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Exercise 13.1 · Q6
Q.

Find the mean deviation about the mean for the following data:

xix_ifif_i
104
3024
5028
7016
908
Haryana BsehTextbookSubjective· 3mImportance★★★★★est
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Mean deviation about the mean measures the average absolute distance of each data point from the mean. For this grouped data, the mean is 5050, and the mean deviation is 1616.

The mean deviation about the mean tells you, on average, how far each observation is from the central value — but without caring whether it’s above or below (that’s why we take absolute values). For grouped data, we first find the mean, then compute the weighted sum of absolute deviations, and finally divide by the total frequency.

Let’s work through it step by step.

  1. Find the total number of observations Add up all the frequencies:

N=4+24+28+16+8=80N = 4 + 24 + 28 + 16 + 8 = 80

  1. Compute the mean (xˉ\bar{x})

    The mean for grouped data is xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}.

    First, calculate each fixif_i x_i:

    • 10×4=4010 \times 4 = 40
    • 30×24=72030 \times 24 = 720
    • 50×28=140050 \times 28 = 1400
    • 70×16=112070 \times 16 = 1120
    • 90×8=72090 \times 8 = 720

    Sum them: ∑fixi=40+720+1400+1120+720=4000\sum f_i x_i = 40 + 720 + 1400 + 1120 + 720 = 4000

    So the mean is:

xˉ=400080=50\bar{x} = \frac{4000}{80} = 50

Tip

Notice the data is symmetric around 50 — the frequencies rise then fall evenly. That’s a quick check that the mean is indeed 50.

  1. Find the absolute deviations from the mean

    For each xix_i, compute ∣xi−xˉ∣|x_i - \bar{x}|:

    • ∣10−50∣=40|10 - 50| = 40
    • ∣30−50∣=20|30 - 50| = 20
    • ∣50−50∣=0|50 - 50| = 0
    • ∣70−50∣=20|70 - 50| = 20
    • ∣90−50∣=40|90 - 50| = 40
  2. Multiply each absolute deviation by its frequency …

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