Statistics: Application Problems (5)

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Question

The mean of 10 observations is 25. After computing, it was discovered that one observation was misread as 35 instead of the correct value 53. Find the corrected mean. Also, if the original variance was 16, find the corrected variance.

Solution — Step by Step

Original sum =10×25=250= 10 \times 25 = 250. Wrong observation contributed 3535 instead of 5353 — a deficit of 1818.

Corrected sum =25035+53=268= 250 - 35 + 53 = 268.

xˉnew=26810=26.8\bar{x}_{\text{new}} = \frac{268}{10} = 26.8

Original σ2=16\sigma^2 = 16, so xi2/nxˉ2=16\sum x_i^2/n - \bar{x}^2 = 16.

xi210=16+252=16+625=641\frac{\sum x_i^2}{10} = 16 + 25^2 = 16 + 625 = 641

So xi2=6410\sum x_i^2 = 6410 (with the wrong value).

Replace 352=122535^2 = 1225 with 532=280953^2 = 2809.

xi2 (corrected)=64101225+2809=7994\sum x_i^2 \text{ (corrected)} = 6410 - 1225 + 2809 = 7994
σnew2=799410(26.8)2=799.4718.24=81.16\sigma^2_{\text{new}} = \frac{7994}{10} - (26.8)^2 = 799.4 - 718.24 = 81.16

Corrected mean = 26.8, corrected variance = 81.16.

Why This Works

Mean and variance are summary statistics — they only depend on xi\sum x_i and xi2\sum x_i^2. So when one observation changes, we don’t need to recompute from scratch. Just adjust the two sums and recompute.

The variance jumps from 16 to 81 because the corrected observation 53 is much further from the new mean than the original 35 was from the original mean — pulling out the spread.

Alternative Method

Recompute from scratch by listing all 10 observations and using σ2=(xixˉ)2/n\sigma^2 = \sum (x_i - \bar{x})^2 / n. Possible if the original data is given, but slow for an exam — the sum-update approach is much faster.

Common Mistake

Students update the mean correctly but forget to update xˉ\bar{x} in the variance formula — they compute σnew2=7994/10252=174.4\sigma^2_{\text{new}} = 7994/10 - 25^2 = 174.4, which is wrong. Always use the new mean (26.8) when recomputing variance after a correction.

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