A2 June 2019 Q7
7. A manufacturer makes two versions of a toy. One version is made out of wood and the other is made out of plastic.
The weights, \(W\) kg, of the wooden toys are normally distributed with mean 2.5 kg and standard deviation 0.7 kg. The weights, \(X\) kg, of the plastic toys are normally distributed with mean 1.27 kg and standard deviation 0.4 kg. The random variables \(W\) and \(X\) are independent.
The manufacturer packs \(n\) of these wooden toys and \(2n\) of these plastic toys into the same container. The maximum weight the container can hold is 252 kg.
The probability of the contents of this container being overweight is 0.2119 to 4 decimal places.
| Scheme | Marks | AO |
|---|---|---|
| Let \(T = W - 2X\) then \(\mathrm{E}(T) = 2.5 - 2 \times 1.27\) | M1 | 3.3 |
| \(= -0.04\) | A1 | 1.1b |
| \(\mathrm{Var}(T) = 0.7^2 + 2^2 \times 0.4^2\) | M1 | 2.1 |
| \(= 1.13\) | A1 | 1.1b |
| \(\mathrm{P}\left(Z \gt \dfrac{0 - \text{“}-0.04\text{”}}{\sqrt{\text{“}1.13\text{”}}}\right) = \mathrm{P}(Z \gt 0.0376\ldots)\) | M1 | 2.1 |
| = awrt 0.484/0.485 | A1 | 1.1b |
| (6) |
Notes
M1: selecting and using an appropriate model. ie \(\pm(W - 2X)\) May be implied by \(-0.04\)
A1: \(-0.04\) oe
M1: for realising the need to use \(\mathrm{Var}(W) + 4\,\mathrm{Var}(X)\). Allow use of 0.7 for \(\mathrm{Var}(W)\) instead of 0.72 and/or 0.4 for \(\mathrm{Var}(X)\) instead of 0.42. May be implied by 1.13
A1: 1.13 only
M1: For realising the \(\mathrm{P}(T \gt 0)\) is required and an attempt to find it. \(\dfrac{0 - \text{“}\text{their } -0.04\text{”}}{\sqrt{\text{“}\text{their } 1.13\text{”}}}\) may be implied by a correct answer. If \(\mathrm{E}(T)\) and \(\mathrm{Var}(T)\) have not been given they must be correct here
A1: awrt 0.484/0.485
| Scheme | Marks | AO |
|---|---|---|
| \(B = W_1 + W_2 + \ldots + W_n + X_1 + X_2 + \ldots + X_{2n}\) | M1 | 3.3 |
| \(\mathrm{E}(B) = 5.04n\) | B1 | 1.1b |
| \(\mathrm{Var}(B) = n \times 0.7^2 + 2n \times 0.4^2\) \(= 0.81n\) | A1 | 1.1b |
| \(\pm\dfrac{252 - \text{“}5.04n\text{”}}{\sqrt{\text{“}0.81n\text{”}}}\) | M1 | 1.1b |
| \(\dfrac{252 - \text{“}5.04n\text{”}}{\sqrt{\text{“}0.81n\text{”}}} = 0.8\) | M1 | 2.1 |
| \(5.04n + 0.72\sqrt{n} - 252 = 0\) oe | ||
| \(\sqrt{n} = -7.14\ldots\) or 7 | M1 | 1.1b |
| \(n = 7^2\) | M1 | 1.1b |
| \(= 49\) | A1cso | 1.1b |
| (8) | ||
| (14 marks) |
Notes
M1: Selecting and using appropriate model. May be implied by 0.81
B1: \(5.04n\) only
A1: \(0.81n\)
M1: For standardising using their mean and sd \(\pm\dfrac{252 - \text{“}5.04n\text{”}}{\sqrt{\text{“}0.81n\text{”}}}\) If mean and sd not given they must be correct here
M1: For constructing an equation and equate their standardisation to 0.8 or awrt 0.7998. Must be of form \(\dfrac{252 - an}{b\sqrt{n}} = 0.8\) or \(\dfrac{252 - an}{bn} = 0.8\)
M1: Correctly solving their 3 term quadratic equation. Condone \(n = 7\)
M1: for realising the need to square their answer or for attempting to square their quadratic equation
A1cso: 49 only