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Construct a grouped frequency distribution table and calculate the mean mark and standard deviation given a set of data and an assumed mean.

Mathematics

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here's the solution for question 12:

a) Construct a grouped frequency distribution table using the class interval 20-29, 30-39, 40-49...

  1. Define class intervals and tally the given marks.

    The marks are:
    29 52 37 79 63 48 65 46 54 32 51 66 74 24 35 61 58 39 49 53
    45 58 56 38 59 28 50 49 67 56 37 45 79 68 43 66 26 47 55 71

    The class intervals are:
    20-29
    30-39
    40-49
    50-59
    60-69
    70-79

  2. Count the frequency for each class interval.

    • 20-29: 29, 24, 28, 26 (4 marks)
    • 30-39: 37, 32, 35, 39, 38, 37 (6 marks)
    • 40-49: 48, 46, 49, 45, 49, 43, 45, 47 (8 marks)
    • 50-59: 52, 54, 51, 53, 58, 58, 56, 59, 50, 56, 55 (11 marks)
    • 60-69: 63, 65, 66, 61, 67, 68, 66 (7 marks)
    • 70-79: 79, 74, 79, 71 (4 marks)
  3. Construct the frequency distribution table.

    Class IntervalTallyFrequency (f)20−29∣∣∣∣430−39∣∣∣∣∣∣640−49∣∣∣∣∣∣∣∣850−59∣∣∣∣∣∣∣∣∣∣∣1160−69∣∣∣∣∣∣∣770−79∣∣∣∣4Total40\begin{array}{|c|c|c|} \hline \textbf{Class Interval} & \textbf{Tally} & \textbf{Frequency (f)} \\ \hline 20-29 & |||| & 4 \\ 30-39 & ||||| | & 6 \\ 40-49 & ||||| ||| & 8 \\ 50-59 & ||||| ||||| | & 11 \\ 60-69 & ||||| || & 7 \\ 70-79 & |||| & 4 \\ \hline \textbf{Total} & & \textbf{40} \\ \hline \end{array}

    b) Using an assumed mean of 54.5, calculate, correct to the nearest whole number, the:

    (i) mean mark;

    (ii) standard deviation of the distribution.

  4. Create an extended frequency table to calculate class marks (xx), deviations (dd), fdfd, and fd2fd^2.

    The assumed mean A=54.5A = 54.5.
    The class mark xx is the midpoint of each interval. For example, for 20-29, x=20+292=24.5x = \frac{20+29}{2} = 24.5.
    The deviation d=x−Ad = x - A.

    Class Intervalfxd = x - 54.5fdd2fd220−29424.5−30−120900360030−39634.5−20−120400240040−49844.5−10−8010080050−591154.5000060−69764.5107010070070−79474.520804001600Total∑f=40∑fd=−170∑fd2=9100\begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Class Interval} & \textbf{f} & \textbf{x} & \textbf{d = x - 54.5} & \textbf{fd} & \textbf{d}^2 & \textbf{fd}^2 \\ \hline 20-29 & 4 & 24.5 & -30 & -120 & 900 & 3600 \\ 30-39 & 6 & 34.5 & -20 & -120 & 400 & 2400 \\ 40-49 & 8 & 44.5 & -10 & -80 & 100 & 800 \\ 50-59 & 11 & 54.5 & 0 & 0 & 0 & 0 \\ 60-69 & 7 & 64.5 & 10 & 70 & 100 & 700 \\ 70-79 & 4 & 74.5 & 20 & 80 & 400 & 1600 \\ \hline \textbf{Total} & \sum f = 40 & & & \sum fd = -170 & & \sum fd^2 = 9100 \\ \hline \end{array}

    (i) Calculate the mean mark.
    The formula for the mean using an assumed mean is xˉ=A+∑fd∑f\bar{x} = A + \frac{\sum fd}{\sum f}.

    xˉ=54.5+−17040\bar{x} = 54.5 + \frac{-170}{40} xˉ=54.5−4.25\bar{x} = 54.5 - 4.25 xˉ=50.25\bar{x} = 50.25

    Correct to the nearest whole number, the mean mark is 50\boxed{\text{50}}.

    (ii) Calculate the standard deviation of the distribution.
    The formula for standard deviation using an assumed mean is σ=∑fd2∑f−(∑fd∑f)2\sigma = \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}.

    σ=910040−(−17040)2\sigma = \sqrt{\frac{9100}{40} - \left(\frac{-170}{40}\right)^2} σ=227.5−(−4.25)2\sigma = \sqrt{227.5 - (-4.25)^2} σ=227.5−18.0625\sigma = \sqrt{227.5 - 18.0625} σ=209.4375\sigma = \sqrt{209.4375} σ≈14.472\sigma \approx 14.472

    Correct to the nearest whole number, the standard deviation is 14\boxed{\text{14}}.

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