Edexcel S1 2022 June — Question 6 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2022
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeConditional probability with normal
DifficultyStandard +0.8 Parts (a) and (b) are routine normal distribution calculations requiring standardization and expected value. Part (c) involves conditional probability with normal distribution, requiring students to apply P(A|B) = P(A∩B)/P(B), work backwards from a probability to find a z-score, then convert to the original scale—this multi-step reasoning with conditional probability elevates it above standard S1 questions.
Spec2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
  1. A manufacturer fills bottles with oil. The volume of oil in a bottle, \(V \mathrm { ml }\), is normally distributed with \(V \sim \mathrm {~N} \left( 100,2.5 ^ { 2 } \right)\)
    1. Find \(\mathrm { P } ( V > 104.9 )\)
    2. In a pack of 150 bottles, find the expected number of bottles containing more than 104.9 ml
    3. Find the value of \(v\), to 2 decimal places, such that \(\mathrm { P } ( V > v \mid V < 104.9 ) = 0.2801\)
Question 6:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(P(V > 104.9) = P\left(Z > \dfrac{104.9 - 100}{2.5}\right)\)M1 Standardising with 104.9, 100 and 2.5 (allow \(\pm\)); implied by 1.96 seen
\(= 1 - 0.975\)M1 For use of \(1-p\) with \(0.9 < p < 1\); condone answer of 0.0249 for this mark
\(= 0.0250\)A1 Allow 0.025 or awrt 0.0250 (NB calculator answer is 0.02499…); answer only scores M1M1A1
(3 marks)
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Expected number \(= 150 \times\) "0.025"M1 For \(150 \times\) "their part (a)"
\(= 3.75\)A1 awrt 3.75; if 3.75 not seen, allow 4 if method mark awarded
(2 marks)
Part (c):
AnswerMarks Guidance
Working/AnswerMark Guidance
\([P(V > v \mid V < 104.9)] = \dfrac{[P(v < V < 104.9)]}{P(V < 104.9)} = 0.2801\)M1 For writing or using a ratio of probabilities with denominator \(P(V < 104.9)\) and equating to 0.2801; \(\dfrac{p}{P(V<104.9)} = 0.2801\) implied by awrt 0.273; use of independence is M0
\(P(v < V < 104.9) = (1 - \text{"0.025"}) - P(V < v)\)M1 For writing or using \(P(V > v \cap V < 104.9) = P(V < 104.9) - P(V < v)\) \([= (1-\text{"0.025"}) - P(V < v)]\)
\(P(V < v) = (1 - \text{"0.025"}) - (1 - \text{"0.025"}) \times 0.2801\ [= 0.7019]\)dM1 Dependent on previous M1; allow equivalent oe \((1 - \text{"their}(a)\text{")}(1 - 0.2801)\); NB: \([P(Vv)=]\) awrt 0.298 implies M1M1M1
\(\dfrac{v - 100}{2.5} = 0.53\)M1A1 Standardising with 100, 2.5 and equating to a \(z\)-value, \(\dfrac{v-100}{2.5} = z\), \(0.4 < \
\(v = 101.325\)A1 awrt 101.33 (allow awrt 101.32 from use of calculator)
ALT for first two M marks: \(\dfrac{P(V
(6 marks)
Total: 11 marks
## Question 6:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(V > 104.9) = P\left(Z > \dfrac{104.9 - 100}{2.5}\right)$ | M1 | Standardising with 104.9, 100 and 2.5 (allow $\pm$); implied by 1.96 seen |
| $= 1 - 0.975$ | M1 | For use of $1-p$ with $0.9 < p < 1$; condone answer of 0.0249 for this mark |
| $= 0.0250$ | A1 | Allow 0.025 or awrt 0.0250 (NB calculator answer is 0.02499…); answer only scores M1M1A1 |

**(3 marks)**

---

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Expected number $= 150 \times$ "0.025" | M1 | For $150 \times$ "their part (a)" |
| $= 3.75$ | A1 | awrt 3.75; if 3.75 not seen, allow 4 if method mark awarded |

**(2 marks)**

---

### Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $[P(V > v \mid V < 104.9)] = \dfrac{[P(v < V < 104.9)]}{P(V < 104.9)} = 0.2801$ | M1 | For writing or using a ratio of probabilities with denominator $P(V < 104.9)$ and equating to 0.2801; $\dfrac{p}{P(V<104.9)} = 0.2801$ implied by awrt 0.273; use of independence is M0 |
| $P(v < V < 104.9) = (1 - \text{"0.025"}) - P(V < v)$ | M1 | For writing or using $P(V > v \cap V < 104.9) = P(V < 104.9) - P(V < v)$ $[= (1-\text{"0.025"}) - P(V < v)]$ |
| $P(V < v) = (1 - \text{"0.025"}) - (1 - \text{"0.025"}) \times 0.2801\ [= 0.7019]$ | dM1 | Dependent on previous M1; allow equivalent oe $(1 - \text{"their}(a)\text{")}(1 - 0.2801)$; NB: $[P(V<v)=]$ awrt 0.702 implies M1M1M1 or $[P(V>v)=]$ awrt 0.298 implies M1M1M1 |
| $\dfrac{v - 100}{2.5} = 0.53$ | M1A1 | Standardising with 100, 2.5 and equating to a $z$-value, $\dfrac{v-100}{2.5} = z$, $0.4 < \|z\| < 0.6$; A1 correct equation with compatible signs; watch out for $\dfrac{v-100}{2.5} =$ probability which is M0 |
| $v = 101.325$ | A1 | awrt 101.33 (allow awrt 101.32 from use of calculator) |

**ALT for first two M marks:** $\dfrac{P(V<v)}{P(V<104.9)} = 1 - 0.2801$ scores M1M1, then follow scheme.

**(6 marks)**

**Total: 11 marks**
\begin{enumerate}
  \item A manufacturer fills bottles with oil. The volume of oil in a bottle, $V \mathrm { ml }$, is normally distributed with $V \sim \mathrm {~N} \left( 100,2.5 ^ { 2 } \right)$\\
(a) Find $\mathrm { P } ( V > 104.9 )$\\
(b) In a pack of 150 bottles, find the expected number of bottles containing more than 104.9 ml\\
(c) Find the value of $v$, to 2 decimal places, such that $\mathrm { P } ( V > v \mid V < 104.9 ) = 0.2801$
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2022 Q6 [11]}}
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Q1 11 Q2 14 Q3 14 Q4 11 Q5 14 Q6 11