Question
The table below shows the number of words in the extended essays of an IB class.
a.Draw a histogram on the grid below for the data in this table.
[3]
b.Write down the modal group.[1]
c.The maximum word count is \(4000\) words.
Write down the probability that a student chosen at random is on or over the word count.[2]
▶️Answer/Explanation
Markscheme
(A3) (C3)
Notes: (A3) for correct histogram, (A2) for one error, (A1) for two errors, (A0) for more than two errors.
Award maximum (A2) if lines do not appear to be drawn with a ruler.
Award maximum (A2) if a frequency polygon is drawn.[3 marks]
\({\text{Modal group}} = 3800 \leqslant w < 4000\) (A1) (C1)[1 mark]
\({\text{Probability}} = \frac{3}{{35}}{\text{ }}(0.0857{\text{, }}8.57\% )\) (A1)(A1) (C2)
Note: (A1) for correct numerator (A1) for correct denominator.[2 marks]
Question
The following histogram shows the weights of a number of frozen chickens in a supermarket. The weights are grouped such that \(1 \leqslant {\text{weight}} < 2\), \(2 \leqslant {\text{weight}} < 3\) and so on.
b.Find the total number of chickens.[1]
c.Write down the modal group.[1]
d.Gabriel chooses a chicken at random.
Find the probability that this chicken weighs less than \(4{\text{ kg}}\).[2]
▶️Answer/Explanation
Markscheme
\(96\) (A1) (C1)[1 mark]
\(3 \leqslant {\text{weight}} < 4{\text{ kg}}\) . Accept \(3 – 4{\text{ kg}}\) (A1) (C1)[1 mark]
For adding three heights or subtracting \(14\) from \(96\) (M1)
\(\frac{{82}}{{96}}{\text{ }}(0.854{\text{ or }}\frac{{41}}{{48}}{\text{, }}85.4\% )\) (ft) from (b). (A1)(ft) (C2)[2 marks]
Question
a.The distribution of the weights, correct to the nearest kilogram, of the members of a football club is shown in the following table.
On the grid below draw a histogram to show the above weight distribution.
[2]
b.Write down the mid-interval value for the \(40 – 49\) interval.[1]
c.Find an estimate of the mean weight of the members of the club.[2]
d.Write down an estimate of the standard deviation of their weights.[1]
▶️Answer/Explanation
Markscheme
(A1)(A1) (C2)
Notes: (A1) for all correct heights, (A1) for all correct end points (\(39.5\), \(49.5\) etc.).
Histogram must be drawn with a ruler (straight edge) and endpoints must be clear.
Award (A1) only if both correct histogram and correct frequency polygon drawn. [2 marks]
\(44.5\) (A1) (C1)
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)
[1 mark]
Unit penalty (UP) applies in this question.
\({\text{Mean}} = \frac{{44.5 \times 6 + 54.5 \times 18 + \ldots }}{{42}}\) (M1)
Note: (M1) for a sum of frequencies multiplied by midpoint values divided by \(42\).
\( = 58.3{\text{ kg}}\) (A1)(ft) (C2)
Note: Award (A1)(A0)(AP) for \(58\).
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)[2 marks]
\({\text{Standard deviation}} = 8.44\) (A1) (C1)
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)[1 mark]
Question
120 Mathematics students in a school sat an examination. Their scores (given as a percentage) were summarized on a cumulative frequency diagram. This diagram is given below.
a.Complete the grouped frequency table for the students.
[3]
b.Write down the mid-interval value of the \(40 < x \leqslant 60\) interval.[1]
c.Calculate an estimate of the mean examination score of the students.[2]
▶️Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)[3 marks]
50 (A1) (C1)[1 mark]
\({\text{Mean}} = \frac{{10 \times 14 + ……. + 90 \times 6}}{{120}}\) (M1)
Note: Award (M1) for correct substitution of their values from (a) in mean formula.
\( = 45\frac{2}{3}(45.7)\) (A1)(ft) (C2)[2 marks]
Question
The weights of 90 students in a school were recorded. The information is displayed in the following table.
a.Write down the mid interval value for the interval \(50 \leqslant w \leqslant 60\).[1]
b.i.Use your graphic display calculator to find an estimate for the mean weight.[2]
b.ii.Use your graphic display calculator to find an estimate for the standard deviation.[1]
c.Find the weight that is 3 standard deviations below the mean.[2]
▶️Answer/Explanation
Markscheme
55 (A1) (C1)[1 mark]
\(62.\bar 5{\text{ }}\left( {62.6} \right)\) (A2)(ft) (C2)[2 marks]
8.86 (A1) (C1)
Note: Follow through from their answer to part (a).[1 mark]
62.6 – 3 × 8.86 = 36.0 (M1)(A1)(ft) (C2)
Note: Accept 36.
Follow through from their values in part (b) only if working is seen.[2 marks]