S1 June 2017 Q5
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.
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.
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.
| Scheme | Marks |
|---|---|
| \([\mathrm{P}(T \gt 20) =]\ \mathrm{P}\left(Z \gt \dfrac{20 - 18}{5}\right)\) | M1 |
| \(\mathrm{P}(Z \gt 0.4) = 1 - 0.6554\) | M1 |
| \(= \)0.3446 or awrt 0.345 | A1 |
| (3) |
Notes
1st M1 for standardising with 20, 18 and 5. Accept \(\pm\)
2nd M1 for attempting 1 – \(p\) [where \(0.5 \lt p \lt 0.7\)]. Beware 1 – 0.4 (or their \(z\) value) is M0
A1 for awrt 0.345 (Correct ans only 3/3)
| Scheme | Marks |
|---|---|
| Require \(\mathrm{P}(T \gt 20 \mid T \gt 15)\) or \(\dfrac{\mathrm{P}(T \gt 20)}{\mathrm{P}(T \gt 15)}\) | M1 |
| \(\dfrac{\text{"(a)"}}{\mathrm{P}\left(Z \gt \frac{15 - 18}{5}\right)} = \dfrac{\text{"(a)"}}{\mathrm{P}(Z \gt -0.6)},\ = \dfrac{\text{"}0.3446\text{"}}{0.7257}\) or \(\dfrac{\text{"}0.345\text{"}}{0.726}\) | M1, A1ft |
| \(= 0.47485\ldots\) = awrt 0.475 | A1 |
| (4) |
Notes
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 \(\mathrm{P}(T \gt 15)\) (no \(\pm\) )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
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 \(\dfrac{\text{"}0.3446\text{"}}{0.7257 \text{ or } 0.726}\) will score M1M1A1ft
2nd A1 for awrt 0.475
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(T \gt d \mid T \gt 15) = 0.5\) or \(\mathrm{P}(T \lt d \mid T \gt 15) = 0.5\) | M1 |
| \(\mathrm{P}(T \gt d)\) or \(\mathrm{P}(15 \lt T \lt d) = 0.5 \times\)”0.7257” =[0.36285] | A1ft |
| \(\mathrm{P}(T \lt d)\) = “0.63715” | M1 |
| So \(\dfrac{d - 18}{5} = 0.35\) (calculator gives 0.35085…) | A1 |
| \(d = 19.754\ldots\) = awrt 19.8 (Accept 19 mins 45(secs) or 19:45 but 19.45 is A0) | A1cso |
| (5) | |
| (12 marks) |
Notes
1st M1 for a correct conditional probability statement that includes the 0.5
1st A1ft for \(\mathrm{P}(T \gt d)\) or \(\mathrm{P}(15 \lt T \lt d) = 0.5 \times\) their \(\mathrm{P}(T \gt 15)\) [provided \(\mathrm{P}(T \gt 15) \gt 0.5\)]
Follow through (3sf) their \(\mathrm{P}(T \gt 15) = 0.7257\) or better from part (b). (Allow 0.726)
Sight of \(0.5 \times\) their 0.7257 = “0.36285” or better scores 1st M1 and 1st A1ft (Allow 0.363)
2nd M1 (dep on 1st M1) for \(\mathrm{P}(T \lt d)\) = 1 – “0.36285” or “0.36285” + 1 – “0.7257” = [0.6371~0.6372]
Sight of their 0.63715 or better (calc: 0.637126… ) scores first 3 marks (Allow 0.637)
2nd A1 for \(\dfrac{d - 18}{5} = 0.35\) (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! \(0.5 \times 0.7257 = 0.36285\) and using this (instead of 0.35) as \(z\) value leads to 19.8 but is A0A0