The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $\mu$ and $\sigma^{2}$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^{2}+\sigma^{2}\right)$ is equal to ______

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Statistics

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If

μ

and

σ^{2}

denote the mean and variance of the correct observations respectively, then

15 (μ + μ^{2} + σ^{2})

is equal to ______

Answer: 2521

Solution

Let the 15 observations have correct sum

Σ x_{i}

and correct sum of squares

Σ x_{i}^{2}

.

Using the wrongly recorded value 10 in place of the correct 12, the reported mean was 12:

\frac{Σ x _{i}^{'}}{15} = 12 \Rightarrow Σ x_{i}^{'} = 180.

Correcting the sum:

Σ x_{i} = 180 - 10 + 12 = 182

, so

μ = \frac{182}{15}

.

From the reported standard deviation 3:

\frac{Σ ( x _{i}^{'} ) ^{2}}{15} - 1 2^{2} = 9 \Rightarrow Σ (x_{i}^{'})^{2} = 15 (153) = 2295.

Correcting the sum of squares:

Σ x_{i}^{2} = 2295 - 1 0^{2} + 1 2^{2} = 2295 - 100 + 144 = 2339

σ^{2} = \frac{2339}{15} - μ^{2} = \frac{2339}{15} - (\frac{182}{15})^{2} .

Now

15 (μ + μ^{2} + σ^{2}) = 15 μ + 15 μ^{2} + 15 σ^{2} = 15 μ + 15 μ^{2} + (2339 - 15 μ^{2}) = 15 μ + 2339.

Since

15 μ = 182

, this equals

182 + 2339 = 2521

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer

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JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Statistics

Solution

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