# Question 15:

## Part (a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $337$ | B1 | |
**[1]**

## Part (a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{\frac{1}{54}(6247066.6 - 55 \times 337^2)}$ | B1 | NB $\sqrt{14.289} = 3.7800\ldots$; may see $\sqrt{\frac{6247066.6}{55} - 337^2} \times \sqrt{\frac{55}{54}}$ or $\sqrt{\frac{6247066.6}{54} - \frac{55 \times 337^2}{54}}$ |
| $\approx 3.78$ | | AG; must see substitution of at least three of 6247066.6, 337, 55 and 54 |
**[1]**

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| allow any two reasons eg distribution is (approximately) symmetrical | E1 | |
| eg distribution is (approximately) bell-shaped / eg distribution is unimodal / eg data is continuous | E1 | |
**[2]**

## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 330)$ found from $N(\text{their } 337, \ 3.78^2)$ | M1 | may be implied by 0.032; allow a more precise value for 3.78 if found in part (a)(ii); NB eg $N(-9\text{E}999, 330, 337, 3.78)$; NB may see $\sigma^2 = \frac{643}{45}$ or $z = \frac{330-337}{3.78}(= -1.85\ldots)$ |
| $100 \times \text{their } 0.032$ | M1 | |
| awrt $3.2$ www | A1 | mark the final answer |
**[3]**

## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(z =) \pm 2.3263\ldots$ seen | B1 | to 2 or more dp |
| their $z = \frac{330 - \mu}{3.78}$ | M1 | |
| $338.79 \approx 339$ | A1 | must be correct to 3 sf; A0 for $\mu \geq 339$ or $\mu > 339$ |
| *Alternatively using calculator:* cdfNormal$(-9.999\times10^{999}, 330, 338, 3.78) = 0.017(155\ldots) > 0.01$ or cdfNormal$(-9.999\times10^{999}, 330, 338.5, 3.78) = 0.012(266\ldots) > 0.01$ | 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 |
| cdfNormal$(-9.999\times10^{999}, 330, 339, 3.78) = 0.0086(339\ldots) < 0.01$ | M1 | allow for any value between 338.8 and 339.5; NB critical value is 338.79 |
| hence minimum value for $\mu$ is $339$ | A1 | must be correct to 3 sf; A0 for $\mu \geq 339$ or $\mu > 339$ |
**[3]**

## Question 15(e):

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 340$, $H_1: \mu < 340$ | B1 | Do not allow $\bar{X}$ or $X$, but allow other symbol if defined as [population] mean volume; allow equivalent in words |
| $\mu$ is the **population mean** (volume of drink in a bottle of Fizzipop) | B1 | |
| $N\left(340, \frac{3.78^2}{100}\right)$ oe seen | M1* | May be implied by $0.0477 - 0.048$; may see $N(340, 0.378^2)$; may see $\sigma^2 = \frac{643}{4500}$ |
| $[P(\bar{X} < 339.37) =] 0.0477 - 0.048$ | A1 | Allow slip such as $X$ for $\bar{X}$ but do not allow $\mu$ |
| Their $0.048$ correctly compared with $0.05$ | 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 | Dependent on award of all other marks apart from second B1; do not allow e.g. conclude/prove/indicate or other assertive statement instead of suggest |
| **Total: [7]** | | |

**Alternative (critical region method):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 340$, $H_1: \mu < 340$ | B1 | As above |
| $\mu$ is the **population mean** | B1 | |
| $N\left(340, \frac{3.78^2}{100}\right)$ oe seen | M1* | May be implied by $339.378 - 339.38$; may see $N(340, 0.378^2)$; may see $\sigma^2 = \frac{643}{4500}$ |
| Critical region is $\bar{X} <$ ]$339.378 - 339.38$ | A1 | Or critical value is $\bar{X} =$ ]$339.378 - 339.38$; allow slip such as $X$ for $\bar{X}$ but do not allow $\mu$ |
| $339.37$ correctly compared with their $339.378$ | M1dep* | Allow e.g. so $339.37$ is in the critical region if critical region explicitly identified |
| Do not accept $H_0$ **or** reject $H_0$ **or** accept $H_1$ **or** significant | A1FT | A0 if $339.37 >$ their $339.378$ |
| Sufficient evidence at 5% level to suggest mean volume less than 340 ml oe | A1 | Dependent on award of all other marks apart from second B1 |

**Alternative (standard normal distribution):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu = 340$, $H_1: \mu < 340$ | B1 | As above |
| $\mu$ is the **population mean** | B1 | |
| $N\left(340, \frac{3.78^2}{100}\right)$ oe seen | M1* | May be implied by $z = -1.667$; may see $N(340, 0.378^2)$; may see $\sigma^2 = \frac{643}{4500}$ |
| $[z =] - 1.667$ | 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 | A0 if their $z > -1.64485$ |
| Sufficient evidence at 5% level to suggest mean volume less than 340 ml oe | A1 | Dependent on award of all other marks apart from second B1 |

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