Edexcel S2 2015 June — Question 4 12 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeMultiple observations or trials
DifficultyModerate -0.3 This is a straightforward S2 question testing standard uniform distribution formulas and binomial/normal approximation. Parts (a)-(c) are direct formula applications, part (d) is routine binomial calculation, part (e) requires reading from a CDF, and part (f) is standard normal approximation to binomial. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec2.04c Calculate binomial probabilities2.04d Normal approximation to binomial5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration
The continuous random variable \(L\) represents the error, in metres, made when a machine cuts poles to a target length. The distribution of \(L\) is a continuous uniform distribution over the interval [0, 0.5]
  1. Find P(\(L < 0.4\)). [1]
  2. Write down E(\(L\)). [1]
  3. Calculate Var(\(L\)). [2]
A random sample of 30 poles cut by this machine is taken.
  1. Find the probability that fewer than 4 poles have an error of more than 0.4 metres from the target length. [3]
When a new machine cuts poles to a target length, the error, \(X\) metres, is modelled by the cumulative distribution function F(\(x\)) where $$\text{F}(x) = \begin{cases} 0 & x < 0 \\ 4x - 4x^2 & 0 \leq x \leq 0.5 \\ 1 & \text{otherwise} \end{cases}$$
  1. Using this model, find P(\(X > 0.4\)) [2]
A random sample of 100 poles cut by this new machine is taken.
  1. Using a suitable approximation, find the probability that at least 8 of these poles have an error of more than 0.4 metres. [3]
Question 4:
AnswerMarks Guidance
4(a)0.8 B1: cao
(b)0.25 B1: cao
(c)0.502
1
 or awrt 0.0208
AnswerMarks
12 480.502
M1: for or for
12
0.5
 2x2dx(their (b))2 with some
0
xn xn1
AnswerMarks
integrationM1A1
1
A1: or awrt 0.0208or
48
awrt 2.08 102
AnswerMarks Guidance
(d)P( L > 0.4) = 0.2 P( L < 0.4) = 0.8
dM1A1
AnswerMarks Guidance
Y~ B(30, 0.2)Y~ B(30, 0.8) B1: using or writing
B(30, their P(L < 0.4) or
B(30, their P(L > 0.4) . If they have
not written these probabilities in this
part use answer from part (a) ie P(L <
0.4) = (a) or
P( L > 0.4) = 1- (a)
AnswerMarks Guidance
P(Y 3) = 0.1227P(Y 4) = 0.1227 M1: dependent on previous B mark
being awarded. Using
B(30,P(L>0.4) with P(Y 3) written
or used
Or
B(30 P(L<0.4)) with P(Y 4) written
or used
A1: awrt 0.123
AnswerMarks
(e)1
140.440.42  or 0.04
 
AnswerMarks
25M1: Using 1- F(0.4)
or F(0.5) – F(0.4)
or P(X 0.5)PX 0.4.
Must see some substitution of 0.4
1
A1: or 0.04 only
AnswerMarks Guidance
25M1A1
(f)Po(4) B1ft: using or writing Po(4)
NB for ft they must either write 100 
“their 0.04” and use Poison or write
AnswerMarks
Po(“their ”) Allow P instead of PoB1ft
P(X 8) = 1- P( X7)
= 1 – 0.9489
AnswerMarks
= 0.0511M1 using or writing 1- P( X7)
If using normal approximation, they
7.54
must either write this or or
2
7.54 7.54 7.520
or or
AnswerMarks
3.84 awrt 1.96 16M1
A1
A1 awrt 0.0511
Notes
Question 4:
--- 4(a) ---
4(a) | 0.8 | B1: cao | B1
(b) | 0.25 | B1: cao | B1
(c) | 0.502
1
 or awrt 0.0208
12 48 | 0.502
M1: for or for
12
0.5
 2x2dx(their (b))2 with some
0
xn xn1
integration | M1A1
1
A1: or awrt 0.0208or
48
awrt 2.08 102
(d) | P( L > 0.4) = 0.2 | P( L < 0.4) = 0.8 | An awrt 0.123 award B1 M1 A1 | B1
dM1A1
Y~ B(30, 0.2) | Y~ B(30, 0.8) | B1: using or writing
B(30, their P(L < 0.4) or
B(30, their P(L > 0.4) . If they have
not written these probabilities in this
part use answer from part (a) ie P(L <
0.4) = (a) or
P( L > 0.4) = 1- (a)
P(Y 3) = 0.1227 | P(Y 4) = 0.1227 | M1: dependent on previous B mark
being awarded. Using
B(30,P(L>0.4) with P(Y 3) written
or used
Or
B(30 P(L<0.4)) with P(Y 4) written
or used
A1: awrt 0.123
(e) | 1
140.440.42  or 0.04
 
25 | M1: Using 1- F(0.4)
or F(0.5) – F(0.4)
or P(X 0.5)PX 0.4.
Must see some substitution of 0.4
1
A1: or 0.04 only
25 | M1A1
(f) | Po(4) | B1ft: using or writing Po(4)
NB for ft they must either write 100 
“their 0.04” and use Poison or write
Po(“their ”) Allow P instead of Po | B1ft
P(X 8) = 1- P( X7)
= 1 – 0.9489
= 0.0511 | M1 using or writing 1- P( X7)
If using normal approximation, they
7.54
must either write this or or
2
7.54 7.54 7.520
or or
3.84 awrt 1.96 16 | M1
A1
A1 awrt 0.0511
Notes
The continuous random variable $L$ represents the error, in metres, made when a machine cuts poles to a target length. The distribution of $L$ is a continuous uniform distribution over the interval [0, 0.5]

\begin{enumerate}[label=(\alph*)]
\item Find P($L < 0.4$). [1]

\item Write down E($L$). [1]

\item Calculate Var($L$). [2]
\end{enumerate}

A random sample of 30 poles cut by this machine is taken.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the probability that fewer than 4 poles have an error of more than 0.4 metres from the target length. [3]
\end{enumerate}

When a new machine cuts poles to a target length, the error, $X$ metres, is modelled by the cumulative distribution function F($x$) where

$$\text{F}(x) = \begin{cases}
0 & x < 0 \\
4x - 4x^2 & 0 \leq x \leq 0.5 \\
1 & \text{otherwise}
\end{cases}$$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Using this model, find P($X > 0.4$) [2]
\end{enumerate}

A random sample of 100 poles cut by this new machine is taken.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{5}
\item Using a suitable approximation, find the probability that at least 8 of these poles have an error of more than 0.4 metres. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2015 Q4 [12]}}
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