Question 2 - A-Level Maths

Edexcel S2 2022 June — Question 2 11 marks

Exam Board	Edexcel
Module	S2 (Statistics 2)
Year	2022
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Find expectation E(X)
Difficulty	Standard +0.3 This is a standard S2 continuous probability question requiring routine integration to find E(T), verification of a probability via integration, and a normal approximation to binomial. All techniques are textbook exercises with no novel insight required, making it slightly easier than average for A-level.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.05a Sample mean distribution: central limit theorem

The time, in minutes, spent waiting for a call to a call centre to be answered is modelled by the random variable $T$ with probability density function

$$f ( t ) = \left\{ \begin{array} { l c } \frac { 1 } { 192 } \left( t ^ { 3 } - 48 t + 128 \right) & 0 \leqslant t \leqslant 4 \\ 0 & \text { otherwise } \end{array} \right.$$

Use algebraic integration to find, in minutes and seconds, the mean waiting time.
Show that $\mathrm { P } ( 1 < T < 3 ) = \frac { 7 } { 16 }$ A supervisor randomly selects 256 calls to the call centre.
Use a suitable approximation to find the probability that more than 125 of these calls take between 1 and 3 minutes to be answered.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) $E(T) = \int_0^4 \frac{1}{192}(t^3 - 48t + 128) dt$	M1
$= \frac{1}{192}\left[\frac{t^5}{5} - 16t^3 + 64t^2\right]_0^4$ or $\left[\frac{t^5}{960} - \frac{1}{12}t^3 + \frac{1}{3}t^2\right]_0^4$ oe	dM1
$= \frac{1}{192}\left(\frac{4^5}{5} - 16(4^3) + 64(4^2) - 0\right) = \frac{16}{15}$ min $→ 1$ minute $4$ seconds	A1	1st M1 for using $\int tf(t)dt$ ignore limits. $t^4 → t^5$ or $t^2 → t^3$ or $t → t^2$ for at least one term, ignore coefficients. Implied by an answer of $\frac{16}{15}$ or $1$ minute $4$ seconds (allow 64s) or awrt 1.067. NB an answer of $\frac{16}{15}$ or $1$ minute $4$ seconds or 64 or awrt 1.067 with no working gains M1M0A0.
(b) $P(\text{call takes between 1 and 3 minutes}) = \int_1^3 \frac{1}{192}(t^3 - 48t + 128) dt$	M1
mor $\left[\frac{t^4}{768} - \frac{1}{8}t^2 + \frac{2}{3}t\right]_1^3$ oe	M1
$= \frac{1}{192}\left(\left(\frac{3^4}{4} - 24(3^2) + 128(3)\right) - \left(\frac{1^4}{4} - 24(1^2) + 128(1)\right)\right) = \frac{7}{16}$ *	dM1, A1*cso	1st M1 attempt to integrate $\int f(t)dt$ $t^n → t^{n+1}$ for at least one term. Ignore limits. If they have integrated $f(t)$ in part (a) and used this in part (b) we will allow this mark. 2nd M1 (dep on 1st M1) for use of correct limits. Must see substitution into their expression. If integration correct allow $\frac{1}{192}\left(\frac{81}{4} - 216 + 384\right) - \frac{1}{192}\left(\frac{1}{4} - 24 + 128\right)$ or $\frac{1}{192}\left(\frac{753}{4} - \frac{417}{4}\right)$ or $\frac{251}{139}$ or $\frac{256}{256}$. 1st A1* cso $\frac{7}{16} = [0.4375]$ fully correct solution (correct integration and substitution). Answer is given so both method marks must be awarded.
(c) $C \sim B(256, \frac{7}{16}) \approx N(112, 63)$	M1, A1
$P(C > 125) \approx P\left(Z > \frac{125.5 - 112}{\sqrt{63}}\right)$	M1M1
$P(Z > 1.70) = 1 - 0.9554 = 0.0446$	A1	1st M1 use or sight of Normal approximation with mean 112. 1st A1 correct mean and variance (condone 63² if used $\sqrt{63}$ in the standardisation). 2nd M1 standardising using their mean and variance. Allow use of 124.5, 125, 125.5, 126, 126.5 or on the numerator 12.5, 13, 13.5, 14, 14.5. 3rd M1 use of continuity correction 125 ± 0.5 Implied by numerator of 12.5 or 13.5. 2nd A1 awrt 0.0445/0.0446 [calc 0.0444865...]. [Exact binomial gives 0.0448518... and gains no marks]

[11 marks]

**(a)** $E(T) = \int_0^4 \frac{1}{192}(t^3 - 48t + 128) dt$ | M1 |

$= \frac{1}{192}\left[\frac{t^5}{5} - 16t^3 + 64t^2\right]_0^4$ or $\left[\frac{t^5}{960} - \frac{1}{12}t^3 + \frac{1}{3}t^2\right]_0^4$ oe | dM1 |

$= \frac{1}{192}\left(\frac{4^5}{5} - 16(4^3) + 64(4^2) - 0\right) = \frac{16}{15}$ min $→ 1$ minute $4$ seconds | A1 | 1st M1 for using $\int tf(t)dt$ ignore limits. $t^4 → t^5$ or $t^2 → t^3$ or $t → t^2$ for at least one term, ignore coefficients. Implied by an answer of $\frac{16}{15}$ or $1$ minute $4$ seconds (allow 64s) or awrt 1.067. NB an answer of $\frac{16}{15}$ or $1$ minute $4$ seconds or 64 or awrt 1.067 with no working gains M1M0A0.

**(b)** $P(\text{call takes between 1 and 3 minutes}) = \int_1^3 \frac{1}{192}(t^3 - 48t + 128) dt$ | M1 |

mor $\left[\frac{t^4}{768} - \frac{1}{8}t^2 + \frac{2}{3}t\right]_1^3$ oe | M1 |

$= \frac{1}{192}\left(\left(\frac{3^4}{4} - 24(3^2) + 128(3)\right) - \left(\frac{1^4}{4} - 24(1^2) + 128(1)\right)\right) = \frac{7}{16}$ * | dM1, A1*cso | 1st M1 attempt to integrate $\int f(t)dt$ $t^n → t^{n+1}$ for at least one term. Ignore limits. If they have integrated $f(t)$ in part (a) and used this in part (b) we will allow this mark. 2nd M1 (dep on 1st M1) for use of correct limits. Must see substitution into their expression. If integration correct allow $\frac{1}{192}\left(\frac{81}{4} - 216 + 384\right) - \frac{1}{192}\left(\frac{1}{4} - 24 + 128\right)$ or $\frac{1}{192}\left(\frac{753}{4} - \frac{417}{4}\right)$ or $\frac{251}{139}$ or $\frac{256}{256}$. 1st A1* cso $\frac{7}{16} = [0.4375]$ fully correct solution (correct integration and substitution). Answer is given so both method marks must be awarded.

**(c)** $C \sim B(256, \frac{7}{16}) \approx N(112, 63)$ | M1, A1 |

$P(C > 125) \approx P\left(Z > \frac{125.5 - 112}{\sqrt{63}}\right)$ | M1M1 |

$P(Z > 1.70) = 1 - 0.9554 = 0.0446$ | A1 | 1st M1 use or sight of Normal approximation with mean 112. 1st A1 correct mean and variance (condone 63² if used $\sqrt{63}$ in the standardisation). 2nd M1 standardising using their mean and variance. Allow use of 124.5, 125, 125.5, 126, 126.5 or on the numerator 12.5, 13, 13.5, 14, 14.5. 3rd M1 use of continuity correction 125 ± 0.5 Implied by numerator of 12.5 or 13.5. 2nd A1 awrt 0.0445/0.0446 [calc 0.0444865...]. [Exact binomial gives 0.0448518... and gains no marks]

**[11 marks]**

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Show LaTeX source

\begin{enumerate}
  \item The time, in minutes, spent waiting for a call to a call centre to be answered is modelled by the random variable $T$ with probability density function
\end{enumerate}

$$f ( t ) = \left\{ \begin{array} { l c } 
\frac { 1 } { 192 } \left( t ^ { 3 } - 48 t + 128 \right) & 0 \leqslant t \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$

(a) Use algebraic integration to find, in minutes and seconds, the mean waiting time.\\
(b) Show that $\mathrm { P } ( 1 < T < 3 ) = \frac { 7 } { 16 }$

A supervisor randomly selects 256 calls to the call centre.\\
(c) Use a suitable approximation to find the probability that more than 125 of these calls take between 1 and 3 minutes to be answered.

\hfill \mbox{\textit{Edexcel S2 2022 Q2 [11]}}

This paper (7 questions)

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Q1 10 Q2 11 Q3 10 Q4 9 Q5 10 Q6 13 Q7 12

Answer	Marks	Guidance
(a) \(E(T) = \int_0^4 \frac{1}{192}(t^3 - 48t + 128) dt\)	M1
\(= \frac{1}{192}\left[\frac{t^5}{5} - 16t^3 + 64t^2\right]_0^4\) or \(\left[\frac{t^5}{960} - \frac{1}{12}t^3 + \frac{1}{3}t^2\right]_0^4\) oe	dM1
\(= \frac{1}{192}\left(\frac{4^5}{5} - 16(4^3) + 64(4^2) - 0\right) = \frac{16}{15}\) min \(→ 1\) minute \(4\) seconds	A1	1st M1 for using \(\int tf(t)dt\) ignore limits. \(t^4 → t^5\) or \(t^2 → t^3\) or \(t → t^2\) for at least one term, ignore coefficients. Implied by an answer of \(\frac{16}{15}\) or \(1\) minute \(4\) seconds (allow 64s) or awrt 1.067. NB an answer of \(\frac{16}{15}\) or \(1\) minute \(4\) seconds or 64 or awrt 1.067 with no working gains M1M0A0.
(b) \(P(\text{call takes between 1 and 3 minutes}) = \int_1^3 \frac{1}{192}(t^3 - 48t + 128) dt\)	M1
mor \(\left[\frac{t^4}{768} - \frac{1}{8}t^2 + \frac{2}{3}t\right]_1^3\) oe	M1
\(= \frac{1}{192}\left(\left(\frac{3^4}{4} - 24(3^2) + 128(3)\right) - \left(\frac{1^4}{4} - 24(1^2) + 128(1)\right)\right) = \frac{7}{16}\) *	dM1, A1*cso	1st M1 attempt to integrate \(\int f(t)dt\) \(t^n → t^{n+1}\) for at least one term. Ignore limits. If they have integrated \(f(t)\) in part (a) and used this in part (b) we will allow this mark. 2nd M1 (dep on 1st M1) for use of correct limits. Must see substitution into their expression. If integration correct allow \(\frac{1}{192}\left(\frac{81}{4} - 216 + 384\right) - \frac{1}{192}\left(\frac{1}{4} - 24 + 128\right)\) or \(\frac{1}{192}\left(\frac{753}{4} - \frac{417}{4}\right)\) or \(\frac{251}{139}\) or \(\frac{256}{256}\). 1st A1* cso \(\frac{7}{16} = [0.4375]\) fully correct solution (correct integration and substitution). Answer is given so both method marks must be awarded.
(c) \(C \sim B(256, \frac{7}{16}) \approx N(112, 63)\)	M1, A1
\(P(C > 125) \approx P\left(Z > \frac{125.5 - 112}{\sqrt{63}}\right)\)	M1M1
\(P(Z > 1.70) = 1 - 0.9554 = 0.0446\)	A1	1st M1 use or sight of Normal approximation with mean 112. 1st A1 correct mean and variance (condone 63² if used \(\sqrt{63}\) in the standardisation). 2nd M1 standardising using their mean and variance. Allow use of 124.5, 125, 125.5, 126, 126.5 or on the numerator 12.5, 13, 13.5, 14, 14.5. 3rd M1 use of continuity correction 125 ± 0.5 Implied by numerator of 12.5 or 13.5. 2nd A1 awrt 0.0445/0.0446 [calc 0.0444865...]. [Exact binomial gives 0.0448518... and gains no marks]