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Worked Examples · Example 6
Q.

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

Marks obtainedNumber of students
10-202
20-303
30-408
40-5014
50-608
60-703
70-802
Rajasthan RbseTextbookSubjective· 5mImportance★★★★★est
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Using class midpoints, the mean is xˉ=45\bar{x}=45, and the mean deviation about the mean is 40040=10\dfrac{400}{40}=10.

For a continuous frequency distribution we treat every observation in a class as sitting at the class midpoint (class mark), then weight each absolute deviation by the class frequency — a standard NCERT Class 11 Maths Statistics technique.

Step 1 — Class marks and ∑fixi\sum f_i x_i.

Midpoint =lower+upper2=\dfrac{\text{lower}+\text{upper}}{2}.

Marksxix_ifif_ifixif_i x_i
10–2015230
20–3025375
30–40358280
40–504514630
50–60558440
60–70653195
70–80752150
Total401800

Step 2 — Mean.

xˉ=∑fixi∑fi=180040=45.\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{1800}{40}=45.

Step 3 — Frequency-weighted absolute deviations.

xix_ifif_i∣xi−45∣\lvert x_i-45\rvertfi∣xi−45∣f_i\lvert x_i-45\rvert
1523060
2532060
3581080
451400

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