S2 June 2015 Q4

EdexcelOld spec12 marksContinuous Random Variables

4. 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]

(a) Find \(\mathrm{P}(L \lt 0.4)\). (1)

(b) Write down \(\mathrm{E}(L)\). (1)

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

(d) 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 \(\mathrm{F}(x)\) where

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 0 \\ 4x - 4x^2 & 0 \leqslant x \leqslant 0.5 \\ 1 & \text{otherwise} \end{cases}\]

(e) Using this model, find \(\mathrm{P}(X \gt 0.4)\) (2)

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

(f) Using a suitable approximation, find the probability that at least 8 of these poles have an error of more than 0.4 metres. (3)

Scheme	Marks
0.8	B1

Scheme	Marks
0.25	B1

Mark scheme (c)

Scheme	Marks
\(\dfrac{(0.5 - 0)^2}{12} = \dfrac{1}{48}\) or awrt 0.0208	M1A1

Notes

M1: for \(\dfrac{(0.5 \pm 0)^2}{12}\) or for \(\displaystyle\int_0^{0.5} 2x^2\,\mathrm{d}x - (\text{their (b)})^2\) with some integration \(x^n \to x^{n+1}\)

A1: \(\dfrac{1}{48}\) or awrt 0.0208 or awrt \(2.08 \times 10^{-2}\)

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(L \gt 0.4) = 0.2\) \| \(\mathrm{P}(L \lt 0.4) = 0.8\) \(Y \sim \mathrm{B}(30, 0.2)\) \| \(Y \sim \mathrm{B}(30, 0.8)\) \(\mathrm{P}(Y \leqslant 3) = 0.1227\) \| \(\mathrm{P}(Y \geqslant 4) = 0.1227\)	B1 dM1A1

Notes

An awrt 0.123 award B1 M1 A1

B1: using or writing B(30, their \(\mathrm{P}(L \lt 0.4)\)) or B(30, their \(\mathrm{P}(L \gt 0.4)\)). If they have not written these probabilities in this part use answer from part (a) ie \(\mathrm{P}(L \lt 0.4)\) = (a) or \(\mathrm{P}(L \gt 0.4)\) = 1 - (a)

M1: dependent on previous B mark being awarded. Using B(30, P(\(L\) > 0.4)) with \(\mathrm{P}(Y \leqslant 3)\) written or used Or B(30 P(\(L\) < 0.4)) with \(\mathrm{P}(Y \geqslant 4)\) written or used

A1: awrt 0.123

Mark scheme (e)

Scheme	Marks
\(1 - \left[4 \times 0.4 - 4 \times 0.4^2\right] = \dfrac{1}{25}\) or 0.04	M1A1

Notes

M1: Using 1 - F(0.4) or F(0.5) – F(0.4) or \(\mathrm{P}(X \leqslant 0.5) - \mathrm{P}(X \leqslant 0.4)\). Must see some substitution of 0.4

A1: \(\dfrac{1}{25}\) or 0.04 only

Mark scheme (f)

Scheme	Marks
Po(4)	B1ft
\(\mathrm{P}(X \geqslant 8) = 1 - \mathrm{P}(X \leqslant 7)\)	M1
\(= 1 - 0.9489\) \(= 0.0511\)	A1

Notes

B1ft: using or writing Po(4) NB for ft they must either write 100 × “their 0.04” and use Poison or write Po(“their \(\lambda\)”) Allow P instead of Po

M1 using or writing 1 - \(\mathrm{P}(X \leqslant 7)\) If using normal approximation, they must either write this or \(\dfrac{7.5 - 4}{2}\) or \(\dfrac{7.5 - 4}{\sqrt{3.84}}\) or \(\dfrac{7.5 - 4}{\text{awrt } 1.96}\) or \(\dfrac{7.5 - 20}{\sqrt{16}}\)

A1 awrt 0.0511