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Calculate the number of years, months and days from 1st Jan 2015 to 23rd July 2023, with 365 days to a year and 30 days to a month.

Mathematics

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Answers

3 parts
a

17\textbf{17}

b

17.17\textbf{17.17}

c

17.17\textbf{17.17}

Step-by-step solution

10 steps

Here is the solution for Question 2:

The table shows the distribution of marks obtained by 40 students:

Marks (x)Number of students (f)
1-52
6-104
11-1510
16-2012
21-258
26-304

Part a· 3 steps

Calculate the mean mark.

  1. Determine the midpoint (xx) for each class interval and calculate fxfx.

    MarksfMidpoint (x)fx
    1-521+52=3\frac{1+5}{2} = 32×3=62 \times 3 = 6
    6-1046+102=8\frac{6+10}{2} = 84×8=324 \times 8 = 32
    11-151011+152=13\frac{11+15}{2} = 1310×13=13010 \times 13 = 130
    16-201216+202=18\frac{16+20}{2} = 1812×18=21612 \times 18 = 216
    21-25821+252=23\frac{21+25}{2} = 238×23=1848 \times 23 = 184
    26-30426+302=28\frac{26+30}{2} = 284×28=1124 \times 28 = 112
  2. Calculate the sum of frequencies (∑f\sum f) and the sum of fxfx (∑fx\sum fx).

    ∑f=2+4+10+12+8+4=40\sum f = 2+4+10+12+8+4 = 40 ∑fx=6+32+130+216+184+112=680\sum fx = 6+32+130+216+184+112 = 680
  3. Apply the formula for the mean of grouped data.

    Mean(xˉ)=∑fx∑f\text{Mean} (\bar{x}) = \frac{\sum fx}{\sum f} xˉ=68040\bar{x} = \frac{680}{40} xˉ=17\bar{x} = \boxed{\textbf{17}}

Part b

17.17\textbf{17.17}

Construct a cumulative frequency table and use it to determine the median mark.

  1. Construct the cumulative frequency table, including class boundaries.

    MarksfClass BoundariesCumulative Frequency (cf)
    1-520.5 - 5.52
    6-1045.5 - 10.52+4=62+4=6
    11-151010.5 - 15.56+10=166+10=16
    16-201215.5 - 20.516+12=2816+12=28
    21-25820.5 - 25.528+8=3628+8=36
    26-30425.5 - 30.536+4=4036+4=40
  2. Determine the median position and identify the median class.

    Total frequency N=40N = 40.
    The median position is N2=402=20th\frac{N}{2} = \frac{40}{2} = 20^{th} value.
    The cumulative frequency first exceeds 20 in the class 16-20 (cf = 28). So, the median class is 16-20.

  3. Identify the values for the median formula.

    The median formula for grouped data is: Median=L+(N2−cfbfm)×wMedian = L + \left(\frac{\frac{N}{2} - cf_b}{f_m}\right) \times w
    Where:
    LL = Lower class boundary of the median class = 15.5
    NN = Total frequency = 40
    cfbcf_b = Cumulative frequency of the class before the median class = 16
    fmf_m = Frequency of the median class = 12
    ww = Class width = 5.5−0.5=55.5 - 0.5 = 5

  4. Apply the median formula.

    Median=15.5+(402−1612)×5Median = 15.5 + \left(\frac{\frac{40}{2} - 16}{12}\right) \times 5 Median=15.5+(20−1612)×5Median = 15.5 + \left(\frac{20 - 16}{12}\right) \times 5 Median=15.5+(412)×5Median = 15.5 + \left(\frac{4}{12}\right) \times 5 Median=15.5+13×5Median = 15.5 + \frac{1}{3} \times 5 Median=15.5+53Median = 15.5 + \frac{5}{3} Median=15.5+1.666...Median = 15.5 + 1.666... Median≈17.17Median \approx \boxed{\textbf{17.17}}

Part c

17.17\textbf{17.17}

Determine the mode mark.

  1. Identify the modal class.

    The modal class is the class with the highest frequency. From the table, the highest frequency is 12, which corresponds to the class interval 16-20. So, the modal class is 16-20.

  2. Identify the values for the mode formula.

    The mode formula for grouped data is: Mode=L+(fm−f1(fm−f1)+(fm−f2))×wMode = L + \left(\frac{f_m - f_1}{(f_m - f_1) + (f_m - f_2)}\right) \times w
    Where:
    LL = Lower class boundary of the modal class = 15.5
    fmf_m = Frequency of the modal class = 12
    f1f_1 = Frequency of the class before the modal class (11-15) = 10
    f2f_2 = Frequency of the class after the modal class (21-25) = 8
    ww = Class width = 5

  3. Apply the mode formula.

    Mode=15.5+(12−10(12−10)+(12−8))×5Mode = 15.5 + \left(\frac{12 - 10}{(12 - 10) + (12 - 8)}\right) \times 5 Mode=15.5+(22+4)×5Mode = 15.5 + \left(\frac{2}{2 + 4}\right) \times 5 Mode=15.5+(26)×5Mode = 15.5 + \left(\frac{2}{6}\right) \times 5 Mode=15.5+13×5Mode = 15.5 + \frac{1}{3} \times 5 Mode=15.5+53Mode = 15.5 + \frac{5}{3} Mode=15.5+1.666...Mode = 15.5 + 1.666... Mode≈17.17Mode \approx \boxed{\textbf{17.17}}
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