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प्रश्न
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
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उत्तर
Given that `n = 10, barx = 45` and `sigma^2 = 16`.
∴ `barx = (sumx_i)/n`
⇒ 45 = `(sumx_i)/10`
⇒ `sumx_i` = 450
Corrected `sumx_i` = 450 – 52 + 25
= 423
∴ Correct Mean `barx = 423/10` = 42.3
And `sigma^2 = (sumx_i^2)/n - ((sumx_i)/n)^2`
⇒ 16 = `(sumx_i^2)/10 - (45)^2`
⇒ 16 = `(sumx_i^2)/10 - 2025`
⇒ `(sumx_i^2)/10` = 2041
∴ `sumx_i^2` = 20410
∴ Correct `sumx_i^2` = 20410 – (52)2 + (25)2
= 20410 – 2704 + 625
= 18331
And corrected variance `sigma^2 = 18331/10 - (42.3)^2`
= 1833.1 – 1789.3
= 43.8
Hence the required mean = 42.3 and variance = 43.8
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