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The table below shows the marks obtained by 278 candidates in an examination. a) State the modal class of the distribution. b) Construct a cumulative frequency table and draw a cumulative frequency cu

Mathematics

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3 parts
a

41-50\text{41-50}

c
45.81
42.38
d
0.1295

Step-by-step solution

10 steps

Here are the solutions to your question:

Part a· 2 steps

State the modal class of the distribution.

  1. Identify the highest frequency in the "Number of candidates" column.

    The highest frequency is 81.

  2. Determine the class interval corresponding to this frequency.

    The class interval for a frequency of 81 is 41-50.

    The modal class is 41-50\boxed{\text{41-50}}.

Part b· 2 steps

Construct a cumulative frequency table and draw a cumulative frequency curve.

  1. Create a cumulative frequency table by adding the frequencies sequentially.

    Also, determine the upper class boundaries for plotting the curve.

    MarksUpper Class BoundaryNumber of candidates (f)Cumulative Frequency (cf)1−1010.52211−2020.562+6=821−3030.5308+30=3831−4040.55838+58=9641−5050.58196+81=17751−6060.565177+65=24261−7070.528242+28=27071−8080.58270+8=278\begin{array}{|c|c|c|c|} \hline \text{Marks} & \text{Upper Class Boundary} & \text{Number of candidates (f)} & \text{Cumulative Frequency (cf)} \\ \hline 1-10 & 10.5 & 2 & 2 \\ 11-20 & 20.5 & 6 & 2+6=8 \\ 21-30 & 30.5 & 30 & 8+30=38 \\ 31-40 & 40.5 & 58 & 38+58=96 \\ 41-50 & 50.5 & 81 & 96+81=177 \\ 51-60 & 60.5 & 65 & 177+65=242 \\ 61-70 & 70.5 & 28 & 242+28=270 \\ 71-80 & 80.5 & 8 & 270+8=278 \\ \hline \end{array}

  2. To draw the cumulative frequency curve (ogive), plot the upper class boundaries against their corresponding cumulative frequencies.

    The points to plot are: $(10.5, 2)$, $(20.5, 8)$, $(30.5, 38)$, $(40.5, 96)$, $(50.5, 177)$, $(60.5, 242)$, $(70.5, 270)$, $(80.5, 278)$.
    Start the curve from $(0.5, 0)$ and connect the plotted points with a smooth curve.

Part c45.81

Use your curve to estimate

  • i) the median of the distribution

  1. Calculate the position of the median.

    The total number of candidates (NN) is 278.
    The median position is N2=2782=139\frac{N}{2} = \frac{278}{2} = 139.

  2. Locate the median on the cumulative frequency curve (or by interpolation).

    Find the mark corresponding to a cumulative frequency of 139.
    The 139th value falls within the 41-50 class (cumulative frequency 177).
    Using interpolation:

    Median=L+(N2−cfbfm)×c\text{Median} = L + \left(\frac{\frac{N}{2} - cf_b}{f_m}\right) \times c

    Where L=40.5L = 40.5 (lower boundary of median class), cfb=96cf_b = 96 (cumulative frequency before median class), fm=81f_m = 81 (frequency of median class), c=10c = 10 (class width).

    Median=40.5+(139−9681)×10\text{Median} = 40.5 + \left(\frac{139 - 96}{81}\right) \times 10 Median=40.5+(4381)×10\text{Median} = 40.5 + \left(\frac{43}{81}\right) \times 10 Median=40.5+5.3086\text{Median} = 40.5 + 5.3086 Median≈45.81\text{Median} \approx \boxed{45.81}
    • ii) the 40th percentile.

  3. Calculate the position of the 40th percentile.

    The position is 0.40×N=0.40×278=111.20.40 \times N = 0.40 \times 278 = 111.2.

  4. Locate the 40th percentile on the cumulative frequency curve (or by interpolation).

    Find the mark corresponding to a cumulative frequency of 111.2.
    The 111.2th value falls within the 41-50 class (cumulative frequency 177).
    Using interpolation:

    40th Percentile=L+(Position−cfbfp)×c\text{40th Percentile} = L + \left(\frac{\text{Position} - cf_b}{f_p}\right) \times c

    Where L=40.5L = 40.5, cfb=96cf_b = 96, fp=81f_p = 81, c=10c = 10.

    40th Percentile=40.5+(111.2−9681)×10\text{40th Percentile} = 40.5 + \left(\frac{111.2 - 96}{81}\right) \times 10 40th Percentile=40.5+(15.281)×10\text{40th Percentile} = 40.5 + \left(\frac{15.2}{81}\right) \times 10 40th Percentile=40.5+1.8765\text{40th Percentile} = 40.5 + 1.8765 40th Percentile≈42.38\text{40th Percentile} \approx \boxed{42.38}

Part d0.1295

What is the probability that a candidate chosen at random scored a mark of at least 60.5%?

  1. Identify the number of candidates who scored at least 60.5%.

    A mark of at least 60.5% means marks ≥60.5\ge 60.5. This includes candidates in the 61-70 class and the 71-80 class.
    Number of candidates with mark ≥60.5=(frequency of 61-70)+(frequency of 71-80)\ge 60.5 = (\text{frequency of 61-70}) + (\text{frequency of 71-80})
    Number of candidates =28+8=36= 28 + 8 = 36.

  2. Calculate the probability.

    The total number of candidates is 278.

    P(mark≥60.5%)=Number of candidates with mark≥60.5Total number of candidatesP(\text{mark} \ge 60.5\%) = \frac{\text{Number of candidates with mark} \ge 60.5}{\text{Total number of candidates}} P(mark≥60.5%)=36278P(\text{mark} \ge 60.5\%) = \frac{36}{278} P(mark≥60.5%)=18139P(\text{mark} \ge 60.5\%) = \frac{18}{139} P(mark≥60.5%)≈0.1295P(\text{mark} \ge 60.5\%) \approx \boxed{0.1295}
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