S3 June 2007 Q7
7. A set of scaffolding poles come in two sizes, long and short. The length \(L\) of a long pole has the normal distribution \(\mathrm{N}(19.7, 0.5^2)\). The length \(S\) of a short pole has the normal distribution \(\mathrm{N}(4.9, 0.2^2)\). The random variables \(L\) and \(S\) are independent.
A long pole and a short pole are selected at random.
Four short poles are selected at random and placed end to end in a row. The random variable \(T\) represents the length of the row.
| Scheme | Marks |
|---|---|
| Let \(X = L - 4S\) then \(\mathrm{E}(X) = 19.7 - 4 \times 4.9,\ = 0.1\) | M1, A1 |
| \(\mathrm{Var}(X) = \mathrm{Var}(L) + 4^2\,\mathrm{Var}(S) = 0.5^2 + 16 \times 0.2^2\) | M1, M1 |
| \(= 0.89\) | A1 |
| \(\mathrm{P}(X \gt 0) = [\mathrm{P}(Z \gt -0.10599\ldots)]\) | M1 |
| \(=\) AWRT (0.542 – 0.544) | A1 |
| (7) |
Notes
1st M1 for defining \(X\) and attempting \(\mathrm{E}(X)\)
1st A1 for 0.1. Answer only will score both marks.
2nd M1 for \(\mathrm{Var}(L)\) +…..
3rd M1 for …. \(4^2\,\mathrm{Var}(S)\). For those who don’t attempt \(L - 4S\) this will be their only mark in (a).
2nd A1 for 0.89
4th M1 for attempting a correct probability, correct expression and attempt to find, which should involve some standardisation: ft their \(\sqrt{0.89}\) and their 0.1.
If 0.1 is used for \(\mathrm{E}(X)\) answer should be > 0.5, otherwise M0.
| Scheme | Marks |
|---|---|
| \(T = S_1 + S_2 + S_3 + S_4\) (May be implied by 0.16) | M1 |
| \(T \sim \mathrm{N}(19.6, 0.16)\) \(\mathrm{E}(T) = 19.6\) | B1 |
| \(\mathrm{Var}(T) = 0.16\) or \(0.4^2\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Let \(Y = L - T\) \(\mathrm{E}(Y) = \mathrm{E}(L) - \mathrm{E}(T) = [\,0.1\,]\) | M1 |
| \(\mathrm{Var}(Y) = \mathrm{Var}(L) + \mathrm{Var}(T) = [\,0.41\,]\) | M1 |
| Require \(\mathrm{P}(-0.1 \lt Y \lt 0.1)\) | M1 |
| \(= \mathrm{P}(Z \lt 0) - \mathrm{P}(Z \lt -0.31..)\) or \(0.5 - \mathrm{P}(Z \lt -0.31..)\) or \(\mathrm{P}(Z \lt 0.31..) - \mathrm{P}(Z \lt 0)\) | M1 |
| \(= 0.1217\) (tables) or \(0.1226..\) (calc) AWRT (0.122 – 0.123) | A1 |
| (5) | |
| (15 marks) |
Notes
1st M1 for a correct method for \(\mathrm{E}(Y)\), ft their \(\mathrm{E}(T)\).
2nd M1 for a correct method for \(\mathrm{Var}(Y)\), ft their \(\mathrm{Var}(T)\). Must have +.
3rd M1 for dealing with the modulus and a correct probability statement. Must be modulus free.
May be implied by e.g. \(\mathrm{P}\left(Z \lt \frac{0.2}{\sqrt{\text{their } 0.41}}\right) - 0.5\), or seeing both 0.378… (or 0.622…) and 0.5
4th M1 for correct expression for the correct probability, as printed or better. E.g. 0.5 + 0.378.. is M0
A1 for AWRT in range.