| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from frequency table |
| Difficulty | Moderate -0.8 This is a standard S1 statistics question testing routine procedures: histogram calculations using frequency density, linear interpolation for median, and mean/standard deviation from grouped data using given formulas. All parts follow textbook methods with no problem-solving or novel insight required. The given value of Σfx² removes computational difficulty. Slightly easier than average due to straightforward application of memorized techniques. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Weight ( \(\boldsymbol { w } \mathbf { ~ k g }\) ) | Frequency (f) | Weight midpoint (x) |
| \(0 \leqslant w < 2\) | 1 | 1 |
| \(2 \leqslant w < 3\) | 8 | 2.5 |
| \(3 \leqslant w < 3.5\) | 17 | 3.25 |
| \(3.5 \leqslant w < 4\) | 17 | 3.75 |
| \(4 \leqslant w < 5\) | 7 | 4.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Width \(= \mathbf{0.5}\) cm | B1 | |
| e.g. \(4\ [\text{cm}^2]\) represents 8 babies or frequency densities are 8 and 34 | M1 | Clear representation of area with frequency or height \(\times\) width \(= 8.5\) |
| Height \(= \mathbf{17}\) cm | A1 | For 17 (cm) [must be clear it is height not frequency] |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_2 = \left\{3\right\} + \frac{(25-9)}{(26-9)} \times 0.5\) or \(\{3.5\} - \frac{(25-24)}{(41-24)} \times 0.5 = \text{awrt } \mathbf{3.47}\) (allow \(\frac{59}{17}\)) | M1, A1 | M1 for \(\frac{16}{17} \times 0.5\) or \(\frac{16.5}{17} \times 0.5\); A1 for awrt 3.47 or \(\frac{59}{17}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum fx = 1\times1 + 2.5\times8 + 3.25\times17 + 3.75\times17 + 4.5\times7 = 171.5\), \(\bar{x} = \frac{171.5}{50} = (3.43)\) | B1cso | For \(\Sigma fx\) (at least 3 correct and no incorrect products) and correct \(\frac{\sum fx}{50}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{\frac{611.375}{50} - 3.43^2} = 0.680147\ldots = \text{awrt } \mathbf{0.680}\) (Accept 0.68) | M1, A1 | M1 for correct expression including square root, must use 3.43; A1 for awrt 0.680 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{P}(W < 3) = \text{P}\!\left(Z < \frac{-0.43}{0.65}\right) = \text{P}(Z < -0.6615\ldots)\) | M1 | Attempt to standardise with 3, 3.43 and 0.65; allow \(\pm\) |
| \(= 1 - 0.7454\) (tables) | M1 | For \(1 - p\) where \(0.74 < p < 0.75\) |
| \(= 0.2546\ \text{awrt } \mathbf{0.254}\)–\(\mathbf{0.255}\) | A1 | For awrt 0.254 or 0.255 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) and (c)(i) mean \(\neq\) median or skew or mean \(\approx\) median or no skew and comment | B1 | Statement about mean/median and compatible comment about normal |
| (d) \(= 0.254\) or \(0.255\) compare data \(= 0.18\) (or 12.7 compared with 9) | B1 | Statement comparing their (d) with data (sight of 0.18 or 12.7 and 9 required) |
| 0.18 different from 0.25 so normal not good or 0.18 similar to 0.25 so normal is OK | dB1 | Dependent on 2nd B1: conclusion about normal compatible with 2nd statement |
| Answer | Marks | Guidance |
|---|---|---|
| No change in mean (since weight is the same) | B1 | For no change in mean |
| Answer | Marks | Guidance |
|---|---|---|
| s.d. will decrease (extra value is at "centre" so data more concentrated) | B1 | For s.d. decreases |
| Both statements correct and correct reasons for each | dB1 | Dependent on 1st and 2nd B1s; correct reason for both mean and s.d. e.g. "new mean the same so within 1 s.d. of old mean" |
# Question 5:
## Part (a)
| Width $= \mathbf{0.5}$ cm | B1 | |
| e.g. $4\ [\text{cm}^2]$ represents 8 babies or frequency densities are 8 and 34 | M1 | Clear representation of area with frequency or height $\times$ width $= 8.5$ |
| Height $= \mathbf{17}$ cm | A1 | For 17 (cm) [must be clear it is height not frequency] |
## Part (b)
| $Q_2 = \left\{3\right\} + \frac{(25-9)}{(26-9)} \times 0.5$ or $\{3.5\} - \frac{(25-24)}{(41-24)} \times 0.5 = \text{awrt } \mathbf{3.47}$ (allow $\frac{59}{17}$) | M1, A1 | M1 for $\frac{16}{17} \times 0.5$ or $\frac{16.5}{17} \times 0.5$; A1 for awrt 3.47 or $\frac{59}{17}$ |
## Part (c)(i)
| $\sum fx = 1\times1 + 2.5\times8 + 3.25\times17 + 3.75\times17 + 4.5\times7 = 171.5$, $\bar{x} = \frac{171.5}{50} = (3.43)$ | B1cso | For $\Sigma fx$ (at least 3 correct and no incorrect products) and correct $\frac{\sum fx}{50}$ |
## Part (c)(ii)
| $\sqrt{\frac{611.375}{50} - 3.43^2} = 0.680147\ldots = \text{awrt } \mathbf{0.680}$ (Accept 0.68) | M1, A1 | M1 for correct expression including square root, must use 3.43; A1 for awrt 0.680 |
## Part (d)
| $\text{P}(W < 3) = \text{P}\!\left(Z < \frac{-0.43}{0.65}\right) = \text{P}(Z < -0.6615\ldots)$ | M1 | Attempt to standardise with 3, 3.43 and 0.65; allow $\pm$ |
| $= 1 - 0.7454$ (tables) | M1 | For $1 - p$ where $0.74 < p < 0.75$ |
| $= 0.2546\ \text{awrt } \mathbf{0.254}$–$\mathbf{0.255}$ | A1 | For awrt 0.254 or 0.255 |
## Part (e)
| (b) and (c)(i) mean $\neq$ median or skew or mean $\approx$ median or no skew **and** comment | B1 | Statement about mean/median and compatible comment about normal |
| (d) $= 0.254$ or $0.255$ compare data $= 0.18$ (or 12.7 compared with 9) | B1 | Statement comparing their (d) with data (sight of 0.18 or 12.7 and 9 required) |
| 0.18 different from 0.25 so normal not good or 0.18 similar to 0.25 so normal is OK | dB1 | Dependent on 2nd B1: conclusion about normal compatible with 2nd statement |
## Part (f)(i)
| No change in mean (since weight is the same) | B1 | For no change in mean |
## Part (f)(ii)
| s.d. will decrease (extra value is at "centre" so data more concentrated) | B1 | For s.d. decreases |
| Both statements correct **and** correct reasons for each | dB1 | Dependent on 1st and 2nd B1s; correct reason for both mean and s.d. e.g. "new mean the same so within 1 s.d. of old mean" |
---5. A midwife records the weights, in kg , of a sample of 50 babies born at a hospital. Her results are given in the table below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Weight ( $\boldsymbol { w } \mathbf { ~ k g }$ ) & Frequency (f) & Weight midpoint (x) \\
\hline
$0 \leqslant w < 2$ & 1 & 1 \\
\hline
$2 \leqslant w < 3$ & 8 & 2.5 \\
\hline
$3 \leqslant w < 3.5$ & 17 & 3.25 \\
\hline
$3.5 \leqslant w < 4$ & 17 & 3.75 \\
\hline
$4 \leqslant w < 5$ & 7 & 4.5 \\
\hline
\end{tabular}
\end{center}
[You may use $\sum \mathrm { f } x ^ { 2 } = 611.375$ ]
A histogram has been drawn to represent these data.
The bar representing the weight $2 \leqslant w < 3$ has a width of 1 cm and a height of 4 cm .
\begin{enumerate}[label=(\alph*)]
\item Calculate the width and height of the bar representing a weight of $3 \leqslant w < 3.5$
\item Use linear interpolation to estimate the median weight of these babies.
\item \begin{enumerate}[label=(\roman*)]
\item Show that an estimate of the mean weight of these babies is 3.43 kg .
\item Find an estimate of the standard deviation of the weights of these babies.
Shyam decides to model the weights of babies born at the hospital, by the random variable $W$, where $W \sim \mathrm {~N} \left( 3.43,0.65 ^ { 2 } \right)$
\end{enumerate}\item Find $\mathrm { P } ( W < 3 )$
\item With reference to your answers to (b), (c)(i) and (d) comment on Shyam's decision.
A newborn baby weighing 3.43 kg is born at the hospital.
\item Without carrying out any further calculations, state, giving a reason, what effect the addition of this newborn baby to the sample would have on your estimate of the
\begin{enumerate}[label=(\roman*)]
\item mean,
\item standard deviation.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q5 [17]}}