Edexcel S1 2010 January — Question 3 11 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2010
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate statistics from grouped frequency table
DifficultyModerate -0.8 This is a routine S1 statistics question testing standard procedures: finding midpoints (trivial arithmetic), calculating mean and standard deviation from grouped data using given summations (direct formula application), linear interpolation for the median (standard technique), and identifying skewness by comparing mean/median. All parts are textbook exercises requiring only recall and careful arithmetic, with no problem-solving or conceptual insight needed.
Spec2.02a Interpret single variable data: tables and diagrams2.02f Measures of average and spread2.02g Calculate mean and standard deviation

The birth weights, in kg, of 1500 babies are summarised in the table below.
Weight (kg)Midpoint, \(x\)kgFrequency, \(f\)
\(0.0 - 1.0\)\(0.50\)\(1\)
\(1.0 - 2.0\)\(1.50\)\(6\)
\(2.0 - 2.5\)\(2.25\)\(60\)
\(2.5 - 3.0\)\(280\)
\(3.0 - 3.5\)\(3.25\)\(820\)
\(3.5 - 4.0\)\(3.75\)\(320\)
\(4.0 - 5.0\)\(4.50\)\(10\)
\(5.0 - 6.0\)\(3\)
[You may use \(\sum fx = 4841\) and \(\sum fx^2 = 15889.5\)]
  1. Write down the missing midpoints in the table above. [2]
  2. Calculate an estimate of the mean birth weight. [2]
  3. Calculate an estimate of the standard deviation of the birth weight. [3]
  4. Use interpolation to estimate the median birth weight. [2]
  5. Describe the skewness of the distribution. Give a reason for your answer. [2]

Part (a)
Answer: 2.75 or \(2\frac{3}{4}\), 5.5 or 5.50 or \(5\frac{1}{2}\)
AnswerMarks
Marks: B1 B1(2)
Part (b)
Answer: Mean birth weight \(= \frac{4841}{1500} = 3.2273\) awrt 3.23
AnswerMarks
Marks: M1 A1(2)
Guidance:
- M1 for a correct expression for mean
- Answer only scores both
Part (c)
Answer: Standard deviation \(= \sqrt{\frac{15889.5}{1500} - \left(\frac{4841}{1500}\right)^2} = 0.421093...\) or \(s = 0.4212337...\)
AnswerMarks
Marks: M1 A1ft A1(3)
Guidance:
- M1 for a correct expression (fit their mean) for sd or variance. Condone mis-labelling eg sd = ... with no square root or no labelling
- 1st A1ft for a correct expression (fit their mean) including square root and no mis-labelling. Allow 1st A1 for \(\sigma^2 = 0.177... \to \sigma = 0.42...\)
- 2nd A1 for awrt 0.421. Answer only scores 3/3
Part (d)
Answer: \(Q_2 = 3.00 + \frac{403}{820} \times 0.5 = 3.2457...\) (allow 403.5...... → 3.25)
AnswerMarks
Marks: M1 A1(2)
Guidance:
- M1 for a correct expression (allow 403.5 i.e. use of \(n+1\)) but must have 3.00, 820 and 0.5
- A1 for awrt 3.25 provided M1 is scored. NB 3.25 with no working scores 0/2 as some candidates think mode is 3.25
Part (e)
Answer: Mean(3.23) < Median(3.25) (or very close). Negative Skew (or symmetrical)
AnswerMarks
Marks: B1ft dB1ft(2)
Guidance:
- 1st B1ft for a comparison of their mean and median (may be in a formula but if \(\pm\)(mean - median) is calculated that's OK. We are not checking the value but the sign must be consistent.)
- Also allow for use of quartiles provided correct values seen: \(Q_1 = 3.02\), \(Q_3 = 3.47\). [They should get (0.22 =) \(Q_2 - Q_1\) > \(Q_3 - Q_1\) (= 0.23) and say (slight) negative skew or symmetric]
- 2nd dB1ft for a compatible comment based on their comparison. Dependent upon a suitable, correct comparison. Mention of "correlation" rather than "skewness" loses this mark
## Part (a)
**Answer:** 2.75 or $2\frac{3}{4}$, 5.5 or 5.50 or $5\frac{1}{2}$

**Marks:** B1 B1 | (2)

## Part (b)
**Answer:** Mean birth weight $= \frac{4841}{1500} = 3.2273$ awrt 3.23

**Marks:** M1 A1 | (2)

**Guidance:**
- M1 for a correct expression for mean
- Answer only scores both

## Part (c)
**Answer:** Standard deviation $= \sqrt{\frac{15889.5}{1500} - \left(\frac{4841}{1500}\right)^2} = 0.421093...$ or $s = 0.4212337...$

**Marks:** M1 A1ft A1 | (3)

**Guidance:**
- M1 for a correct expression (fit their mean) for sd or variance. Condone mis-labelling eg sd = ... with no square root or no labelling
- 1st A1ft for a correct expression (fit their mean) including square root and no mis-labelling. Allow 1st A1 for $\sigma^2 = 0.177... \to \sigma = 0.42...$
- 2nd A1 for awrt 0.421. Answer only scores 3/3

## Part (d)
**Answer:** $Q_2 = 3.00 + \frac{403}{820} \times 0.5 = 3.2457...$ (allow 403.5...... → 3.25)

**Marks:** M1 A1 | (2)

**Guidance:**
- M1 for a correct expression (allow 403.5 i.e. use of $n+1$) but must have 3.00, 820 and 0.5
- A1 for awrt 3.25 provided M1 is scored. NB 3.25 with no working scores 0/2 as some candidates think mode is 3.25

## Part (e)
**Answer:** Mean(3.23) < Median(3.25) (or very close). Negative Skew (or symmetrical)

**Marks:** B1ft dB1ft | (2)

**Guidance:**
- 1st B1ft for a comparison of their mean and median (may be in a formula but if $\pm$(mean - median) is calculated that's OK. We are not checking the value but the sign must be consistent.)
- Also allow for use of quartiles provided correct values seen: $Q_1 = 3.02$, $Q_3 = 3.47$. [They should get (0.22 =) $Q_2 - Q_1$ > $Q_3 - Q_1$ (= 0.23) and say (slight) negative skew or symmetric]
- 2nd dB1ft for a compatible comment based on their comparison. Dependent upon a suitable, correct comparison. Mention of "correlation" rather than "skewness" loses this mark

---
The birth weights, in kg, of 1500 babies are summarised in the table below.

\begin{tabular}{|c|c|c|}
\hline
Weight (kg) & Midpoint, $x$kg & Frequency, $f$ \\
\hline
$0.0 - 1.0$ & $0.50$ & $1$ \\
$1.0 - 2.0$ & $1.50$ & $6$ \\
$2.0 - 2.5$ & $2.25$ & $60$ \\
$2.5 - 3.0$ & & $280$ \\
$3.0 - 3.5$ & $3.25$ & $820$ \\
$3.5 - 4.0$ & $3.75$ & $320$ \\
$4.0 - 5.0$ & $4.50$ & $10$ \\
$5.0 - 6.0$ & & $3$ \\
\hline
\end{tabular}

[You may use $\sum fx = 4841$ and $\sum fx^2 = 15889.5$]

\begin{enumerate}[label=(\alph*)]
\item Write down the missing midpoints in the table above. [2]

\item Calculate an estimate of the mean birth weight. [2]

\item Calculate an estimate of the standard deviation of the birth weight. [3]

\item Use interpolation to estimate the median birth weight. [2]

\item Describe the skewness of the distribution. Give a reason for your answer. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2010 Q3 [11]}}