| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Challenging +1.2 This is a multi-part normal distribution question requiring standard probability calculations, understanding of independent events, and a hypothesis test. Parts (a)-(c) involve straightforward z-score calculations with the added complexity of considering multiple independent batteries, while part (d) is a routine one-sample z-test. The conceptual challenge of understanding that all 4 batteries must survive (minimum of 4 independent events) elevates this slightly above average, but the calculations themselves are standard A-level techniques with no novel insights required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(L > 16) = 0.69146\ldots\) awrt \(\mathbf{0.691}\) | B1 | For evaluating probability using calculator (awrt 0.691). Accept 0.6915 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(L>20 \mid L>16) = \frac{P(L>20)}{P(L>16)}\) | M1 | First step of identifying suitable conditional probability (either form) |
| \(= \frac{0.308537\ldots}{(a)}\) or \(\frac{1-(a)}{(a)} = 0.44621\ldots\) | A1ft, A1 | For ratio of probabilities with numerator = awrt 0.309 or \(1-(a)\) and denom = their (a); for awrt 0.446 |
| \((0.44621\ldots)^4 = 0.03964\ldots\) awrt \(\mathbf{0.0396}\) | dM1, A1 | Dep on 1st M1; for awrt 0.0396 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Require: \([P(L>4)]^2 \times [P(L>20 \mid L>16)]^2\) | M1 | Correct approach (may be implied by A1ft) |
| \(= (0.99976\ldots)^2 \times (``0.44621\ldots")^2\) | A1ft | For \(P(L>4) =\) awrt 0.9998 used and ft their 0.44621 in correct expression |
| \(= 0.19901\ldots\) awrt \(\mathbf{0.199}\) | A1cso | For 0.199 or better with clear evidence of M1. Must see M1 scored by correct expression in symbols or values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 18\), \(H_1: \mu > 18\) | B1 | For both hypotheses in terms of \(\mu\) |
| \(\bar{L} \sim N\!\left(18,\left(\frac{4}{\sqrt{20}}\right)^2\right)\) | M1 | For selecting suitable model. Sight of normal, mean 18, sd \(\frac{4}{\sqrt{20}}\) or variance \(= 0.8\) |
| \(P(\bar{L} > 19.2) = P(Z > 1.3416\ldots) = 0.089856\ldots\) | A1 | For using model correctly. Allow awrt 0.0899 or 0.09 from correct probability statement |
| \((0.0899 > 5\%)\) or \((19.2 < 19.5)\) or \(1.34 < 1.6449\) so not significant | A1 | |
| Insufficient evidence to support Alice's claim (or belief) | A1 | Dep on M1 and 1st A1 for correct contextual conclusion mentioning Alice's claim/belief or there is insufficient evidence that the mean lifetime is more than 18 hours |
# Question 5:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(L > 16) = 0.69146\ldots$ awrt $\mathbf{0.691}$ | B1 | For evaluating probability using calculator (awrt 0.691). Accept 0.6915 |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(L>20 \mid L>16) = \frac{P(L>20)}{P(L>16)}$ | M1 | First step of identifying suitable conditional probability (either form) |
| $= \frac{0.308537\ldots}{(a)}$ or $\frac{1-(a)}{(a)} = 0.44621\ldots$ | A1ft, A1 | For ratio of probabilities with numerator = awrt 0.309 or $1-(a)$ and denom = their (a); for awrt 0.446 |
| $(0.44621\ldots)^4 = 0.03964\ldots$ awrt $\mathbf{0.0396}$ | dM1, A1 | Dep on 1st M1; for awrt 0.0396 |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Require: $[P(L>4)]^2 \times [P(L>20 \mid L>16)]^2$ | M1 | Correct approach (may be implied by A1ft) |
| $= (0.99976\ldots)^2 \times (``0.44621\ldots")^2$ | A1ft | For $P(L>4) =$ awrt 0.9998 used and ft their 0.44621 in correct expression |
| $= 0.19901\ldots$ awrt $\mathbf{0.199}$ | A1cso | For 0.199 or better with clear evidence of M1. Must see M1 scored by correct expression in symbols or values |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 18$, $H_1: \mu > 18$ | B1 | For both hypotheses in terms of $\mu$ |
| $\bar{L} \sim N\!\left(18,\left(\frac{4}{\sqrt{20}}\right)^2\right)$ | M1 | For selecting suitable model. Sight of normal, mean 18, sd $\frac{4}{\sqrt{20}}$ or variance $= 0.8$ |
| $P(\bar{L} > 19.2) = P(Z > 1.3416\ldots) = 0.089856\ldots$ | A1 | For using model correctly. Allow awrt 0.0899 or 0.09 from correct probability statement |
| $(0.0899 > 5\%)$ or $(19.2 < 19.5)$ or $1.34 < 1.6449$ so not significant | A1 | |
| Insufficient evidence to support Alice's claim (or belief) | A1 | Dep on M1 and 1st A1 for correct contextual conclusion mentioning Alice's claim/belief or there is insufficient evidence that the mean lifetime is more than 18 hours |
---\begin{enumerate}
\item The lifetime, $L$ hours, of a battery has a normal distribution with mean 18 hours and standard deviation 4 hours.
\end{enumerate}
Alice's calculator requires 4 batteries and will stop working when any one battery reaches the end of its lifetime.\\
(a) Find the probability that a randomly selected battery will last for longer than 16 hours.
At the start of her exams Alice put 4 new batteries in her calculator. She has used her calculator for 16 hours, but has another 4 hours of exams to sit.\\
(b) Find the probability that her calculator will not stop working for Alice's remaining exams.
Alice only has 2 new batteries so, after the first 16 hours of her exams, although her calculator is still working, she randomly selects 2 of the batteries from her calculator and replaces these with the 2 new batteries.\\
(c) Show that the probability that her calculator will not stop working for the remainder of her exams is 0.199 to 3 significant figures.
After her exams, Alice believed that the lifetime of the batteries was more than 18 hours. She took a random sample of 20 of these batteries and found that their mean lifetime was 19.2 hours.\\
(d) Stating your hypotheses clearly and using a $5 \%$ level of significance, test Alice's belief.
\hfill \mbox{\textit{Edexcel Paper 3 2018 Q5 [14]}}