| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Direct frequency calculation from histogram |
| Difficulty | Easy -1.3 This is a standard S1 histogram question requiring frequency density calculations and basic probability. Part (a) uses the given information to establish the scale, then applies it mechanically. Parts (b)-(f) involve routine frequency calculations, simple probability, and textbook interpretations of mean/median relationships for skewness. No problem-solving insight required—purely procedural application of histogram reading skills taught in every S1 course. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(25 \text{ small sq}^2 = 5 \text{ tomatoes or } 1 \text{ large square} = 5 \text{ tomatoes or } fd=5 \text{ for } 2\text{-}3\) or \(\frac{5}{25} \times 20\) or \(5 \times 0.8\) or \(2 \times 2 = \underline{4}\) | M1, A1 |
| (b) | \(100 - (5 + \text{'a'}) \text{ or } 16 + 32 + 25 + 10 + 8\), so probability \(= \frac{91}{100}\) (condone 91%) | M1, A1 |
| (c) | \(\frac{(7-6.25) \times 16 + 25 + 10 + 8}{100}\) or \(1 - \frac{(a) + 5 + 16 + (6.25-5) \times 16}{100} = \frac{55}{100}\) | M1, A1 |
| (d) | Since '\(0.55' \geq 0.5\) (or equivalent reason) and state median \(> 6.25\) | B1 |
| (e) | Median \(>\) mean, so negative skew | B1 |
| (f) | Freq. for \((5.5 < \text{weight} < 7) = (7-5.5) \times '16' \text{ or } \frac{3}{4} \times '32'\), probability \(= \frac{24}{100}\) | M1, A1 |
| \(P(\text{both weight between } 5.5 \text{ and } 7) = \frac{24}{100} \times \frac{23}{99} = \frac{46}{825}\) (o.e.) or awrt \(\underline{0.056}\) | M1 A1 | 2nd M1 for \(\frac{\text{'24'} \times \text{'24}-1}{100 \times 99}\) ft their 24 but must have numerator \(<\) denominator of \(100 \times 99\). 2nd A1 for \(\frac{46}{825}\) (o.e.) or awrt 0.056 NB \(\frac{24}{100} \times \frac{24}{100}\) scores M1A1M0A0 [0.0576 alone 0/4] |
| M1, A1 (4) | ||
| [12 marks] |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $25 \text{ small sq}^2 = 5 \text{ tomatoes or } 1 \text{ large square} = 5 \text{ tomatoes or } fd=5 \text{ for } 2\text{-}3$ or $\frac{5}{25} \times 20$ or $5 \times 0.8$ or $2 \times 2 = \underline{4}$ | M1, A1 | M1 for a correct statement linking area with frequency or calculation or at least 2 frequencies on/in histogram bars. A1 for an answer of 4 (if not in script, can be awarded if 4 seen correctly on histogram). If answers on both diagram and script contradict, the script has preference. |
| (b) | $100 - (5 + \text{'a'}) \text{ or } 16 + 32 + 25 + 10 + 8$, so probability $= \frac{91}{100}$ (condone 91%) | M1, A1 | M1 for $100 - (5 + \text{'a'})$ ft $0 < \text{'their (a)'} < 10$ or for a correct method for finding the sum of the areas of all the bars above 3 (condone one slip if 5 terms seen) |
| (c) | $\frac{(7-6.25) \times 16 + 25 + 10 + 8}{100}$ or $1 - \frac{(a) + 5 + 16 + (6.25-5) \times 16}{100} = \frac{55}{100}$ | M1, A1 | M1 fully correct expression (possibly ft their (a)) and need division by 100 (o.e.). A1 for $\frac{11}{20}$ or $0.55$ (o.e.) [Allow 55% or ratio 55:100] |
| (d) | Since '$0.55' \geq 0.5$ (or equivalent reason) and state median $> 6.25$ | B1 | B1 for $Q_2 > 6.25$ with reason based on (c) where $0.5 < \text{'their (c)'} < 1$ [comparison of "55" & 50] |
| (e) | Median $>$ mean, so negative skew | B1 | B1 for stating "median $>$ mean" and "negative skew" (independent of (d)) |
| (f) | Freq. for $(5.5 < \text{weight} < 7) = (7-5.5) \times '16' \text{ or } \frac{3}{4} \times '32'$, probability $= \frac{24}{100}$ | M1, A1 | 1st M1 for method to find the frequency between 5.5 and 7 (Implied by the 24 used) e.g. $(4+5+16+16 \times 2) - (4+5+16+16 \times 0.5) = 57-33$ based on $(\leq 7) - (\leq 5.5)$. 1st A1 for $\frac{24}{100}$ (o.e.) |
| | $P(\text{both weight between } 5.5 \text{ and } 7) = \frac{24}{100} \times \frac{23}{99} = \frac{46}{825}$ (o.e.) or awrt $\underline{0.056}$ | M1 A1 | 2nd M1 for $\frac{\text{'24'} \times \text{'24}-1}{100 \times 99}$ ft their 24 but must have numerator $<$ denominator of $100 \times 99$. 2nd A1 for $\frac{46}{825}$ (o.e.) or awrt 0.056 NB $\frac{24}{100} \times \frac{24}{100}$ scores M1A1M0A0 [0.0576 alone 0/4] |
| | | M1, A1 (4) | |
| | | [12 marks] | |\begin{enumerate}
\item Ralph records the weights, in grams, of 100 tomatoes. This information is displayed in the histogram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{1130517e-33d0-41b1-9303-2d981379954d-02_981_1268_338_274}
\end{enumerate}
Given that 5 of the tomatoes have a weight between 2 and 3 grams,\\
(a) find the number of tomatoes with a weight between 0 and 2 grams.
One of the tomatoes is selected at random.\\
(b) Find the probability that it weighs more than 3 grams.\\
(c) Estimate the proportion of the tomatoes with a weight greater than 6.25 grams.\\
(d) Using your answer to part (c), explain whether or not the median is greater than 6.25 grams.
Given that the mean weight of these tomatoes is 6.25 grams and using your answer to part (d),\\
(e) describe the skewness of the distribution of the weights of these tomatoes. Give a reason for your answer.
Two of these 100 tomatoes are selected at random.\\
(f) Estimate the probability that both tomatoes weigh within 0.75 grams of the mean.
\hfill \mbox{\textit{Edexcel S1 2017 Q1 [12]}}