## Question 3:

**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| fd scale of $1\text{cm} = 5$ deduced from bars 3–4 | M1 | For deducing a correct fd scale (seen on graph or in text); may be implied by 25 or 7 |
| $2\sim3$ has freq $= \mathbf{25}$ **and** $4\sim6$ has freq $= 112-(64+13+3+``25") = \mathbf{7}$ | A1 | For both 25 **and** 7 |

**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| For a bar between $4\sim6$ of height $\frac{"7"}{2} = 3.5$ small squares **or** bar between $6\sim8$ of height 1.5 small squares | M1 | Ignore their fd scale in part (b). For a correct bar over $4\sim6$ follow through their "7" from table, or for a correct bar over $6\sim8$ |
| For a fully correct histogram with all 3 bars plotted correctly | A1ft | For a fully correct histogram (all 3 bars correct height and correct width). Allow ft on their $4\sim6$ bar |

**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Require $\int_{(1)}^{(8)} \frac{k}{x^2}\,dx = 112$ | M1 | For correct integral expression $= 112$ (condone missing $dx$ and ignore limits), or attempt to integrate ($x^{-2} \to x^{-1}$) and set area $= 112$ (ignore limits) |
| $= \left[\frac{-k}{x}\right]_1^8 = \left(-\frac{k}{8}\right)-\left(-\frac{k}{1}\right)[=112]$ | M1 | For correct integration and some use of limits of 1 and 8 (condone missing 112). $\frac{7}{8}k = 112$ implies M1M1 |
| $\left[\frac{7}{8}k = 112 \Rightarrow\right] \quad k = \mathbf{128}$ | A1 | For 128 |

---