9709/63

Mathematics 9709/63May/June 2018

Cambridge AS Level · Probability & Statistics 1 (S1) · worked solutions for every part, with the mark scheme

7
questions
50
marks
75
minutes

Topics Discrete Random Variables · Representation of Data · The Normal Distribution · Probability · Permutations and Combinations

Q1Representation of DataFree sample

The masses in kilograms of 50 children having a medical check-up were recorded correct to the nearest kilogram. The results are shown in the table.

Mass (kg)101410-14151915-19202420-24253425-34355935-59
Frequency61214108
(i)

Find which class interval contains the lower quartile.

1M
DifficultyMedium-Easy
Worked solution

Approach

Find the lower quartile position using the total frequency, then use cumulative frequencies to identify which class interval contains this value.

Working

Total frequency:

N=6+12+14+10+8=50N = 6 + 12 + 14 + 10 + 8 = 50

Lower quartile position:

N4=504=12.5\frac{N}{4} = \frac{50}{4} = 12.5

Cumulative frequencies:

Mass (kg)<14.5:6<19.5:6+12=18<24.5:18+14=32<34.5:32+10=42<59.5:42+8=50\begin{aligned} \text{Mass (kg)} &< 14.5: & 6 \\ &< 19.5: & 6 + 12 = 18 \\ &< 24.5: & 18 + 14 = 32 \\ &< 34.5: & 32 + 10 = 42 \\ &< 59.5: & 42 + 8 = 50 \end{aligned}

The 12.5th value falls in the class interval 15151919 since 6<12.5186 < 12.5 \leq 18.

Answer

151915\text{–}19
Final answer

15-19

Detailed explanation

Walkthrough

First, we calculate the total number of children recorded, which is the sum of all frequencies: 6+12+14+10+8=506 + 12 + 14 + 10 + 8 = 50. The lower quartile (Q1) is the value below which 25% of the data falls. For 50 data points, the lower quartile position is 504=12.5\frac{50}{4} = 12.5.

Next, we build up the cumulative frequency table to see where the 12.5th value lies. The first class (10–14) has 6 children, so cumulative frequency is 6. Adding the second class (15–19) with 12 children gives a cumulative frequency of 18. Since 6<12.5186 < 12.5 \leq 18, the lower quartile must lie in the 15–19 class interval.

Key Takeaways

  • The lower quartile position for NN data points is N/4N/4.
  • Cumulative frequencies help identify which class interval contains a given percentile.
  • For grouped data, we can only identify the class interval, not the exact value, of a quartile.

Common Mistakes

  • Using N/4+1N/4 + 1 instead of N/4N/4 for the lower quartile position.
  • Forgetting that class boundaries for continuous data are at the half-integers (e.g., 14.5, 19.5), so the class 15–19 has upper boundary 19.5.
  • Miscounting cumulative frequencies.

Things to Be Careful About

  • The class intervals are 10101414, 15151919, 20202424, 25253434, 35355959. These are continuous intervals with boundaries at 9.5,14.5,19.5,24.5,34.5,59.59.5, 14.5, 19.5, 24.5, 34.5, 59.5.
  • The question accepts 15151919 or 14.514.519.519.5 as the answer.
  • The unit (kg) is not required for the mark.
Techniques used
calculate total frequencyfind lower quartile positionbuild cumulative frequency tableidentify class interval
(ii)

On the grid, draw a histogram to illustrate the data in the table.

4M
DifficultyMedium
Worked solution

Approach

For a histogram with unequal class widths, the vertical axis must represent frequency density (fd = frequency / class width), and bar widths must be proportional to class widths.

Working

Step 1: Calculate class boundaries and class widths.

The class boundaries are at the half-integers:

10149.5 to 14.5,width=5151914.5 to 19.5,width=5202419.5 to 24.5,width=5253424.5 to 34.5,width=10355934.5 to 59.5,width=25\begin{aligned} 10\text{–}14 &\Rightarrow 9.5 \text{ to } 14.5, \quad \text{width} = 5 \\ 15\text{–}19 &\Rightarrow 14.5 \text{ to } 19.5, \quad \text{width} = 5 \\ 20\text{–}24 &\Rightarrow 19.5 \text{ to } 24.5, \quad \text{width} = 5 \\ 25\text{–}34 &\Rightarrow 24.5 \text{ to } 34.5, \quad \text{width} = 10 \\ 35\text{–}59 &\Rightarrow 34.5 \text{ to } 59.5, \quad \text{width} = 25 \end{aligned}

Step 2: Calculate frequency densities.

fd for 1014=65=1.2fd for 1519=125=2.4fd for 2024=145=2.8fd for 2534=1010=1.0fd for 3559=825=0.32\begin{aligned} \text{fd for } 10\text{–}14 &= \frac{6}{5} = 1.2 \\ \text{fd for } 15\text{–}19 &= \frac{12}{5} = 2.4 \\ \text{fd for } 20\text{–}24 &= \frac{14}{5} = 2.8 \\ \text{fd for } 25\text{–}34 &= \frac{10}{10} = 1.0 \\ \text{fd for } 35\text{–}59 &= \frac{8}{25} = 0.32 \end{aligned}

Step 3: Draw the histogram.

  • Horizontal axis: Mass (kg), from 9.59.5 to 59.559.5, with linear scale.
  • Vertical axis: Frequency density (fd), from 00 to at least 33, with linear scale.
  • Bar widths in ratio 5:5:5:10:25=1:1:1:2:55 : 5 : 5 : 10 : 25 = 1 : 1 : 1 : 2 : 5.
  • Bar heights equal to the frequency densities: 1.2,2.4,2.8,1.0,0.321.2, 2.4, 2.8, 1.0, 0.32.
  • Bars are drawn without gaps, starting from x=9.5x = 9.5.

Answer

Histogram with frequency densities 1.2,2.4,2.8,1.0,0.321.2, 2.4, 2.8, 1.0, 0.32 and class widths 5,5,5,10,255, 5, 5, 10, 25.

Frequency densities: 1.2, 2.4, 2.8, 1.0, 0.32\text{Frequency densities: } 1.2,\ 2.4,\ 2.8,\ 1.0,\ 0.32
Final answer

Histogram with fd = 1.2, 2.4, 2.8, 1.0, 0.32 and bar widths 5, 5, 5, 10, 25

Detailed explanation

Walkthrough

When drawing a histogram for grouped data with unequal class widths, the key principle is that the area of each bar must be proportional to the frequency. This means:

  • Bar width = class width
  • Bar height = frequency density = frequency / class width
  • Area = width × height = class width × (frequency / class width) = frequency

Step 1: Identify the class boundaries. Since the data is continuous (masses recorded to the nearest kg), the boundaries are at the half-integers: 9.5, 14.5, 19.5, 24.5, 34.5, 59.5.

Step 2: Calculate the class widths:

  • 10–14: 14.59.5=514.5 - 9.5 = 5
  • 15–19: 19.514.5=519.5 - 14.5 = 5
  • 20–24: 24.519.5=524.5 - 19.5 = 5
  • 25–34: 34.524.5=1034.5 - 24.5 = 10
  • 35–59: 59.534.5=2559.5 - 34.5 = 25

Step 3: Calculate the frequency densities (fd = frequency / class width):

  • 6/5=1.26 / 5 = 1.2
  • 12/5=2.412 / 5 = 2.4
  • 14/5=2.814 / 5 = 2.8
  • 10/10=1.010 / 10 = 1.0
  • 8/25=0.328 / 25 = 0.32

Step 4: Draw the histogram on the grid:

  • The horizontal axis must be labelled "Mass (kg)" and range from at least 9.5 to 59.5 with a linear scale.
  • The vertical axis must be labelled "fd" (or "frequency density") and range from 0 to at least 3, with a linear scale and at least 3 equally spaced values marked.
  • Draw five bars with no gaps between them, starting at x = 9.5.
  • The first four bars have equal width (proportional to 5), and the last two have widths in ratio 10:25 = 2:5 relative to the first.
  • The heights are 1.2, 2.4, 2.8, 1.0, and 0.32 respectively.

Key Takeaways

  • For histograms with unequal class widths, always use frequency density on the vertical axis.
  • Frequency density = frequency / class width.
  • Bar widths must be proportional to class widths, and bars must have no gaps.
  • The horizontal axis must start at the lower class boundary (9.5), not at zero, unless a break is indicated.

Common Mistakes

  • Using frequency instead of frequency density on the vertical axis (this gives incorrect bar heights for unequal class widths).
  • Forgetting that the class 35–59 has width 25, not 24 or 20.
  • Drawing bars with gaps between them (histograms for continuous data have no gaps).
  • Starting the horizontal axis at 0 without indicating a break, when the data starts at 9.5.
  • Not labelling both axes correctly (must include "Mass (kg)" and "fd" or "frequency density").

Things to Be Careful About

  • The class widths are 5, 5, 5, 10, 25 — these are not equal, so frequency density must be used.
  • The horizontal axis must range from at least 9.5 to 59.5. If it starts from zero, a break in the scale must be indicated.
  • The vertical axis must be linear with at least 3 equally spaced values marked.
  • At least 3 linearly spaced values must appear on each axis for full marks.
Techniques used
calculate class widthscalculate frequency densitiesdetermine class boundariesdraw histogram with correct proportions

The rest of this paper

6 more questions
  • Q2The Normal Distribution · Discrete Random Variables6M
  • Q3Probability6M
  • Q4Representation of Data7M
  • Q5Discrete Random Variables · Probability8M
  • Q6The Normal Distribution · Discrete Random Variables8M
  • Q7Permutations and Combinations10M
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