Edexcel S2 2010 June — Question 5 15 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2010
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson hypothesis test
DifficultyStandard +0.3 This is a straightforward application of standard Poisson distribution techniques: part (a) is bookwork recall of conditions, part (b) involves routine probability calculations with parameter scaling, and part (c) is a standard hypothesis test using normal approximation. All steps follow textbook procedures with no novel problem-solving required, making it slightly easier than average for S2.
Spec2.04d Normal approximation to binomial2.05a Hypothesis testing language: null, alternative, p-value, significance

A company has a large number of regular users logging onto its website. On average 4 users every hour fail to connect to the company's website at their first attempt.
  1. Explain why the Poisson distribution may be a suitable model in this case. [1]
Find the probability that, in a randomly chosen 2 hour period,
    1. all users connect at their first attempt,
    2. at least 4 users fail to connect at their first attempt.
    [5]
The company suffered from a virus infecting its computer system. During this infection it was found that the number of users failing to connect at their first attempt, over a 12 hour period, was 60.
  1. Using a suitable approximation, test whether or not the mean number of users per hour who failed to connect at their first attempt had increased. Use a 5\% level of significance and state your hypotheses clearly. [9]

AnswerMarks Guidance
(a) Connecting occurs at random/independently, singly or at a constant rateB1 (1)
(b) \(Po(8)\)B1 (5)
(i) \(P(X = 0) = 0.0003\)M1A1; M1
(ii) \(P(X > 4) = 1 - P(X \leq 3)\)
AnswerMarks
\(= 1 - 0.0424\)M1; A1
\(= 0.9576\)
(c) \(H_0: \lambda = 4\) (48) \(H_1: \lambda > 4\) (48)
AnswerMarks Guidance
\(N(48, 48)\)B1; M1 A1 (9)
Method 1Method 2
\(P(X \geq 59.5) = P\left(Z \geq \frac{59.5 - 48}{\sqrt{48}}\right)\)\(\frac{x - 0.5 - 48}{\sqrt{48}} = 1.6449\)
\(= P(Z \geq 1.66)\)\(x = 59.9\)
\(= 1 - 0.9515\)
\(= 0.0485\)
M1 M1 A1; M1 M1 A1 M1 for attempting a continuity correction (Method 1: 60 + 0.5 / Method 2: \(x + 0.5\)); M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5: accept ± z.] Method 2 [ (x±0.5) and equal to a ± z value]; A1 correct z value awrt ±1.66 or \(\frac{59.5 - 48}{\sqrt{48}}\) or \(\frac{x - 0.5 - 48}{\sqrt{48}}\) = 1.6449; A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59.9
0.0485 < 0.05M1; A1 ft (9)
Reject \(H_0\). Significant. 60 lies in the Critical regionM1 A1 ft
The number of failed connections at the first attempt has increased.
Notes
(a) B1 Any one of randomly/independently/singly/constant rate. Must have context of connection/logging on/fail
(b) B1 Writing or using \(Po(8)\) in (i) or (ii)
(i) M1 for writing or finding \(P(X = 0)\); A1 awrt 0.0003
(ii) M1 for writing or finding \(1 - P(X \leq 3)\); A1 awrt 0.958
(c) B1 both hypotheses correct. Must use \(\lambda\) or \(\mu\); M1 identifying normal; A1 using or seeing mean and variance of 48. These first two marks may be given if the following are seen in the standardisation formula: 48 and \(\sqrt{48}\) or awrt 6.93; M1 for attempting a continuity correction (Method 1: 60 ± 0.5 / Method 2: \(x \pm 0.5\)); M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5: accept ± z.] Method 2 [ (x±0.5) and equal to a ± z value]; A1 correct z value awrt ±1.66 or \(\frac{59.5 - 48}{\sqrt{48}}\), or \(\frac{x - 0.5 - 48}{\sqrt{48}} = 1.6449\); A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59.9; M1 for "reject \(H_0\)" or "significant" maybe implied by "correct contextual comment"; If one tail hypotheses given follow through "their prob" and 0.05, \(p < 0.5\); If two tail hypotheses given follow through "their prob" with 0.025, \(p < 0.5\); If one tail hypotheses given follow through "their prob" and 0.95, \(p > 0.5\); If two tail hypotheses given follow through "their prob" with 0.975, \(p > 0.5\); If no \(H_1\) given they get M0; A1 ft correct contextual statement followed through from their prob and \(H_1\). Need faulty bolts and decreased. Allow "there are more failed connections". NB A correct contextual statement alone followed through from their prob and \(H_1\) gets M1 A1
**(a)** Connecting occurs at random/independently, singly or at a constant rate | B1 | (1) | Any one of randomly/independently/singly/constant rate. Must have context of connection/logging on/fail

**(b)** $Po(8)$ | B1 | (5) | B1 Writing or using $Po(8)$ in (i) or (ii)

**(i)** $P(X = 0) = 0.0003$ | M1A1; M1 | | M1 for writing or finding $P(X = 0)$; A1 awrt 0.0003

**(ii)** $P(X > 4) = 1 - P(X \leq 3)$

$= 1 - 0.0424$ | M1; A1 | |

$= 0.9576$ | | |

**(c)** $H_0: \lambda = 4$ (48) $H_1: \lambda > 4$ (48)

$N(48, 48)$ | B1; M1 A1 | (9) | B1 both hypotheses correct. Must use $\lambda$ or $\mu$; M1 identifying normal; A1 using or seeing mean and variance of 48. These first two marks may be given if the following are seen in the standardisation formula: 48 and $\sqrt{48}$ or awrt 6.93

**Method 1** | **Method 2** |
|---|---|
| $P(X \geq 59.5) = P\left(Z \geq \frac{59.5 - 48}{\sqrt{48}}\right)$ | $\frac{x - 0.5 - 48}{\sqrt{48}} = 1.6449$ |
| $= P(Z \geq 1.66)$ | $x = 59.9$ |
| $= 1 - 0.9515$ | |
| $= 0.0485$ | |

| M1 M1 A1; M1 M1 A1 | | M1 for attempting a continuity correction (Method 1: 60 + 0.5 / Method 2: $x + 0.5$); M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5: accept ± z.] Method 2 [ (x±0.5) and equal to a ± z value]; **A1** correct z value awrt ±1.66 or $\frac{59.5 - 48}{\sqrt{48}}$ or $\frac{x - 0.5 - 48}{\sqrt{48}}$ = 1.6449; A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59.9

0.0485 < 0.05 | M1; A1 ft | (9) |

Reject $H_0$. Significant. 60 lies in the Critical region | M1 A1 ft | |

The number of failed connections at the first attempt has increased. | | |

**Notes**

**(a)** B1 Any one of randomly/independently/singly/constant rate. Must have context of connection/logging on/fail

**(b)** B1 Writing or using $Po(8)$ in (i) or (ii)

**(i)** M1 for writing or finding $P(X = 0)$; A1 awrt 0.0003

**(ii)** M1 for writing or finding $1 - P(X \leq 3)$; A1 awrt 0.958

**(c)** B1 both hypotheses correct. Must use $\lambda$ or $\mu$; M1 identifying normal; A1 using or seeing mean and variance of 48. These first two marks may be given if the following are seen in the standardisation formula: 48 and $\sqrt{48}$ or awrt 6.93; M1 for attempting a continuity correction (Method 1: 60 ± 0.5 / Method 2: $x \pm 0.5$); M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5: accept ± z.] Method 2 [ (x±0.5) and equal to a ± z value]; **A1** correct z value awrt ±1.66 or $\frac{59.5 - 48}{\sqrt{48}}$, or $\frac{x - 0.5 - 48}{\sqrt{48}} = 1.6449$; A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59.9; M1 for "reject $H_0$" or "significant" maybe implied by "correct contextual comment"; If one tail hypotheses given follow through "their prob" and 0.05, $p < 0.5$; If two tail hypotheses given follow through "their prob" with 0.025, $p < 0.5$; If one tail hypotheses given follow through "their prob" and 0.95, $p > 0.5$; If two tail hypotheses given follow through "their prob" with 0.975, $p > 0.5$; If no $H_1$ given they get M0; **A1 ft** correct contextual statement followed through from their prob and $H_1$. Need faulty bolts and decreased. Allow "there are more failed connections". **NB** A correct contextual statement alone followed through from their prob and $H_1$ gets M1 A1

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A company has a large number of regular users logging onto its website. On average 4 users every hour fail to connect to the company's website at their first attempt.

\begin{enumerate}[label=(\alph*)]
\item Explain why the Poisson distribution may be a suitable model in this case. [1]
\end{enumerate}

Find the probability that, in a randomly chosen 2 hour period,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item 
\begin{enumerate}[label=(\roman*)]
\item all users connect at their first attempt,
\item at least 4 users fail to connect at their first attempt.
\end{enumerate}
[5]
\end{enumerate}

The company suffered from a virus infecting its computer system. During this infection it was found that the number of users failing to connect at their first attempt, over a 12 hour period, was 60.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Using a suitable approximation, test whether or not the mean number of users per hour who failed to connect at their first attempt had increased. Use a 5\% level of significance and state your hypotheses clearly. [9]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2010 Q5 [15]}}