Skip to content
NCERT Exemplar · Q1
Q.

Find the mean deviation about the mean of the distribution:

Size2021222324
Frequency64514
Tripura TbseShort· 5mImportance★★★★★est
50% · 45/90 Questions
✓ Free question

For this distribution the mean is xˉ=21.65\bar{x}=21.65 and the mean deviation about the mean is 1.251.25.

What we are finding

Mean deviation about the mean measures, on average, how far each value sits from the mean. For a frequency distribution it is

M.D.(xˉ)=∑fi ∣xi−xˉ∣∑fi.\text{M.D.}(\bar{x}) = \frac{\sum f_i\,\lvert x_i - \bar{x}\rvert}{\sum f_i}.

Step 1 — Total frequency and mean

N=∑fi=6+4+5+1+4=20N = \sum f_i = 6+4+5+1+4 = 20

∑fixi=20(6)+21(4)+22(5)+23(1)+24(4)=120+84+110+23+96=433\sum f_i x_i = 20(6)+21(4)+22(5)+23(1)+24(4) = 120+84+110+23+96 = 433

xˉ=43320=21.65\bar{x} = \frac{433}{20} = 21.65

Step 2 — Absolute deviations, weighted by frequency

xix_ifif_i∣xi−xˉ∣\lvert x_i-\bar{x}\rvertfi∣xi−xˉ∣f_i\lvert x_i-\bar{x}\rvert
2061.659.90
2140.652.60
2250.351.75
2311.351.35
2442.359.40
Total2025.00

Step 3 — Divide by NN

M.D.(xˉ)=∑fi∣xi−xˉ∣N=25.0020=1.25\text{M.D.}(\bar{x}) = \frac{\sum f_i\lvert x_i-\bar{x}\rvert}{N} = \frac{25.00}{20} = 1.25

✓Final answer

Mean =21.65=21.65 and the mean deviation about the mean =1.25=\boxed{1.25}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.