# Question 6:

## Part (a)
| $\frac{200-\mu}{7.8} = -1.2816$ [calc gives $1.28155156\ldots$] | M1 B1 | For attempt to standardise with 200 and 7.8 set $= \pm$ any $z$ value ($|z|>1$); for $z = \pm1.2816$ or better |
| $\mu = 209.996\ldots$, awrt $\mathbf{210}$ | A1 | The 210 must follow from correct working |

## Part (b)
| $P(X > 225) = P\!\left(Z > \frac{225 - \text{"210''}}{7.8}\right)$ | M1 | For attempting to standardise with 225, their mean and 7.8; allow $\pm$ |
| $= P(Z > 1.92)$ or $1 - P(Z < 1.92)$ (allow 1.93) | A1 | For $Z >$ awrt 1.92/3; must have correct area indicated |
| $= 1 - 0.9726 = 0.0274$ (or better) [calc gives $0.0272037\ldots$] | | |
| $= $ awrt $\mathbf{2.7\%}$, allow $\mathbf{0.027}$ | A1 | For 2.7% or better; correct ans scores 3/3 |

## Part (c)
| $\frac{210-205}{\sigma} = 2.3263$ or $\frac{200-205}{\sigma} = -2.3263$ [calc gives $2.3263478\ldots$] | M1 B1 | For attempt to standardise with 200 or 210, 205 and $\sigma$; for $z = 2.3263$ or better and compatible signs |
| $\sigma = \frac{5}{2.3263}$ | dM1 | Dependent on first M1; for correctly rearranging to make $\sigma = \ldots$; must have $\sigma > 0$ |
| $\sigma = 2.15$ $(2.14933\ldots)$ | A1 | For awrt 2.15; must follow from correct working; range $2.320 < z \leq 2.331$ will give awrt 2.15 |

The image you've shared appears to be a back/cover page of an Edexcel publication from Summer 2013, containing only publisher information, contact details, and logos (Ofqual, Welsh Assembly Government, and GEA).

There is **no mark scheme content** visible on this page to extract. It contains only:

- Publisher contact details (Edexcel Publications, Mansfield)
- Telephone/Fax/Email information
- Order Code UA036993 Summer 2013
- Pearson Education Limited registration details
- Three accreditation logos

If you have additional pages from the mark scheme, please share those and I'll extract the content as requested.