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Worked Examples · Example 10
Q.

Calculate the mean, variance and standard deviation for the following distribution:

ClassFrequency
30-403
40-507
50-6012
60-7015
70-808
80-903
90-1002
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36% · 32/90 Questions
✓ Free question

Using class midpoints, the mean is xˉ=62\bar{x}=62, the variance is σ2=201\sigma^2=201, and the standard deviation is σ=201≈14.18\sigma=\sqrt{201}\approx 14.18.

For grouped data we compute the mean from the class marks, then the variance as the frequency-weighted average of the squared deviations from the mean (dividing by NN, the population form used in NCERT Class 11 Maths Statistics), and the standard deviation as its square root.

Step 1 — Class marks, ∑fixi\sum f_i x_i and ∑fixi2\sum f_i x_i^2.

Classxix_ifif_ifixif_i x_ifixi2f_i x_i^2
30–403531053675
40–5045731514175
50–60551266036300
60–70651597563375
70–8075860045000
80–9085325521675
90–10095219018050
Total503100202250

Step 2 — Mean.

xˉ=∑fixi∑fi=310050=62.\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{3100}{50}=62.

Step 3 — Variance.

Using σ2=∑fixi2N−xˉ2\sigma^2=\dfrac{\sum f_i x_i^2}{N}-\bar{x}^2:

σ2=20225050−622=4045−3844=201.\sigma^2=\frac{202250}{50}-62^2=4045-3844=201.

(Equivalently, ∑fi(xi−xˉ)2=10050\sum f_i (x_i-\bar{x})^2 = 10050, giving σ2=1005050=201\sigma^2=\dfrac{10050}{50}=201.)

Step 4 — Standard deviation.

σ=201≈14.18.\sigma=\sqrt{201}\approx 14.18.

Watch out

Variance uses the sum of the squared deviations weighted by frequency, divided by NN (not N−1N-1) for this population-style NCERT calculation.

✓Final answer

Mean =62=62, variance =201=201, and standard deviation =201≈14.18=\sqrt{201}\approx 14.18.

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