S3 June 2017 Q7
7. Sugar is packed into medium bags and large bags. The weights of the medium bags of sugar are normally distributed with mean 520 grams and standard deviation 10 grams. The weights of the large bags of sugar are normally distributed with mean 1510 grams and standard deviation 20 grams.
A random sample of 5 medium bags of sugar is taken.
| Scheme | Marks |
|---|---|
| \(L \sim \mathrm{N}(1510, 20^2)\) and \(M \sim \mathrm{N}(520, 10^2)\) | |
| \(W = L - (M_1 + M_2 + M_3)\) | B1 |
| \(\mathrm{E}(W) = 1510 - 3 \times 520 = -50\) | B1 |
| \(\mathrm{Var}(W) = 20^2 + 10^2 + 10^2 + 10^2 = 700\) | M1,A1 |
| \(\mathrm{P}(W \gt 15) \quad = \mathrm{P}\left(Z \gt \dfrac{15 - -50}{\sqrt{700}}\right)\) | dM1 |
| \(= \mathrm{P}(Z \gt 2.456769\ldots)\) | |
| \(= 0.0069\) | A1 |
| (6) |
Notes
1st B1 Allow \(L - (M + M + M)\) but not \(L - 3M\) Can be implied by correct \(\mathrm{Var}(W)\).
May use \(W = L - (M_1 + M_2 + M_3) - 15\) for B1.
2nd B1 Accept 50 if definition reversed. Accept \(\mathrm{E}(W) = 1510 - 3 \times 520 - 15 = -65\)
M1,A1 Attempt \(\mathrm{Var}(W) = \mathrm{Var}(L) + 3\mathrm{Var}(M)\). Do not condone missing squares, cao.
dM1 Attempting the correct probability and standardising with their mean and sd dependent on 1st M1. If values for \(W\) is not being used or not their variance score M0. Must use 15.
Accept \(\mathrm{P}(W \gt 0) \quad = \mathrm{P}\left(Z \gt \dfrac{0 - -65}{\sqrt{700}}\right)\)
A1 0.0071 by calc. awrt 0.007
| Scheme | Marks |
|---|---|
| \(X = 3M - L\) | |
| \(\mathrm{E}(X) = 3 \times 520 - 1510 = 50\) | B1 |
| \(\mathrm{Var}(X) = 3^2 \times 10^2 + 20^2 = 1300\) | M1,A1 |
| \(\mathrm{P}(X \gt 0) = \mathrm{P}\left(Z \gt \dfrac{-50}{\sqrt{1300}}\right)\) | dM1 |
| \(= \mathrm{P}(Z \gt -1.38675\ldots) = 0.9177\) | A1 |
| (5) |
Notes
\(X = 3M - L\) Can be implied by correct variance.
B1 Accept -50 if reversed.
M1,A1 Attempt \(\mathrm{Var}(X) = 3^2\mathrm{Var}(M) + \mathrm{Var}(L)\). Do not condone missing squares, cao. Condone \(10^2 + 3^2 \times 20^2\) for M1A0. (corrected from the printed mark scheme: printed as \(\mathrm{Var}(W) = 3^2\mathrm{Var}(M) + \mathrm{Var}(S)\))
dM1 Attempting the correct probability and standardising with their mean and sd.
A1 0.9172 by calc. awrt 0.917-0.918
| Scheme | Marks |
|---|---|
| P(all 5 bags weigh more than 520 grams) \(= \left(\dfrac{1}{2}\right)^5 = \dfrac{1}{32} = 0.03125\) | B1 |
| \(\bar{M} \sim \mathrm{N}\left(520, \dfrac{10^2}{5}\right)\) or \(\displaystyle\sum_{i=1}^{5} M_i \sim \mathrm{N}(2600, 500)\) | B1 |
| \(\mathrm{P}(\bar{M} \gt d) = \mathrm{P}\left(Z \gt \dfrac{d - 520}{\frac{10}{\sqrt{5}}}\right) = 0.03125\) or \(\mathrm{P}(T \gt 5d) = \mathrm{P}\left(Z \gt \dfrac{5d - 2600}{\sqrt{500}}\right) = 0.03125\) | M1 |
| \(\Rightarrow \dfrac{d - 520}{\frac{10}{\sqrt{5}}} = 1.86(27\ldots)\) or \(\dfrac{5d - 2600}{\sqrt{500}} = 1.86(27\ldots)\) | M1 |
| \(d = 528.3\) | A1 |
| (5) | |
| (16 marks) |
Notes
1st B1 0.03125
2nd B1 Both mean and variance required in either case. Can be implied below.
1st M1 Standardise using \(d\), 520 and 10 or \(5d\), 2600 and \(\sqrt{500}\).
2nd M1 Equate to \(z\) value
A1 awrt 528.3
ALT (c)
Accept use \(d\) as difference to 520 provided 520 added to final answer:
| Scheme | Marks |
|---|---|
| P(all 5 bags weigh more than 520 grams) \(= \left(\dfrac{1}{2}\right)^5 = \dfrac{1}{32} = 0.03125\) | B1 |
| \(\bar{M} \sim \mathrm{N}\left(0, \dfrac{10^2}{5}\right)\) or \(\displaystyle\sum_{i=1}^{5} M_i \sim \mathrm{N}(0, 500)\) | B1 |
| \(\mathrm{P}(\bar{M} \gt d) = \mathrm{P}\left(Z \gt \dfrac{d}{\frac{10}{\sqrt{5}}}\right) = 0.03125\) or \(\mathrm{P}(T \gt 5d) = \mathrm{P}\left(Z \gt \dfrac{5d}{\sqrt{500}}\right) = 0.03125\) | M1 |
| \(\Rightarrow \dfrac{d}{\frac{10}{\sqrt{5}}} = 1.86(27\ldots)\) or \(\dfrac{5d}{\sqrt{500}} = 1.86(27\ldots)\) | M1 |
| \(d = 520 + 8.3 = 528.3\) | A1 |
| (5) |
1st B1 0.03125
2nd B1 Both mean and variance required in either case. Can be implied below.
1st M1 Standardise using \(d\) and 10 or \(5d\) and \(\sqrt{500}\).
2nd M1 Equate to \(z\) value
A1 awrt 528.3