June 2024 Paper 2 Q15
15 Bottles of Fizzipop nominally contain 330 ml of drink. A consumer affairs researcher collects a random sample of 55 bottles of Fizzipop and records the volume of drink in each bottle.
Summary statistics for the researcher’s sample are shown in the table.
| \(n\) | 55 |
|---|---|
| \(\sum x\) | 18 535 |
| \(\sum x^2\) | 6 247 066.6 |
The researcher uses software to produce a histogram with equal class intervals, which is shown below.
In order to comply with new regulations, no more than 1% of bottles of Fizzipop should contain less than 330 ml.
The manufacturer decides to meet the new regulations by adjusting the manufacturing process so that the mean volume of drink in a bottle of Fizzipop is increased.
The standard deviation is unaltered.
The mean volume of drink in a bottle of Fizzipop is set to 340 ml. After several weeks the quality control manager suspects the mean volume may have reduced. She collects a random sample of 100 bottles of Fizzipop.
The mean volume of drink in a bottle in the sample is found to be 339.37 ml.
| Scheme | Marks | AO |
|---|---|---|
| (i) 337 | B1 | 1.1 |
| [1] | ||
| (ii) \(\sqrt{\frac{1}{54}\left(6247066.6 - 55 \times 337^2\right)}\) | B1 | 1.1 |
| \(\approx 3.78\) | ||
| [1] |
Notes
(ii) B1: NB \(\sqrt{14.289} = 3.7800\ldots\)
may see \(\sqrt{\frac{6247066.6}{55} - 337^2} \times \sqrt{\frac{55}{54}}\) or \(\sqrt{\frac{6247066.6}{54} - \frac{55 \times 337^2}{54}}\) oe;
AG
must see substitution of at least three of 6247066.6, 337, 55 and 54
| Scheme | Marks | AO |
|---|---|---|
| allow any two reasons eg distribution is (approximately) symmetrical eg distribution is (approximately) bell-shaped eg distribution is unimodal eg data is continuous | E1 E1 | 2.4 2.2b |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{P}(X \lt 330)\) found from N(their 337, \(3.78^2\)) | M1 | 3.3 |
| \(100 \times\) their 0.032 | M1 | 3.4 |
| awrt 3.2 www | A1 | 1.1 |
| [3] |
Notes
M1: may be implied by 0.032; allow a more precise value for 3.78 if found in part (a)(ii)
NB eg N(−9E999,330,337,3.78);
NB may see \(\sigma^2 = \frac{643}{45}\) or \(z = \frac{330-337}{3.78}\ (= -1.85\ldots)\)
A1: mark the final answer
| Scheme | Marks | AO |
|---|---|---|
| \((z =)\ \pm 2.3263\ldots\) seen | B1 | 3.1a |
| their \(z = \frac{330 - \mu}{3.78}\) | M1 | 2.1 |
| \(338.79 \approx 339\) | A1 | 3.2a |
| [3] |
Notes
B1: to 2 or more dp
A1: must be correct to 3 sf
A0 for \(\mu \geqslant 339\) or \(\mu \gt 339\)
Alternatively using calculator
| Scheme | Marks | AO |
|---|---|---|
| eg cdfNormal\((-9.999 \times 10^{999}, 330, 338, 3.78) = 0.017(155\ldots) \gt 0.01\) or eg cdfNormal\((-9.999 \times 10^{999}, 330, 338.5, 3.78) = 0.012(266\ldots) \gt 0.01\) | M1 | |
| eg cdfNormal\((-9.999 \times 10^{999}, 330, 339, 3.78) = 0.0086(339\ldots) \lt 0.01\) | M1 | |
| hence minimum value for \(\mu\) is 339 | A1 | |
| 3 |
M1: allow slip in calculation if intent is clear; allow for any value between 338 and 338.78 inclusive
must see correct distributions if probabilities are wrong
M1: allow for any value between 338.8 and 339.5
NB critical value is 338.79
A1: must be correct to 3 sf; A0 for \(\mu \geqslant 339\) or \(\mu \gt 339\)
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(\mu = 340\) H\(_1\): \(\mu \lt 340\) | B1 | 1.1 |
| \(\mu\) is the population mean (volume of drink in a bottle of Fizzipop) | B1 | 2.5 |
| \(\mathrm{N}\left(340, \frac{3.78^2}{100}\right)\) oe seen | M1* | 3.3 |
| \([\mathrm{P}(\bar{X} \lt 339.37) =]\ 0.0477 - 0.048\) | A1 | 3.4 |
| their 0.048 correctly compared with 0.05 | M1dep* | 3.4 |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | 1.1 |
| there is sufficient evidence at the 5% level to suggest that the mean volume of drink in a bottle of Fizzipop is less than 340 ml oe | A1 | 3.5a |
| [7] |
Notes
B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words
M1*: may be implied by \(0.0477 - 0.048\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)
A1: allow slip such as \(X\) for \(\bar{X}\) but do not allow \(\mu\)
A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest
Alternatively
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(\mu = 340\) H\(_1\): \(\mu \lt 340\) | B1 | 1.1 |
| \(\mu\) is the population mean (volume of drink in a bottle of Fizzipop) | B1 | 2.5 |
| \(\mathrm{N}\left(340, \frac{3.78^2}{100}\right)\) oe seen | M1* | 3.3 |
| [critical region is \(\bar{X} \lt]\ 339.378 - 339.38\) | A1 | 3.4 |
| 339.37 correctly compared with their 339.378 | M1dep* | 3.4 |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | 1.1 |
| there is sufficient evidence at the 5% level to suggest that the mean volume of drink in a bottle of Fizzipop is less than 340 ml oe | A1 | 3.5a |
B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words
M1*: may be implied by \(339.378 - 339.38\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)
A1: or [critical value is \(\bar{X} =]\ 339.378 - 339.38\);
allow slip such as \(X\) for \(\bar{X}\) but do not allow \(\mu\)
M1dep*: allow eg so 339.37 is in the critical region if critical region explicitly identified
A1FT: A0 if \(339.37 \gt\) their 339.378
A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest
Alternatively, using standard Normal distribution
| Scheme | Marks | AO |
|---|---|---|
| H\(_0\): \(\mu = 340\) H\(_1\): \(\mu \lt 340\) | B1 | |
| \(\mu\) is the population mean (volume of drink in a bottle of Fizzipop) | B1 | |
| \(\mathrm{N}\left(340, \frac{3.78^2}{100}\right)\) oe seen | M1* | |
| \([z =]\ -1.667\) | A1 | |
| their \(z\) correctly compared with \(-1.64485\) to 2 or more dp oe; must come from N\((340, \sigma)\) | M1dep* | |
| do not accept H\(_0\) or reject H\(_0\) or accept H\(_1\) or significant | A1FT | |
| there is sufficient evidence at the 5% level to suggest that the mean volume of drink in a bottle of Fizzipop is less than 340 ml oe | A1 |
B1: do not allow \(\bar{X}\) or \(X\), but allow other symbol if defined as [population] mean volume;
allow equivalent in words
M1*: may be implied by \(z = -1.667\)
may see N\((340, 0.378^2)\)
may see \(\sigma^2 = \frac{643}{4500}\)
A1FT: A0 if their \(z \gt -1.64485\)
A1: dependent on award of all other marks apart from second B1
do not allow eg conclude / prove / indicate or other assertive statement instead of suggest