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Find the mean number of mangoes kept in a packing box. Which method did you choose?

Mathematics

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Answer

57.1875 mangoes\text{57.1875 mangoes}

Step-by-step solution

5 steps
  1. Calculate the class mark (xix_i) for each class interval.

    The class mark is the midpoint of the class interval.
    For 50 - 52: x1=50+522=51x_1 = \frac{50+52}{2} = 51
    For 53 - 55: x2=53+552=54x_2 = \frac{53+55}{2} = 54
    For 56 - 58: x3=56+582=57x_3 = \frac{56+58}{2} = 57
    For 59 - 61: x4=59+612=60x_4 = \frac{59+61}{2} = 60
    For 62 - 64: x5=62+642=63x_5 = \frac{62+64}{2} = 63

  2. Choose an assumed mean (AA) and calculate the class size (hh).

    Let's choose the middle class mark as the assumed mean, A=57A = 57.
    The class size hh is the difference between consecutive class marks, h=54−51=3h = 54 - 51 = 3.

  3. Calculate the deviation (di=xi−Ad_i = x_i - A) and the step-deviation (ui=dihu_i = \frac{d_i}{h}) for each class.

  4. Calculate the product of frequency and step-deviation (fiuif_i u_i).

    We can organize these calculations in a table:

    Number of mangoesNumber of boxes (fi)Class Mark (xi)di=xi−57ui=di3fiui50−521551−6−2−3053−5511054−3−1−11056−581355700059−61115603111562−6425636250Total∑fi=400∑fiui=25\begin{array}{|c|c|c|c|c|c|} \hline \text{Number of mangoes} & \text{Number of boxes } (f_i) & \text{Class Mark } (x_i) & d_i = x_i - 57 & u_i = \frac{d_i}{3} & f_i u_i \\ \hline 50 - 52 & 15 & 51 & -6 & -2 & -30 \\ 53 - 55 & 110 & 54 & -3 & -1 & -110 \\ 56 - 58 & 135 & 57 & 0 & 0 & 0 \\ 59 - 61 & 115 & 60 & 3 & 1 & 115 \\ 62 - 64 & 25 & 63 & 6 & 2 & 50 \\ \hline \textbf{Total} & \sum f_i = 400 & & & & \sum f_i u_i = 25 \\ \hline \end{array}
  5. Apply the formula for the mean using the step-deviation method.

    The formula for the mean (xˉ\bar{x}) using the step-deviation method is:

    xˉ=A+(∑fiui∑fi)h\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right) h

    Substitute the values:

    xˉ=57+(25400)×3\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3 xˉ=57+(116)×3\bar{x} = 57 + \left(\frac{1}{16}\right) \times 3 xˉ=57+316\bar{x} = 57 + \frac{3}{16} xˉ=57+0.1875\bar{x} = 57 + 0.1875 xˉ=57.1875\bar{x} = 57.1875

    The method chosen is the Step-Deviation Method.

    The mean number of mangoes kept in a packing box is 57.1875 mangoes\text{57.1875 mangoes}.

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