem-solving required, making it easier than average but not trivial due to the multi-part structure and the reverse calculation in part (c). |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| 17(a)(i) | States correct probability OE | 1.2 |
| Answer | Marks |
|---|---|
| 17(a)(ii) | Calculates the correct |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW [0.817, 0.82] | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 17(b) | Standardises correctly | |
| PI by correct answer | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 17(c) | Obtains z value from inverse |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer or equation | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝜎𝜎 𝑐𝑐𝑎𝑎 𝜎𝜎 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝜎𝜎 𝑐𝑐𝑎𝑎 𝜎𝜎 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | A1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 17:
--- 17(a)(i) ---
17(a)(i) | States correct probability OE | 1.2 | B1 | 1
--- 17(a)(ii) ---
17(a)(ii) | Calculates the correct
probability
AWFW [0.817, 0.82] | 1.1b | B1 | 0.818
--- 17(b) ---
17(b) | Standardises correctly
PI by correct answer | 3.1b | M1 | x – = x – 8
1.5 1.5
𝜇𝜇
x = 6.08
Obtains correct value of x
AWFW [6.07, 6.08]
ISW | 1.1b | A1
--- 17(c) ---
17(c) | Obtains z value from inverse
normal distribution
AWFW [0.67, 0.68] or
[–0.68, –0.67]
PI by correct answer or equation | 1.1b | B1 | z = –0.6745
5 – 7 = –0.6745
σ
σ = 2.97
Uses
7−5 5−7
PI by correct equation
𝜎𝜎 𝑐𝑐𝑎𝑎 𝜎𝜎 | 3.1b | M1
Forms an equation using
and their z value
7−5 5−7
Accept z = [−4, 4] except ±0.25
𝜎𝜎 𝑐𝑐𝑎𝑎 𝜎𝜎 | 1.1a | M1
Obtains correct value of
standard deviation
AWFW [2.94, 2.99]
CAO
ISW | 1.1b | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionThe lifetime of Zaple smartphone batteries, $X$ hours, is normally distributed with mean 8 hours and standard deviation 1.5 hours.
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item Find P($X \neq 8$)
[1 mark]
\item Find P($6 < X < 10$)
[1 mark]
\end{enumerate}
\item Determine the lifetime exceeded by 90\% of Zaple smartphone batteries.
[2 marks]
\item A different smartphone, Kaphone, has its battery's lifetime, $Y$ hours, modelled by a normal distribution with mean 7 hours and standard deviation $\sigma$.
25\% of randomly selected Kaphone batteries last less than 5 hours.
Find the value of $\sigma$, correct to three significant figures.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2020 Q17 [8]}}