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Draw a histogram and calculate the mean score from the given frequency distribution.

Mathematics

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Answers

2 parts
ii
66.34
iii
65.21

Step-by-step solution

8 steps

It looks like you're asking for the solution to the first question on the page, which involves the distribution table.

Part 1

(Distribution Table)

Given the frequency distribution:

Class IntervalFrequency (ff)
40-497
50-596
60-6910
70-797
80-894
90-994

Part i· 2 steps

Draw a histogram for the distribution.

  1. Determine the class boundaries.

    For continuous data, the class boundaries are found by taking the average of the upper limit of one class and the lower limit of the next class.
    Class boundaries:

    39.5 - 49.5 (Frequency: 7)

    49.5 - 59.5 (Frequency: 6)

    59.5 - 69.5 (Frequency: 10)

    69.5 - 79.5 (Frequency: 7)

    79.5 - 89.5 (Frequency: 4)

    89.5 - 99.5 (Frequency: 4)

  2. To draw a histogram, you would plot the class boundaries on the x-axis and the frequencies on the y-axis.

    Draw bars for each class interval, with the height of each bar corresponding to its frequency. The bars should touch each other.
    (As an AI, I cannot physically draw the histogram, but the data above provides the necessary information for you to draw it.)

Part ii66.34

Calculate the mean score.

  1. Find the midpoint (xx) for each class interval.

    Class IntervalFrequency (ff)Midpoint (xx)fxfx
    40-49744.5311.5
    50-59654.5327.0
    60-691064.5645.0
    70-79774.5521.5
    80-89484.5338.0
    90-99494.5378.0

  2. Calculate the sum of frequencies (∑f\sum f) and the sum of fxfx (∑fx\sum fx).

    ∑f=7+6+10+7+4+4=38\sum f = 7+6+10+7+4+4 = 38 ∑fx=311.5+327.0+645.0+521.5+338.0+378.0=2521\sum fx = 311.5+327.0+645.0+521.5+338.0+378.0 = 2521

  3. Calculate the mean using the formula xˉ=∑fx∑f\bar{x} = \frac{\sum fx}{\sum f}.

    xˉ=252138≈66.342\bar{x} = \frac{2521}{38} \approx 66.342

    The mean score is 66.34\boxed{66.34} (rounded to two decimal places).

Part iii65.21

Find the mode.

  1. Identify the modal class, which is the class interval with the highest frequency.

    The highest frequency is 10, corresponding to the class interval 60-69. So, the modal class is 60-69.

  2. Use the formula for the mode of grouped data

    Mode=L+(fm−f12fm−f1−f2)×w\text{Mode} = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times w

    Where:
    LL = lower class boundary of the modal class = 59.5
    fmf_m = frequency of the modal class = 10
    f1f_1 = frequency of the class before the modal class = 6
    f2f_2 = frequency of the class after the modal class = 7
    ww = class width = 69.5−59.5=1069.5 - 59.5 = 10

  3. Substitute the values into the formula.

    Mode=59.5+(10−62(10)−6−7)×10\text{Mode} = 59.5 + \left( \frac{10 - 6}{2(10) - 6 - 7} \right) \times 10 Mode=59.5+(420−13)×10\text{Mode} = 59.5 + \left( \frac{4}{20 - 13} \right) \times 10 Mode=59.5+(47)×10\text{Mode} = 59.5 + \left( \frac{4}{7} \right) \times 10 Mode=59.5+407\text{Mode} = 59.5 + \frac{40}{7} Mode=59.5+5.71428...\text{Mode} = 59.5 + 5.71428... Mode≈65.214\text{Mode} \approx 65.214

    The mode is 65.21\boxed{65.21} (rounded to two decimal places).

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