Pre-U Pre-U 9795/2 2016 June — Question 5 10 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2016
SessionJune
Marks10
TopicApproximating Binomial to Normal Distribution
TypeFind parameter from normal approximation
DifficultyChallenging +1.2 This question requires knowledge of normal approximation conditions, applying continuity correction, and using inverse normal tables to set up simultaneous equations. While it involves multiple steps and working backwards from probabilities to find parameters, the techniques are standard for this topic and the algebraic manipulation is straightforward once the equations are established.
Spec5.02d Binomial: mean np and variance np(1-p)5.05a Sample mean distribution: central limit theorem

5 The random variable \(R\) has the distribution \(\mathrm { B } ( n , p )\).
  1. State two conditions that \(n\) and \(p\) must satisfy if the distribution of \(R\) can be well approximated by a normal distribution. Assume now that these conditions hold. Using the normal approximation, it is given that \(\mathrm { P } ( R < 25 ) = 0.8282\) and \(\mathrm { P } ( R \geqslant 28 ) = 0.0393\), correct to 4 decimal places.
  2. Find the mean and standard deviation of the approximating normal distribution.
  3. Hence find the value of \(p\) and the value of \(n\).

(i) \(n\) large; \(p\) close to \(\tfrac{1}{2}\)
- B1: Or \(np > 5\)
- B1: \(nq > 5\) [*not* \(npq > 5\)]
[2]
(ii) \(\dfrac{24.5-\mu}{\sigma} = \Phi^{-1}(0.8282) = 0.947\)
\(\dfrac{27.5-\mu}{\sigma} = \Phi^{-1}(0.9697) = 1.759\)
\(\mu = \mathbf{21},\ \sigma = \mathbf{3.69}\)
- M1: One standardised, \(= \Phi^{-1}\), allow \(\sigma^2\), cc, \(1-\) errors
- A1: LHS of both equations correct including signs and cc
- B1: Both \(z\)-values correct to 3 sf, \(\pm 1\) in third dp
- M1: Solve to find both \(\mu\) and \(\sigma\)
- A1: \(\mu\), a.r.t. 21.0; \(\sigma\), in range \([3.69, 3.70]\)
[5]
(iii) \(q = npq/np = 21/3.694^2\ [= 0.65]\)
\(p = \mathbf{0.35},\ n = \mathbf{60}\)
- M1: Correct method of solution for \(n\), \(p\) or \(q\), allow \(\sqrt{npq}\)
- A1ft: \(npq = \sigma^2\) [not \(\sigma\)], ft on their \(npq\) [13.65]
- A1: \(p\), a.r.t. 0.350 and \(n = 60\) [integer] *only* [not 60.0]
[3]
(i) $n$ large; $p$ close to $\tfrac{1}{2}$

- B1: Or $np > 5$
- B1: $nq > 5$ [*not* $npq > 5$]

**[2]**

(ii) $\dfrac{24.5-\mu}{\sigma} = \Phi^{-1}(0.8282) = 0.947$

$\dfrac{27.5-\mu}{\sigma} = \Phi^{-1}(0.9697) = 1.759$

$\mu = \mathbf{21},\ \sigma = \mathbf{3.69}$

- M1: One standardised, $= \Phi^{-1}$, allow $\sigma^2$, cc, $1-$ errors
- A1: LHS of both equations correct including signs and cc
- B1: Both $z$-values correct to 3 sf, $\pm 1$ in third dp
- M1: Solve to find both $\mu$ and $\sigma$
- A1: $\mu$, a.r.t. 21.0; $\sigma$, in range $[3.69, 3.70]$

**[5]**

(iii) $q = npq/np = 21/3.694^2\ [= 0.65]$

$p = \mathbf{0.35},\ n = \mathbf{60}$

- M1: Correct method of solution for $n$, $p$ or $q$, allow $\sqrt{npq}$
- A1ft: $npq = \sigma^2$ [not $\sigma$], ft on their $npq$ [13.65]
- A1: $p$, a.r.t. 0.350 and $n = 60$ [integer] *only* [not 60.0]

**[3]**
5 The random variable $R$ has the distribution $\mathrm { B } ( n , p )$.\\
(i) State two conditions that $n$ and $p$ must satisfy if the distribution of $R$ can be well approximated by a normal distribution.

Assume now that these conditions hold. Using the normal approximation, it is given that $\mathrm { P } ( R < 25 ) = 0.8282$ and $\mathrm { P } ( R \geqslant 28 ) = 0.0393$, correct to 4 decimal places.\\
(ii) Find the mean and standard deviation of the approximating normal distribution.\\
(iii) Hence find the value of $p$ and the value of $n$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q5 [10]}}