Aspire Faculty ID #10448 · Topic: NIMCET 2014 · Just now
NIMCET 2014

In a group of 200 students, the mean and the standard deviation of scores were found to be 40 and 15, respectively. Later on it was found that the two scores 43 and 35 were misread as 34 and 53, respectively. The corrected mean of scores is:

Solution

Given: \(n=200\), reported mean \(=40\) ⇒ reported total $$S_{\text{reported}}=200\times 40=8000.$$

Two scores were misread: true \((43,35)\) but used as \((34,53)\).

Correct the total: $$S_{\text{correct}}=S_{\text{reported}}-\big(34+53\big)+\big(43+35\big) =8000-87+78=7991.$$

Corrected mean: $$\bar{x}_{\text{correct}}=\frac{S_{\text{correct}}}{n} =\frac{7991}{200}=39.955\;\approx\;39.96.$$

Answer: \(39.955\) (≈ \(39.96\)).

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