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Worked Examples · Example 32
Q.

Suppose a class of 25 students conducted a quiz and grades obtained are given in following table. Find the mean deviation about mean of data.

Grade510152025
Number of students74635
CBSENCERTSubjective· 3mImportance★★★★★est
77% · 44/57 Questions
✓ Free question

For the 25-student grade data, the mean grade is 1414 and the mean deviation about the mean is 15825=6.32\dfrac{158}{25}=6.32.

xˉ=∑fixiN\bar x=\dfrac{\sum f_ix_i}{N}, MDxˉ=∑fi∣xi−xˉ∣N\quad MD_{\bar x}=\dfrac{\sum f_i\lvert x_i-\bar x\rvert}{N}

where xix_i = grade value, fif_i = number of students, N=∑fiN=\sum f_i.

  1. Build the working table:
Grade xix_i510152025Total
fif_i74635N=25N=25
fixif_ix_i35409060125350350
  1. N=25N=25.
  2. Mean: xˉ=35025=14\bar x=\dfrac{350}{25}=14.
  3. Deviation table:
xix_i510152025
∣xi−14∣\lvert x_i-14\rvert941611
fi∣xi−14∣f_i\lvert x_i-14\rvert631661855
  1. ∑fi∣xi−xˉ∣=63+16+6+18+55=158\sum f_i\lvert x_i-\bar x\rvert = 63+16+6+18+55=158.
  2. MDxˉ=15825=6.32MD_{\bar x}=\dfrac{158}{25}=6.32.
  3. Self-check: 25×6.32=15825\times6.32=158. ✓
✓Final answer

Mean deviation about the mean =6.32=6.32

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