| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Forward transformation: find new statistics |
| Difficulty | Easy -1.2 This is a straightforward S1 question testing standard formulas for mean and variance, unit conversion (linear transformation with multiplier 100), and conceptual understanding of how adding data affects spread. All parts use direct formula application or simple reasoning with no problem-solving required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \frac{58}{40} = \mathbf{1.45}\) | B1 | For a correct mean (accept an exact fraction) |
| \(\sigma^2 = \frac{84.829}{40} - 1.45^2 = 0.018225\) | M1 | For a correct expression for \(\sigma^2\) (or \(s^2\)) (if their mean and condone inside square root) |
| \(= \text{awrt } \mathbf{0.0182}\) | A1 | For awrt 0.0182 (NB \(s^2 = 0.0186923\ldots\) awrt 0.0187) Correct ans only 2/2 [No fraction] |
| New mean = 145 | B1ft | |
| New \(\sigma = \mathbf{13.5}\) | B1 | For new s.d. = awrt 13.5 (accept \(s = 13.6719\ldots\) or awrt 13.7) |
| Reason e.g. mean of two extra children is the same as the original mean Conclusion the mean is therefore unchanged or = 145 | M1, A1 | For a suitable reason. May see recalculation e.g. \(\frac{\text{"145"} \times 40 + 130 + 160}{42}\) (o.e.) e.g. "both 15 away from the mean" or "both same distance from the mean" or "mean of new values is 145 or the same". For 145 or 1.45 or "no change" but M1 must be seen [no further comment needed if answer matches their (b) or (a)] |
| Reason e.g. extra children more than 1 sd from mean so increased spread Conclusion therefore standard deviation will increase | M1, A1 | For a suitable reason but must have idea that the "gap" (= 15) > 1 st. dev. [ft \(\sigma < 15\)]. For stating standard deviation will be greater (o.e.) [M1 must be seen] |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \frac{58}{40} = \mathbf{1.45}$ | B1 | For a correct mean (accept an exact fraction) |
| $\sigma^2 = \frac{84.829}{40} - 1.45^2 = 0.018225$ | M1 | For a correct expression for $\sigma^2$ (or $s^2$) (if their mean and condone inside square root) |
| $= \text{awrt } \mathbf{0.0182}$ | A1 | For awrt 0.0182 (NB $s^2 = 0.0186923\ldots$ awrt 0.0187) **Correct ans only 2/2 [No fraction]** |
| New mean = **145** | B1ft | |
| New $\sigma = \mathbf{13.5}$ | B1 | For new s.d. = awrt 13.5 (accept $s = 13.6719\ldots$ or awrt 13.7) |
| **Reason** e.g. mean of two extra children is the same as the original mean **Conclusion** the mean is therefore unchanged or = **145** | M1, A1 | For a suitable reason. May see recalculation e.g. $\frac{\text{"145"} \times 40 + 130 + 160}{42}$ (o.e.) e.g. "both 15 away from the mean" or "both same distance from the mean" or "mean of new values is 145 or the same". For 145 or 1.45 or "no change" but M1 must be seen [no further comment needed if answer matches their (b) or (a)] |
| **Reason** e.g. extra children more than 1 sd from mean so increased spread **Conclusion** therefore standard deviation will increase | M1, A1 | For a suitable reason but must have idea that the "gap" (= 15) > 1 st. dev. [ft $\sigma < 15$]. For stating standard deviation will be **greater** (o.e.) [M1 must be seen] |
**Notes:**
**(a)** B1 for a correct mean (accept an exact fraction)
M1 for a correct expression for $\sigma^2$ (or $s^2$) (fit their mean and condone inside square root)
A1 for awrt 0.0182 (NB $s^2 = 0.0186923\ldots$ awrt 0.0187)**Correct ans only 2/2 [No fraction]**
**(b)** 1st B1ft for new mean = 145 or $100 \times$ their $\bar{x}$
2nd B1 for new s.d. = awrt 13.5 (accept $s = 13.6719\ldots$ or awrt 13.7)
**(c)(i)** 1st M1 for a suitable reason. May see recalculation e.g. $\frac{\text{"145"} \times 40 + 130 + 160}{42}$ (o.e.)
e.g. "both 15 away from the mean" or "both same distance from the mean" or "mean of new values is 145 or the same"
1st A1 for 145 or 1.45 or "no change" but M1 must be seen [no further comment needed if answer matches their (b) or (a)]
**(c)(ii)** 2nd M1 for a suitable reason but must have idea that the "gap" (= 15) > 1 st. dev. [ft $\sigma < 15$]
2nd A1 for stating standard deviation will be **greater** (o.e.) [M1 must be seen]
**Calculations (You may see):** e.g. $\sum x^2 = 84.829 + 1.3^2 + 1.6^2 = 89.079$ leading to $\sigma = \sqrt{0.01842\ldots} = 0.13575\ldots$ or 13.6 (cm)
or $\frac{89.079}{42} = 2.1209\ldots > \frac{84.829}{40} = 2.1207\ldots$ but $\frac{\sum x}{n}$ stays the same so $\sigma$ greater
**BUT** M0A0 unless we see mention of 15 (cm) or 1.5 (m) being more than 1 sd
---\begin{enumerate}
\item The heights, $x$ metres, of 40 children were recorded by a teacher. The results are summarised as follows
\end{enumerate}
$$\sum x = 58 \quad \sum x ^ { 2 } = 84.829$$
(a) Find the mean and the variance of the heights of these 40 children.
The teacher decided that these statistics would be more useful in centimetres.\\
(b) Find\\
(i) the mean of these heights in centimetres,\\
(ii) the standard deviation of these heights in centimetres.
Two more children join the group. Their heights are 130 cm and 160 cm .\\
(c) (i) State, giving a reason, the mean height of the 42 children.\\
(ii) Without recalculating the standard deviation, state, giving a reason, whether the standard deviation of the heights of the 42 children will be greater than, less than or the same as the standard deviation of the heights of the group of 40 children.\\
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\hfill \mbox{\textit{Edexcel S1 2019 Q1 [9]}}