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Exercise 13.2 · Q7
Q.

Find the mean and variance for the following frequency distribution.

ClassesFrequencies
0-302
30-603
60-905
90-12010
120-1503
150-1805
180-2102
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Using class midpoints, the mean is 107 and the variance is 2276.

Because the data is grouped, each class is represented by its midpoint xix_i. We find the mean as a frequency-weighted average, then the variance as the frequency-weighted average of squared deviations from that mean.

Step 1 — Midpoints, total frequency, and ∑fixi\sum f_i x_i.

Midpoints: 15, 45, 75, 105, 135, 165, 195.

Classfif_ixix_ifixif_i x_i
0–3021530
30–60345135
60–90575375
90–120101051050
120–1503135405
150–1805165825
180–2102195390
Total303210

Step 2 — Mean.

xˉ=∑fixiN=321030=107.\bar{x}=\frac{\sum f_i x_i}{N}=\frac{3210}{30}=107.

Step 3 — Squared deviations from the mean, weighted by frequency.

| xix_i | fif_i | xi−107x_i-107 | (xi−107)2(x_i-107)^2 | fi(xi−107)2f_i(x_i-107)^2 |

|---|---|---|---|---| …

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