S2 June 2016 Q5
5. In a large school, 20% of students own a touch screen laptop. A random sample of \(n\) students is chosen from the school. Using a normal approximation, the probability that more than 55 of these \(n\) students own a touch screen laptop is 0.0401 correct to 3 significant figures.
Find the value of \(n\). (8)
| Scheme | Marks |
|---|---|
| \(\mathrm{N}(0.2n, 0.16n)\) | B1 |
| \(\mathrm{P}\left(Z \gt \dfrac{55.5 - 0.2n}{\sqrt{0.16n}}\right) = 0.0401\) | M1 |
| \(\dfrac{55.5 - 0.2n}{\sqrt{0.16n}} = 1.75\) | B1M1A1 |
| \(0.2n + 0.7\sqrt{n} - 55.5 = 0\) | M1d |
| \(\sqrt{n} = 15\) | A1 |
| \(n = 225\) | A1 |
| (8) |
Notes
B1: Mean = \(0.2n\) and Var = \(0.16n\) oe this may be awarded if they appear in the standardisation as \(0.2n\) and either \(0.16n\) or \(\sqrt{0.16n}\)
M1: Using a continuity correction either 55.5 or 54.5
B1: Using a \(z\) = awrt \(\pm\) 1.75
M1: Standardising using either 55.5, 54.5 or 55 and equal to a \(z\) value. Follow through their mean and variance. If they have not given the mean and Var earlier then they must be correct
A1: A correct equation. May be awarded for \(\dfrac{55.5 - 0.2n}{\sqrt{0.16n}} = 1.75\) Condone use of an inequality sign rather than an equals sign
M1d: This is dependent on the previous method mark being awarded. Using either the quadratic formula or completing the square or factorising or any correct method to solve their 3 term equation. If they write the formula down then allow a slip. If no formula written down then it must be correct for their equation. May be implied by correct answer or \(\sqrt{n} = 15\) or 342.25
NB you may award this mark if they use 54.5 for awrt 14.9, -18.4, 221 or 337; 55 for awrt -18.4, 14.9, 223 or -117
If the answer is not one of these then the method for solving their 3 term equation must be seen.
A1: Allow 15 or -18.5 do not need to see \(n\) or \(\sqrt{n}\). Condone \(n = 15\) or \(n = -18.5\)
A1: cao 225 do not need to see \(n\) or \(\sqrt{n}\)
Alternative method for last 3 marks
| Scheme | Marks |
|---|---|
| \((0.2n - 55.5)^2 = \left(-0.7\sqrt{n}\right)^2\) \(0.04n^2 - 22.69n + 3080.25 = 0\) \(n = 225\) or 1369/4 \(n = 225\) |
M1 solving 3 term quadratic in \(n\) as above
A1 either 225 or 1369/4 or 342.25
A1 must select 225