Pre-U Pre-U 9795/2 2019 Specimen — Question 2 5 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2019
SessionSpecimen
Marks5
TopicApproximating Binomial to Normal Distribution
TypeFind minimum/maximum n for probability condition
DifficultyStandard +0.3 This is a straightforward application of normal approximation to binomial/Poisson distributions with standard continuity correction. Part (a) requires routine calculation with justification of approximation conditions (np and nq both >5), while part (b) involves finding N from a probability inequality using inverse normal tables. Both parts are standard textbook exercises requiring no novel insight, though slightly above average due to the two-part structure and need to work with Poisson approximation in part (b).
Spec5.01a Permutations and combinations: evaluate probabilities5.02n Sum of Poisson variables: is Poisson

2
  1. The probability that a shopper obtains a parking space on the river embankment on any given Saturday morning is 0.2 . Using a suitable normal approximation, find the probability that, over a period of 100 Saturday mornings, the shopper finds a parking space at least 15 times. Justify the use of the normal approximation in this case.
  2. The number of parking tickets that a traffic warden issues on the river embankment during the course of a week has a Poisson distribution with mean 36 . The probability that the traffic warden issues more than \(N\) parking tickets is less than 0.05 . Using a suitable normal approximation, find the least possible value of \(N\).

Question 2(a)
\(np = 100 \times \frac{1}{5} = 20\) and \(npq = 20 \times \frac{4}{5} = 16\) [B1]
Standardisation \(z = \frac{14.5 - 20}{4} = -1.375\) [M1]
\(\Rightarrow P(\geqslant 15) = 0.915\) (within range [0.915, 0.916]) (ft on variance) [A1A1, A1ftA1]
Justified as \(np = 20 > 5\) and \(nq = 80 > 5\) OR \(n\) large, \(p\) not too far from \(\frac{1}{2}\) [B1]
Total: 5 marks
Question 2(b)
mean = variance = \(36 \Rightarrow\) standard deviation \(= 6\) [B1]
\(z = 1.645\) [B1]
\(\dfrac{\left(N + \frac{1}{2}\right) - 36}{6} > 1.645\) [M1A1]
(Allow working with equality, but must be \(\left(N + \frac{1}{2}\right)\))
\(\Rightarrow N > 45.37\) \(\therefore\) least \(N = 46\) [A1]
Total: 5 marks
**Question 2(a)**

$np = 100 \times \frac{1}{5} = 20$ and $npq = 20 \times \frac{4}{5} = 16$ [B1]

Standardisation $z = \frac{14.5 - 20}{4} = -1.375$ [M1]

$\Rightarrow P(\geqslant 15) = 0.915$ (within range [0.915, 0.916]) (**ft** on variance) [A1A1, A1ftA1]

Justified as $np = 20 > 5$ and $nq = 80 > 5$ **OR** $n$ large, $p$ not too far from $\frac{1}{2}$ [B1]

**Total: 5 marks**

**Question 2(b)**

mean = variance = $36 \Rightarrow$ standard deviation $= 6$ [B1]

$z = 1.645$ [B1]

$\dfrac{\left(N + \frac{1}{2}\right) - 36}{6} > 1.645$ [M1A1]

(Allow working with equality, but must be $\left(N + \frac{1}{2}\right)$)

$\Rightarrow N > 45.37$ $\therefore$ least $N = 46$ [A1]

**Total: 5 marks**
2
\begin{enumerate}[label=(\alph*)]
\item The probability that a shopper obtains a parking space on the river embankment on any given Saturday morning is 0.2 . Using a suitable normal approximation, find the probability that, over a period of 100 Saturday mornings, the shopper finds a parking space at least 15 times. Justify the use of the normal approximation in this case.
\item The number of parking tickets that a traffic warden issues on the river embankment during the course of a week has a Poisson distribution with mean 36 . The probability that the traffic warden issues more than $N$ parking tickets is less than 0.05 . Using a suitable normal approximation, find the least possible value of $N$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q2 [5]}}