# Question 5:

## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.9 < x \leqslant 1.1$ group - width $\mathbf{0.5}$ (cm) | B1 | |
| $1.5\text{ cm}^2$ is 2 seeds or $6.75\text{ cm}^2$ is 9 seeds or $0.5c=6.75$ or $\frac{9}{0.2}\times0.3$ | M1 | A correct statement comparing area and number of seeds; allow height $\times$ width $= 6.75$ for M1 |
| $0.9 < x \leqslant 1.1$ group - height $\mathbf{13.5}$ (cm) | A1 | |

## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The data/weights are continuous | B1 | |

## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 1.4125$ awrt $\mathbf{1.41}$ | B1 | Allow $\frac{113}{80}$ but not $\frac{67.8}{48}$ |
| $\sigma=\sqrt{\frac{101.56}{48}-\left(\frac{113}{80}\right)^2}=0.347\ldots$ awrt $\mathbf{0.347}$ ($s=$ awrt 0.351) | M1M1A1 | M1 attempt at $\frac{101.56}{48}-'\mu'^2$; M1 using square root |

## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Median $= [1.3]+\dfrac{3}{11}\times0.2$, allow $(n+1)=[1.3]+\dfrac{3.5}{11}\times0.2$ | M1 | M1 for correct fraction: $[1.3]+\frac{3}{11}\times0.2$ or $m=1.5-\frac{8}{11}\times0.2$; allow $\frac{[m-1.3]}{1.5-1.3}=\frac{24-21}{32-21}$ |
| $=$ awrt $\mathbf{1.35}$ or $\mathbf{1.355}$ (or if using $(n+1)$ allow awrt $\mathbf{1.36}$) | A1 | A1 from correct working |

## Part (e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{27}{48}$ oe 0.5625 (allow 0.563) | B1 | |

## Part (f):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean increases **and** standard deviation decreases | B1 | B1 for both correct comments |
| e.g. '$\sum fy$ increases (so the mean increases) **and** an extreme value has been replaced/data is more concentrated around the mean (so the standard deviation decreases)' | depB1 | dep on previous B1 for **complete** reasoning for both cases |

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