| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Conditional probability with normal |
| Difficulty | Standard +0.8 This S1 question requires understanding of conditional probability with normal distribution (part b), which goes beyond routine standardization. While part (a) is standard, part (b) requires P(T>20|T>15) = P(T>20)/P(T>15), and part (c) needs finding the median of a truncated normal distribution. These are non-trivial applications that require conceptual understanding beyond mechanical z-score calculations. |
| 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) | \([P(T > 20)] = P\left(Z > \frac{20 - 18}{5}\right)\) | M1 |
| \(P(Z > 0.4) = 1 - 0.6554\) | M1 | 2nd M1 for attempting \(1 - p\) [where \(0.5 < p < 0.7\)]. Beware \(1 - 0.4\) (or their \(z\) value) is M0 |
| \(= 0.3446\) or awrt \(0.345\) | A1 | For awrt 0.345 (Correct ans only 3/3) |
| (b) | Require \(P(T > 20 | T > 15)\) or \(\frac{P(T > 20)}{P(T > 15)}\) |
| \(\frac{P(Z > \frac{15-18}{5})}{P(Z > -0.6)} = \frac{"{0.3446"}}{0.7257}\) or \(\frac{"{0.345"}}{0.726}\) | M1, A1ft | Allow one digit transcription errors from (a) e.g. 0.3464 or 0.3466 etc for 2nd M1 and 1st A1ft. For their 0.3446 on numerator and denominator of 0.7257 (or better: 0.7257469...) provided Num \(<\) Denom. Allow 0.726 on the denominator. Sight of \(\frac{"{0.3446"}}{0.7257}\) or 0.726 will score M1M1A1ft |
| \(= 0.47485\ldots = \text{awrt } 0.475\) | A1 | For awrt 0.475 |
| (c) | \(P(T > d | T > 15) = 0.5\) or \(P(T < d |
| \(P(T > d) = 0.5 × "\)0.7257"\( = [0.36285]\) | A1ft | 1st A1ft for \(P(T > d) = 0.5 ×\) their \(P(T > 15)\) [provided \(P(T > 15) > 0.5\)] Follow (3sf) their \(P(T > 15) = 0.7257\) or better from part (b). (Allow 0.726) Sight of \(0.5 × \text{their } 0.7257 = "0.36285"\) or better scores first 3 marks (Allow 0.363) |
| \(P(T < d) = "\)0.63715"\( or \)1 - "\(0.7257"" = [0.6371{-}0.6372]\) | M1 | 2nd M1 (dep on 1st M1) for \(P(T < d) = 1 - "\)0.36285"\( or "\)0.63285"\( + \)1 -\( "\)0.7257$" |
| So \(\frac{d - 18}{5} = 0.35\) (calculator gives 0.35085...) | A1 | |
| \(d = 19.754\ldots = \text{awrt } 19.8\) | A1cso | For \((d = )\) awrt 19.8 (accept 19.7 not awrt 19.7) Must come from correct work. |
| Answer | Marks |
|---|---|
| (a) | 1st M1 for standardising with 20, 18 and 5. Accept \(±\). 2nd M1 for attempting \(1 - p\) [where \(0.5 < p < 0.7\)]. Beware \(1 - 0.4\) (or their \(z\) value) is M0. A1 for awrt 0.345 (Correct ans only 3/3) |
| (b) | 1st M1 for either correct conditional probability statement (allow "in words" or any letter except Z). 1st M1 can be implied by 2nd M1 so a mark of M0M1 should not be given. 2nd M1 for using their (a) on num. and attempting to standardise \(P(T > 15)\) (no \(±\) on denom. Num.>Deno. is M0). Allow one digit transcription errors from (a) e.g. 0.3464 or 0.3466 etc for 2nd M1 and 1st A1ft. For their 0.3446 on numerator and denominator of 0.7257 (or better: 0.7257469...) provided Num \(<\) Denom. Allow 0.726 on the denominator. Sight of \(\frac{"{0.3446"}}{0.7257}\) or 0.726 will score M1M1A1ft. 2nd A1 for awrt 0.475 |
| (c) | 1st M1 for a correct conditional probability statement that includes the 0.5. 1st A1ft for \(P(T > d) = 0.5 ×\) their \(P(T > 15)\) [provided \(P(T > 15) > 0.5\)] Follow (3sf) their \(P(T > 15) = 0.7257\) or better from part (b). (Allow 0.726). Sight of \(0.5 × \text{their } 0.7257 = "0.36285"\) or better scores first 3 marks (Allow 0.363). 2nd M1 (dep on 1st M1) for \(P(T < d) = 1 -\) "\(0.36285"\) or "\(0.63285"\) + \(1 -\) "\(0.7257\)"\(. Sight of their 0.63715 or better (calc: 0.637126...) scores first 3 marks (Allow 0.637). 2nd A1 for \)\frac{d - 18}{5} = 0.35\( (or better) (Calc could give 0.350788...). 3rd A1cso for \)(d = )$ awrt 19.8 (accept 19.7 not awrt 19.7) Must come from correct work. |
| Beware! | \(0.5 × 0.7257 = 0.36285\) and using this (instead of 0.35) as \(z\) value leads to 19.8 but is A0A0 |
(a) | $[P(T > 20)] = P\left(Z > \frac{20 - 18}{5}\right)$ | M1 | 1st M1 for standardising with 20, 18 and 5. Accept $±$ |
| $P(Z > 0.4) = 1 - 0.6554$ | M1 | 2nd M1 for attempting $1 - p$ [where $0.5 < p < 0.7$]. Beware $1 - 0.4$ (or their $z$ value) is M0 |
| $= 0.3446$ or awrt $0.345$ | A1 | For awrt 0.345 (Correct ans only 3/3) |
(b) | Require $P(T > 20 | T > 15)$ or $\frac{P(T > 20)}{P(T > 15)}$ | M1 | 1st M1 for either correct conditional probability statement (allow "in words" or any letter except Z). 1st M1 can be implied by 2nd M1 so a mark of M0M1 should not be given. 2nd M1 for using their (a) on num. and attempting to standardise $P(T > 15)$ (no $±$ on denom. Num.>Deno. is M0 |
| $\frac{P(Z > \frac{15-18}{5})}{P(Z > -0.6)} = \frac{"{0.3446"}}{0.7257}$ or $\frac{"{0.345"}}{0.726}$ | M1, A1ft | Allow one digit transcription errors from (a) e.g. 0.3464 or 0.3466 etc for 2nd M1 and 1st A1ft. For their 0.3446 on numerator and denominator of 0.7257 (or better: 0.7257469...) provided Num $<$ Denom. Allow 0.726 on the denominator. Sight of $\frac{"{0.3446"}}{0.7257}$ or 0.726 will score M1M1A1ft |
| $= 0.47485\ldots = \text{awrt } 0.475$ | A1 | For awrt 0.475 |
(c) | $P(T > d | T > 15) = 0.5$ or $P(T < d | T > 15) = 0.5$ | M1 | 1st M1 for a correct conditional probability statement that includes the 0.5 |
| $P(T > d) = 0.5 × "$0.7257"$ = [0.36285]$ | A1ft | 1st A1ft for $P(T > d) = 0.5 ×$ their $P(T > 15)$ [provided $P(T > 15) > 0.5$] Follow (3sf) their $P(T > 15) = 0.7257$ or better from part (b). (Allow 0.726) Sight of $0.5 × \text{their } 0.7257 = "0.36285"$ or better scores first 3 marks (Allow 0.363) |
| $P(T < d) = "$0.63715"$ or $1 - "$0.7257"" = [0.6371{-}0.6372]$ | M1 | 2nd M1 (dep on 1st M1) for $P(T < d) = 1 - "$0.36285"$ or "$0.63285"$ + $1 -$ "$0.7257$" |
| So $\frac{d - 18}{5} = 0.35$ (calculator gives 0.35085...) | A1 | |
| $d = 19.754\ldots = \text{awrt } 19.8$ | A1cso | For $(d = )$ awrt 19.8 (accept 19.7 not awrt 19.7) Must come from correct work. |
**Notes:**
(a) | 1st M1 for standardising with 20, 18 and 5. Accept $±$. 2nd M1 for attempting $1 - p$ [where $0.5 < p < 0.7$]. Beware $1 - 0.4$ (or their $z$ value) is M0. A1 for awrt 0.345 (Correct ans only 3/3) |
(b) | 1st M1 for either correct conditional probability statement (allow "in words" or any letter except Z). 1st M1 can be implied by 2nd M1 so a mark of M0M1 should not be given. 2nd M1 for using their (a) on num. and attempting to standardise $P(T > 15)$ (no $±$ on denom. Num.>Deno. is M0). Allow one digit transcription errors from (a) e.g. 0.3464 or 0.3466 etc for 2nd M1 and 1st A1ft. For their 0.3446 on numerator and denominator of 0.7257 (or better: 0.7257469...) provided Num $<$ Denom. Allow 0.726 on the denominator. Sight of $\frac{"{0.3446"}}{0.7257}$ or 0.726 will score M1M1A1ft. 2nd A1 for awrt 0.475 |
(c) | 1st M1 for a correct conditional probability statement that includes the 0.5. 1st A1ft for $P(T > d) = 0.5 ×$ their $P(T > 15)$ [provided $P(T > 15) > 0.5$] Follow (3sf) their $P(T > 15) = 0.7257$ or better from part (b). (Allow 0.726). Sight of $0.5 × \text{their } 0.7257 = "0.36285"$ or better scores first 3 marks (Allow 0.363). 2nd M1 (dep on 1st M1) for $P(T < d) = 1 -$ "$0.36285"$ or "$0.63285"$ + $1 -$ "$0.7257$"$. Sight of their 0.63715 or better (calc: 0.637126...) scores first 3 marks (Allow 0.637). 2nd A1 for $\frac{d - 18}{5} = 0.35$ (or better) (Calc could give 0.350788...). 3rd A1cso for $(d = )$ awrt 19.8 (accept 19.7 not awrt 19.7) Must come from correct work. |
**Beware!** | $0.5 × 0.7257 = 0.36285$ and using this (instead of 0.35) as $z$ value leads to 19.8 but is A0A0 |5. Yuto works in the quality control department of a large company. The time, $T$ minutes, it takes Yuto to analyse a sample is normally distributed with mean 18 minutes and standard deviation 5 minutes.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Yuto takes longer than 20 minutes to analyse the next sample. (3)
The company has a large store of samples analysed by Yuto with the time taken for each analysis recorded. Serena is investigating the samples that took Yuto longer than 15 minutes to analyse.
She selects, at random, one of the samples that took Yuto longer than 15 minutes to analyse.
\item Find the probability that this sample took Yuto more than 20 minutes to analyse.
Serena can identify, in advance, the samples that Yuto can analyse in under 15 minutes and in future she will assign these to someone else.
\item Estimate the median time taken by Yuto to analyse samples in future.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2017 Q5 [12]}}