## Question 6:

### Part (a)
| $\mu$ or $\bar{x} = \frac{8360}{10} = 836$ | B1 | |
| $\sigma = \sqrt{\frac{\sum(x-\bar{x})^2}{10}} = \sqrt{6384}$ or $4\sqrt{399} = 79.89993...$ = awrt $79.9$ | M1, A1 (3) | M1 for $\frac{63840}{10}$ with or without $\sqrt{}$. NB $\sum x^2 = 7052800$ but must see $\sigma^2 = \frac{7052800}{10} - (836)^2$ for M1. A1 for awrt 79.9; accept $s =$ awrt 84.2. Correct answer only M1A1 |

### Part (b)
| mean $>$ median so positive skew | B1 | For correct comparison of mean and median (allow just $836 > 815$) |
| | dB1 (2) | Dependent on 1st B1 for positive (skew) only. If mean $< 815$ award B0B1 for comparison and statement of negative skew |

### Part (c)
| $\frac{776 + 896}{2} = 836$ which is same as $\bar{x}$, or one is 60 above $\bar{x}$, one 60 below | B1 | For suitable calculation showing mean of two rabbits (or all 12) is same |
| So no change in the mean | dB1 (2) | Dependent on suitable calculation or reason. If they only say differences are same (but not 1 above and 1 below) and state no change then B0B1 |

### Part (d)
| $(896 - 836)^2 = (776 - 836)^2 = 60^2 = 3600 < 6384$ the average of $\sum(x-\bar{x})^2$ | B1 | For suitable calculation showing 60 or 3600 and comparing with 79.9 or 6384 respectively |
| Or $\sum(x-\bar{x})^2 \to 63840 + 2 \times 60^2 = 71040$ and $\frac{71040}{12} = 5920 < \frac{63840}{10}$ | | |
| So standard deviation will reduce | dB1 (2) | Dependent on 1st B1 for stating s.d. "reduces" |

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