OCR S2 2014 June — Question 1 5 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2014
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeSingle probability inequality
DifficultyModerate -0.8 This is a straightforward application of the normal approximation to the binomial distribution with continuity correction. Students need to check np and nq are both >5, calculate mean and variance, apply continuity correction (P(F>40) becomes P(X>40.5)), and use normal tables. It's a standard S2 textbook exercise requiring routine procedure rather than problem-solving.
Spec2.04d Normal approximation to binomial5.04a Linear combinations: E(aX+bY), Var(aX+bY)
1 The random variable \(F\) has the distribution \(B ( 50,0.7 )\). Use a suitable approximation to find \(\mathbf { P } \boldsymbol { ( } \mathbf { F > } \mathbf { 4 0 } \boldsymbol { ) }\). [5]
Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(N(35, 10.5)\)M1 Normal, mean 35
Both parameters correctA1 Allow \(\sqrt{10.5}\) or \(10.5^2\)
\(1 - \Phi\!\left(\dfrac{40.5-35}{\sqrt{10.5}}\right) = 1 - \Phi(1.697)\)M1 Standardise using \(np\), \(npq\); allow no \(\sqrt{}\) or \(10.5^2\); allow wrong or no cc
Both 40.5 and \(\sqrt{npq}\)A1 [Ans 0.0448 or 0.9552 can imply first 4 marks]
\(= 1 - 0.9552 = \mathbf{0.0448}\)A1 Answer a.r.t. 0.045. [Exact binomial (0.040232): 0/5]
[5] 
## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $N(35, 10.5)$ | M1 | Normal, mean 35 |
| Both parameters correct | A1 | Allow $\sqrt{10.5}$ or $10.5^2$ |
| $1 - \Phi\!\left(\dfrac{40.5-35}{\sqrt{10.5}}\right) = 1 - \Phi(1.697)$ | M1 | Standardise using $np$, $npq$; allow no $\sqrt{}$ or $10.5^2$; allow wrong or no cc |
| Both 40.5 and $\sqrt{npq}$ | A1 | [Ans 0.0448 or 0.9552 can imply first 4 marks] |
| $= 1 - 0.9552 = \mathbf{0.0448}$ | A1 | Answer a.r.t. 0.045. [Exact binomial (0.040232): 0/5] |
| | **[5]** | |

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1 The random variable $F$ has the distribution $B ( 50,0.7 )$. Use a suitable approximation to find $\mathbf { P } \boldsymbol { ( } \mathbf { F > } \mathbf { 4 0 } \boldsymbol { ) }$. [5]

\hfill \mbox{\textit{OCR S2 2014 Q1 [5]}}
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Q1 5 Q2 7 Q3 7 Q4 7 Q5 13 Q6 12 Q7 15 Q8 6