NVAMediumJEE Main 2024 · 27 January, Shift 2Measures of Dispersion

Mathematics Question from JEE Main 2024 · 27 January, Shift 2

The mean and standard deviation of 1515 observations were found to be 1212 and 33 respectively. On rechecking, it was found that an observation was read as 1010 instead of 1212. If μ\mu and σ2\sigma^2 denote the mean and variance of the correct observations, then 15(μ+μ2+σ2)15(\mu + \mu^2 + \sigma^2) is equal to:

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