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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.
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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