GATE CS 2012 Q64 — General Aptitude

Let the original list of $n$ numbers be $x_1, x_2, \dots, x_n$ with mean $\mu$ and standard deviation $\sigma$.

Let's analyze each assertion:

• Assertion P: If we add $7$ to each entry, the new entries are $y_i = x_i + 7$.
  The new mean $\mu_{\text{new}}$ is:$$\mu_{\text{new}} = \frac{\sum_{i=1}^{n} (x_i + 7)}{n} = \frac{\sum_{i=1}^{n} x_i}{n} + \frac{7n}{n} = \mu + 7$$Thus, adding $7$ to each entry adds $7$ to the mean. (CORRECT)


• Assertion Q: The standard deviation measures dispersion from the mean. Since every data point and the mean shift by the same amount ($+7$), their relative distances remain unchanged:$$\sigma_{\text{new}} = \sqrt{\frac{\sum_{i=1}^{n} ((x_i + 7) - (\mu + 7))^2}{n}} = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n}} = \sigma$$Thus, the standard deviation remains unchanged. (INCORRECT)


• Assertion R: If we double each entry, the new entries are $z_i = 2x_i$.
  The new mean is:$$\mu_{\text{new}} = \frac{\sum_{i=1}^{n} 2x_i}{n} = 2 \left(\frac{\sum_{i=1}^{n} x_i}{n}\right) = 2\mu$$Thus, doubling each entry doubles the mean. (CORRECT)


• Assertion S: If we double each entry, the new standard deviation is:$$\sigma_{\text{new}} = \sqrt{\frac{\sum_{i=1}^{n} (2x_i - 2\mu)^2}{n}} = \sqrt{4 \cdot \frac{\sum_{i=1}^{n} (x_i - \mu)^2}{n}} = 2\sigma$$Thus, the standard deviation is doubled, not left unchanged. (INCORRECT)

Hence, only assertions P and R are correct.